point-set-topology

[quartz horizon]

And where there’s a directed edge from a to b iff aRb

(So there may be self-loops)

You can generate an equivalence relation S from this as follows

Pretend the edges are undirected, and say aSb iff a and b are in the same connected component

[prime elbow]

directed edges means direct edge from a to b or via chain of edges?

[quartz horizon]

Direct edge

[prime elbow]

okay

now i got it

[quartz horizon]

You can unfold this to say that aSb iff there’s a sequence x_0 = a, …, x_n = b where either x_k R x_(k + 1) or x_(k + 1) R x_k for all k from 0 to n - 1

(You need a bit of a special case for n = 0 to ensure S is reflexive)

[prime elbow]

so here relation is i1(x) = i2(x) ?

[quartz horizon]

Yeah, and then you generate an equivalence relation from that

[prime elbow]

okay, thank you

[grim thistle]

does anyone else experience a existential terror from topology?

like a feeling you get just like from geometric dreams

[quartz horizon]

If I think too long about spaces like the long line

[desert vortex]

me when I cant make arguments just from using sequences

[bitter turtle]

Hello, I was wondering if ideally I should learn real analysis before doing a dedicated course on point set. I’m reading on abstract algebra right now and find it quite fun, so I wanted to go on and improve to point set topology, then algebraic topology.

[quartz horizon]

I think it helps a lot to have done some real analysis

Topology tends to be more “soft”-style I’d say

[bitter turtle]

Yeah, that’s what I was thinking. What in real analysis do you usually use for topology

[quartz horizon]

Intermediate and extreme value theorems are quite helpful

Also they give lots of metric space intuition

[quick delta]

It’s mostly examples and intuition
Like IVT/EVT are good intuition for things about connected/compact sets
And metric spaces are a “prototypical” example of a topological space

[bitter turtle]

Makes sense. Thank you, and I’m pretty sure abstract algebra isn’t really used in point set? Just asking because I’m learning it right now

[quartz horizon]

Not that I remember, at least

[quick delta]

Not really
Mathematical maturity always helps, but it’s definitely not a requirement

[bitter turtle]

Alright, thanks

[umbral hamlet]

You can use group theory for some good examples of stuff in pointset

But of course it doesn't drive the actual theory, otherwise its just algtop

[normal swift]

@crimson terrace may i ask how you got your dumbass license?

i want one too

[crimson terrace]

Sorry dude
You get invited to get it :/

[normal swift]

😭

[crimson terrace]

I'll put in a good word for you at our next meeting

[normal swift]

tysm, this means so much to me.

[tender halo]

analysis is a much more pleasant subject when you study it from books titled "point-set topology"

[lucid ocean]

tbh yeah, the essence is probably clearer there

[warm cape]

I don't know how to prove this can anyone help??

[lucid ocean]

is this just A inter union_(s in S) s = union_(s in S) A inter s?

anyway, this is probably best in #proofs-and-logic

[junior adder]

Does $\prod \overline{A_{\alpha}} \subseteq \prod \overline{A_{\alpha}}$ require axiom of choice to prove for an infinite collection of sets ?

(Nvm found the answer, gonna leave it here in case anybody ever uses the search function. Equality is equivalent to AoC)

Keywords:Munkres, Theorem 19.5, axiom of choice, closure, Cartesian product

https://math.stackexchange.com/questions/3814769/munkres-thm-19-5-and-axiom-of-choice

[gentle osprey]

phoenixperson

[lament steppe]

If I have a metric space M and a subset S, what exactly does it mean for the metric to be restricted to points in S?

My current thought is that if there is a ball for M that contains points inside and outside of S (center at a point in S), then that same ball using the restricted metric will only contain points in S and the points outside of S are ignored. Is that the correct interpretation?

[unreal stratus]

Nice

[tender halo]

it means looking at S as a metric space with a metric inherited from M

the fact that the subspace topology and the topology from the inherited metric are the same is a (albeit trivial but still) separate fact

i guess thats kind of splitting hairs

[lament steppe]

That is what I'm trying to prove! I guess what was throwing me for a loop is that if S has an isolated point, then there is most assuredly a ball in M such that in the subspace Topology that single point is open. But I guess this "isolated point" from the perspective of S is also open right?

For instance if you make a ball in the restricted space that only contains that isolated point, it doesn't contain any neighbors of said point because they belong in M \ S (which don't exist in the restricted metric)

[quartz horizon]

The metric is a function M x M -> R

You restrict it to a function S x S -> R

[tender halo]

sure

[worn mortar]

Let X and Y be sets. Suppose f is a function from X to Y and S is a subset of X. The restriction of f to the subset S is the function f|S from S to Y, defined by (f|S)(x)=f(x) for all x in S.

[pearl valve]

I am familiar with the following way to describe a topology:

Given a set X and a subset B of P(X), verifying B is a basis thus determines a unique topology.

In one of my classes, my professor wrote “this space X has the topology determined by neighborhood basis: “blablabla”.

I am unfamiliar with “neighborhood basis” and can’t find a good resource to learn about them.

In particular,

1. what axioms does a set B of P(X) need to satisfy to be a neighborhood basis, and

2. why does specifying a valid neighborhood basis determine a unique topology?

I am struggling to answer these questions or find clear definitions

From few of what Ican find online , it starts from a topology T and then defines a neighborhood base given a topology, but I can’t find a clear definitions where you start with X, define a neighborhood base, and show it creates a topology

[opaque scroll]

A basis of neighborhoods at x should be a set of neighborhoods such that every neighborhood of x contains a basis set. The only thing needed here is that the intersection of two basis sets at x contains another basis set at x.

A basis of neighborhoods for the entire space is then probably a basis for each point.

[pearl valve]

But I am confused what the definitions are if we don’t actually specify a topology.

This definition seems to require we first say what the neighborhoods are. But I don’t know what the topology is. I only know the set X

[opaque scroll]

A set is a neighborhood if it contains a basis neighborhood, and is open iff it is a neighborhood of all its points

[opaque scroll]

It is like I said: the intersection of two basis neighborhoods (at x) should contain another basis neighborhood (at x)

[pearl valve]

I am still confused as it seems we are defining basis neighborhoods in terms of neighborhoods, and neighborhoods in terms of basis neighborhoods. But without a verified topology, it’s not clear what neighborhood is

Like with a basis, there are are axioms to verify that are independent of a topology. If a subset of P(X) satisfies these axioms, then the basis corresponds to unique topology

I don’t know what the analogous axioms are for “neighborhood basis” or even what neighborhood basis is if you don’t have an already specified topology

[opaque scroll]

I'm not sure why you're confused, but I can try to spell it out:

A basis of neighborhoods of X is a collection of sets such that

• Each set contains x
• The intersection of two sets contains another set in the collection.

Once you have this basis you define a neighborhood at x to be any set which contains one of the basis sets.

Then once you have a basis of neighborhoods at each point you define a set to be open iff it is a neighborhood of all its points

So the axioms are "the intersection of two basis neighborhoods at x should contain another basis neighborhood at x"

Which is the same axiom as for a usual basis

[pearl valve]

So a couple questions about this construction

1. How do we know that defining open sets in this way actually determines a valid topology? This should probably be a straightforward definition chase

2. Is this definition of neighborhood consistent with the usual definition of neighborhood? Also probably a definition chase right?

3. Is there a reference where these definitions are clearly stated?

[tender halo]

that is not enough for a nbhd basis

[pearl valve]

Just working with these, I’m getting stuck on proving finite intersections of open sets are open

[tender halo]

thats just a collection of bases at points

which need not be a nbhd basis

[opaque scroll]

Then what is the definition?

[pearl valve]

A great mystery

Every textbook I can find only states it

Like so

[opaque scroll]

I mean, what I said uniquely specifies a topology

[pearl valve]

How do you show finite intersection of open sets are open with what is stated from ur message

That’s the only part I can’t verify

[opaque scroll]

Say U and V are open.

Consider z in UnV, then as U and V are neighborhoods of z they contain basis sets. Let's call them Uz and Vz. Then
UnV contains Uz n Vz which contains a basis neighborhood. So UnV is a neighborhood of all its points (aka open)

[pearl valve]

Oh wait

Is it BP2?

Or V-c

[tender halo]

yea, that one makes sure that different points have the same opinion on what nbhds are

[opaque scroll]

BP2 seems to specify that the neighborhoods be open. Which seems kinda against the point of a neighborhood basis, as then it's just a basis

[tender halo]

its not a basis because a basis is just a family of sets, nbhd basis is assigning a family of sets to each point

[opaque scroll]

Different points agreeing on what neighborhoods are seems to kinda go against the point of neighborhoods

[alpine nest]

And indeed the union of neighborhood bases assigned to all the points will be a basis for the topology as a whole

(and conversely, given a basis for the whole topology, you can take "all the basis sets containing x" as a neighborhood basis for every x)

[pearl valve]

The motivating problem I am trying to show that this is actually a topology

I just wanted to understand all the definition being used

[opaque scroll]

Anyway if that's the case then I guess it's easy.

A neighborhood basis is just a basis with some extra information you should ignore

[pearl valve]

So to show a map is continuous, you can verify just on the basis elements. Similarly, you can verify on the neighborhood basis elements?

[tender halo]

sure

[pearl valve]

For trying to show that something is a TVS

[pearl valve]

From this result

[tender halo]

they are useful for constructing intricate topologies

much easier to think about

and check

than the intersection condition

[alpine nest]

Yep, thinking locally is often more convenient than thinking of the basis as a whole.

Sure, if you're saying "take a basic neighborhood of x", you're essentially saying "take a basis set containing x", but having the explicit connection to x is often useful

[tender halo]

also there can be basic sets that contain x that you dont want to take into the nbhd basis (for example, you want to keep it countable)

[alpine nest]

Yeah, thinking in terms of a neighboorhood basis lets you specify certain basic sets containing x

[pearl valve]

I was told they are relevant for functional analysis

Yesterday

Someone here said you can’t always do everything with just a basis

[alpine nest]

Yes, the weak and weak* topologies can be quite conveniently described in terms of neighborhood bases

This is more often phrased by defining them as "the weakest (coarsest) topologies in which certain maps are continuous", but such topologies lend themselves nicely to the neighborhood basis language

[tender halo]

they were talking about nbhd basis, not the regular kind of basis

[desert vortex]

oops

[pearl valve]

this is only true if we restrict the neighborhood bases to open sets right?

[tender halo]

they are usually meant to be open sets yea

otherwise its unclear how to make an appropriate coherence condition

[raw ledge]

Hi I have a question about TDA. If one uses a deterministic rule to make multiple point clouds, can one say the homological features of the point clouds are only due to the deterministic rule applied to make them?

Cooked phrasing holy crap

[opaque scroll]

I mean, the point cloud is due to the process you use to make it yes....

Though different methods of computing e.g. persistent homology can produce sightly different features even for the same data

[rancid umbra]

it seems like all the classically equivalent characterizations of connectedness are no longer equivalent if we reject the law of excluded middle.
my base characterization of connectedness is not disconnected, where a space is disconnected if there are two non-empty, disjoint, open subsets which cover the entire space

for example, chain conectedness implies connectedness, but i don't think you can prove the converse without LEM

if the only clopen subsets are the entire space and the empty space, we can prove that X is not disconnected (so it is connected), but i don't think we can prove the converse without LEM

if the only clopen subsets are the entire space and the empty space, we can prove that the space is chain connected, but i don't think we can prove the converse without LEM

i just find this weird

wondering about anybody else's thoughts

[somber gorge]

i have zero intuition about what dropping lem looks like, but i think it makes sense. like you mentioned, a lot of these characterizations pivot on the fact that connected means not disconnected (or at least the proofs ive seen do). i guess the way i think about this is that mutually exclusive conditions "collapse" and so any "auxiliary" characterization of either condition becomes meaningless.

[rancid umbra]

yea, i think that the usual definition of connectedness being defined “in the negative” is playing a role here.

something i find interesting is like, most of the actual content of the proof can be done constructively. for example, if you set out to prove that connectedness implies chain conntectedness, you will be able to construct a set that is inhabited, open, and whose complement is open, all constructively, but you won’t be able to conclude that this set together with its complement union to the whole space without LEM

[rancid umbra]

do you mind expanding on what you mean by ‘mutually exclusive conditions “collapse”’?

[rugged escarp]

The conditions where if you have one , you don't have the other?

or maybe you mean examples?

[somber gorge]

like, normally you have that "either a space is connected or disconnected". but when that isnt logically true, its like pushing in the base of a house of cards

[rancid umbra]

okay, i see what you mean

and so auxiliary characterizations of connectedness, for example, become meaningless because their proofs of equivalence rely on LEM

[somber gorge]

yeah basically. or at least i think thats the case, since like what it means to be connected is fundamentally not disconnected

[rancid umbra]

doing constructive topology sounds kind of awful

[quartz horizon]

The characterisation I know is “Hom(X, -) preserves coproducts”

[copper quiver]

Wait, I've actually never heard of that.

[quartz horizon]

It’s equivalent to the usual characterisation

[copper quiver]

Are you saying X is connected iff Hom(X, \sqcup Y_i)=\sqcup Hom(X;Y_i)?

Because I don't think that works.

[gaunt linden]

It feels like it ought to work classically. Which direction feels off to you?

[copper quiver]

My issue is that it looks like it's true for all X.

Oh wait.

No.

That's products, not coproducts.

[quartz horizon]

Mhm exactly

[copper quiver]

Ok, now it intuitively makes sense to me.

[gaunt linden]

It does make the empty space not connected, which I suppose is a feature. :-)

[copper quiver]

Shouldn't be hard to prove.

Ok, I proved it to myself, nice.

[quartz horizon]

:D

It works in general categories

[copper quiver]

I'm not following, what's the generalization of this statement to an arbitrary category?

[gaunt linden]

I suppose it amounts to defining what "connected" might mean in other categories (or at least those other categories that have coproducts).

[copper quiver]

Ok, yeah, I see.

[quartz horizon]

Mhm

I first learned this from “topology: a categorical approach”

[copper quiver]

I think I've quickly read that, don't remember coming across this though.

[gaunt linden]

On the other hand that definition would make it pretty hard for a pointed space to be "connected" ...

[copper quiver]

A pointed space is connected iff the space you get by applying the forgetful functor is connected, obviously.

But yeah.

[gaunt linden]

Yeah, and the "preserves coproducts" definition doesn't appear to produce that, unless I'm tripping...

[copper quiver]

I'd be surpised if it did.

Hom(X,Y_1 \sqcup Y_2) is definitely not the union of Hom(X,Y_1) and Hom(X,Y_2).

Because you can "bend" X around the basepoint.

[gaunt linden]

Yes, exactly.

So that suggests we shouldn't be quite so fast to generalize that definition to other categories.

[copper quiver]

Yeah, although it looks like it's in some way the fault of the basepoint, so maybe if there's some fancy way to forget basepoints categorically, then we could adapt this?

I have no idea how you would do this though.

[quartz horizon]

Hm apparently you want to focus on “extensive categories”

https://ncatlab.org/nlab/show/extensive+category

[gaunt linden]

Or perhaps I was just unlucky to pick just the wrong topology-adjacent category to test drive it with.

[worn mortar]

Is that book good

[rugged escarp]

Not a first read for topo

But maybe second or third

[gritty widget]

yeah

Second countable spaces that aren't hereditarily second countable...

Iirc you can have second countable spaces that aren't separable too

[gritty widget]

Wait actually idk this might be true without AC and LEM nvm

<@&268886789983436800>

[cyan pewter]

time to hunt

[sonic crane]

Thats neat

[neat sandal]

got topology by munkres for christmas 😎

very good book so far

[sonic crane]

Yas

[rough siren]

You are so lucky , I cant get my hands on it .

[queen prism]

yea munkres personally went up to his door and hand-delivered the book to him

[rancid umbra]

munkresclaus

[cosmic zodiac]

Moved it to here because it's more top than pure analysis lol

[warped helm]

I think the proof will be better written if you’re more direct about which set is the closed set in K cap Y

Like, say upfront that the closure of E in Y is cl(E) cap Y

[alpine nest]

Also you might want to specify whether the interior and boundary are taken in the relative topology or in the original topology

[cosmic zodiac]

Oh yeah I didn't even see that. I assume for this context its the original topology, right?

[alpine nest]

I mean, it's your proof

Also, how is relative topology defined in your case, and what is corollary 1.2.11?

[cosmic zodiac]

I'll send screenshots, one sec

[warped helm]

I think the proof should avoid the whole int cup boundary thing but thats just imo

[cosmic zodiac]

Sorry about the highlighting lol

[cosmic zodiac]

I guess I included it as a kind of bridge between K and E

[warped helm]

It’s not wrong per se but you’ll have to be crystal clear about this sort of detail

Which I think introduces more of a headache than the proof is worth

[jaunty rune]

On topology of $\mathbb{R}^2$, because of the Heine-Borel theorem, any closed and bounded subset is compact, right? But say I have the unit square and cover it with open sets such that after the $n$th open set there is still $\frac{1}{2^n}$ of the square uncovered. With all these open sets every point of the square is covered, but then there can be no finite subcover of the square so it isn't compact. So is such an open cover just impossible?

[gentle osprey]

TorusField

[hexed steppe]

what exactly do you have in mind? for simplicity just describe the analogous construction with [0,1]

[jaunty rune]

Ah trying to describe it for 1 dimension made me realize why it doesn't work for my original 2 dimensional thought experiment: There's this little "corner" that remains which you can't cover with an open set without then opening the opportunity for a finite subcover

[hexed steppe]

wdym

did you mean (0,1/2), (1/2, 3/4), (3/4, 7/8),…

[jaunty rune]

Not quite

[jaunty rune]

For [0,1] I could have the open set $(0 - \epsilon, \frac{1}{2})$, then $(\frac{1}{2} - \epsilon, \frac{3}{4})$, $(\frac{3}{4} - \epsilon, \frac{7}{8})$, and so on. But what I just realized is that nothing here would cover $1$, and if I ad hoc'd an open set for $1$ I would have to cover a little less than 1, which would make a finite subcover possible.

[gentle osprey]

TorusField

[jaunty rune]

Or I suppose the epsilons are not necessary after the first one, but same idea

[hexed steppe]

sure

[thin scarab]

(From May's Concise) When May says "when this holds," is he saying "when X is weak Hausdorff" or is he saying "when g(K) is closed for an individual K" ?

[thin scarab]

Nevermind, thought of a counterexample for the second interpretation

[safe remnant]

(from munkres) maybe a dumb question but for part (d) is he saying that parts (a) and (b) additionally necessitate that X is/isn't a complete metric space? or is he saying its an alternate condition to it being compact in the case of (a) but not (b)? or is he saying something else?

I think he's saying the second of those because im pretty sure R isnt compact anyways but im not sure and wanted to double check

[rancid umbra]

the result (a) holds even with the weaker assumption that X is complete

if X is only complete, part (b) doesn’t necessarily hold.

[safe remnant]

ah so does compactness also imply completeness then?

[rancid umbra]

in a metric space, yes.

if X is a metric space, X is compact if and only if X is complete and totally bounded

[safe remnant]

makes sense, thanks

[unreal stratus]

Does completeness even make sense outside metric spaces

Or maybe I am thinking the opposite way around lol as in you are saying it works for any metric space

[safe remnant]

i think he specified because compactness does make sense outside of them

[rancid umbra]

i think in uniform spaces there is a notion of completeness using cauchy filters

but like

[unreal stratus]

O inch resting

[rancid umbra]

i hadn’t had that in mind when i made the comment. i suppose what i said was a bit redundant

[unreal stratus]

Nah dw it chill lol

Idk why I said anything lol

[robust sage]

it is an awesome text

[rugged nexus]

can someone explain how a circle can fit the definition of 'topological space'?

[opaque scroll]

There are several ways to describe the topology of the circle. One perhaps simple one could be: an open arc is an arc of the circle (the set of points between two other points) not including the endpoints.

The open sets are then unions of open arcs.

[quick crane]

The circle is a subspace of $\mathbb{R}^2$

[gentle osprey]

L

[opaque scroll]

indeed, so you could maybe start by trying to understand how the plane is a topological space

[rugged nexus]

yeah that sounds like a good idea

how is the plane a topological space

[quartz horizon]

It’s a metric space

And every metric space is a topological space

[rugged nexus]

and what is a metric space

[wide kayak]

if you consider all “open balls” centered at a point p = (x,y) with radius r, B(p,r) = {(a,b) | sqrt((a-x)^2 + (b-y)^2) < r}

where p ranges over all of the plane, and r ranges over all positive real numbers

this forms what’s called a basis for the standard topology of the plane

and this is also how to get the standard metric topology associated with a metric

a metric is like a “distance function” on a set

the standard distance in the plane is the prototypical example

we declare a subset U of the plane to be “open” if for every p in U, there is some open ball B(p,r) contained in U

for example, the half-plane {(x,y) | x > 0} is open

R^3 and R^n in general have a topology in the same way, by using the higher dimensional distance function, which comes from the generalized Pythagorean theorem in the same way

[rugged nexus]

wait, a ball is a set of distances from a given point??

oh b is a basis

is that what B means?

[warped helm]

perhaps begin reading a topology textbook

munkres is a good start

[rugged nexus]

Topology (2nd ed.)?

[warped helm]

a ball of radius r centered at x with respect to a metric d is the set {y : d(x,y) < r}

yes

its the set of all points less than a distance r from x

[rugged nexus]

what is y an element of

or i suppose

what is the set a subset of

if that makes sense

like what's y

[kind marlin]

whatever the metric is defined oon

[rugged nexus]

what is that for R2

[kind marlin]

its R2

[warped helm]

points in the metric space x lives in

[rugged nexus]

uuhh

so y is a point

is that right

[warped helm]

yes

[kind marlin]

start with the set you're interested in

is that set R2

or a subset of R2

[rugged nexus]

i got no clue

i think im js gonna read the book

im still so far from understanding what a topological space is

[kind marlin]

reading a book is definitely a good idea

[rugged nexus]

let alone a category or functor

[rugged nexus]

i already mentioned i was mainly looking for that but it did get buried so

[kind marlin]

munkres presents everything really well

[wide kayak]

no, B(p,r) in the notation I used means a single open ball of radius r centered at p. But I was talking about the collection of all such balls

and that is a basis

[warped helm]

in full generality its just a set and a collection of subsets of this set that meet certain conditions

[kind marlin]

a topological space is just

1. a set S

2. a collection C of subsets of S that satisfies certain properties

we call a subset open if it belongs to C

a basis is a way of representing C in terms of a smaller collection, such that each set in C is a union of sets in your basis

[warped helm]

i highly advise doing analysis in R and metric spaces before jumping into general topology

[rugged nexus]

im just trying to do category theory

[warped helm]

why

[kind marlin]

you dont need topology to do category theory

[wide kayak]

as someone who studied it a lot in grad school, I don't recommend trying to learn it before first doing some abstract algebra and topology. It's possible, but you won't be able to appreciate many important and exciting examples of things

it's true you don't "need" to, but I really wouldn't recommend it

[radiant stone]

I want to learn category theory and need to do topology first rather than abstract algebra

ah sure, a group is an object whose delooping BG is a 1-truncated ∞-groupoid whose higher homotopies are contractible

homotopy theory ahh

[queen prism]

yea it's not strictly necessary but categories make a lot more sense the more math you're familiar with

[crisp lintel]

Tbf the reason they came here is because they wanted to learn about sheaves

[worn mortar]

Exactly

[worn mortar]

I would not try to do this

[rugged nexus]

see blake's message

[warped helm]

my question is still “why”

[rugged nexus]

it sounded interesting idk

i liked the idea of rejecting lem and applying it

[warped helm]

fair enough

[rugged nexus]

wtf am i doing here

[worn mortar]

That’s cool and all, and it should stay interesting, but going into category theory while saying you’re far away from understanding what a topological space is tough

[rugged nexus]

well i can tell that now

this doesn't exactly help

[rugged nexus]

ok so you would recommend using abstract algebra instead of topology?

ohh i know a little abstract algebra

[worn mortar]

Since that’s where category theory came from, yea

[rugged nexus]

thats way easier than topology

[worn mortar]

Awodey has a book

[rugged nexus]

ok but how do i learn category theory from abstract algebra then

the wikipedia page heavily pulls from topology

[rancid umbra]

there are no shortcuts to climbing a mountain

for then you will be able to see, my child

[rugged nexus]

so

[rancid umbra]

https://tenor.com/view/dont-matter-dont-care-peaceful-laid-back-zen-life-gif-11841640

[rugged nexus]

ok yeah i have no idea what you're on about

[rancid umbra]

lol im kinda joking with you. the way that you learn category theory is by going through and learning linear algebra, abstract algebra, topology, etc

there are no short cuts

[rugged nexus]

ok so i DO need topology

also im not learning the entirety of category theory mostly just sheaves

in fact im probably going to stop at sheaves

whatever the case

[rancid umbra]

hm. since you are trying to "speed run" sheaves, take about a week or so, figure out what a topological space is, then come back and digest the wikipedia definition of a a sheaf bit-by-bit

[rugged nexus]

i know basic abstract algebra (ish)
i dont know topology

i want to know sheaves

where do i start

[rugged nexus]

ok

thank you

[rugged nexus]

can C be a set

[rancid umbra]

what else are you thinking it could be?

[rugged nexus]

i have no clue

[source: nLab]

that i guess

[rancid umbra]

if S is a set, then the powerset of S is a set. subsets of sets are sets. since C is a subset of the powerset P(S) of S, then it is a set

[rugged nexus]

ohhh

why did he call it a collection then

[kind marlin]

set is kind of a loaded term

[rugged nexus]

what

how so

[rancid umbra]

collection is sometimes used colloquially to mean set

not technically

[rugged nexus]

ok

anyway

do i have to read through that one book

[rancid umbra]

munkres?

[rugged nexus]

munkres yeah

[warped helm]

at least the first half ideally

[rancid umbra]

hmm. it would probably be a good idea to read the first few chapters

[rugged nexus]

ideally is a fun word

[warped helm]

you can skip the manifolds section if you want

[rugged nexus]

which ones exactly

[warped helm]

the ones before the algebraic topology part of the book

[rugged nexus]

ok cool

[rancid umbra]

the first 2 are sufficient for speed running the defintion of a sheaf, imo

the first chapter will improve your set theory

the second will introduce you to the notion of what a topological space is and what continuous functions are

[rugged nexus]

uhhh ok

[rancid umbra]

this will probably take like 1-2 weeks

maybe longer

[rugged nexus]

oh dear

[rancid umbra]

but like

im just estimating

i could be totally wrong

i don't know your schedule

i don't know how much time you are putting in

[rugged nexus]

ok ok im gonna read that

[rancid umbra]

i don't think you are supposed to send books here

but yes, that is the book

[kind marlin]

collections can kind of be whatever you want, whereas sets mean something more precise

[rancid umbra]

then i think it's more precise to say that collection is a loaded term

[rugged nexus]

what c squared said

[rugged nexus]

do i have to do the practice problems

[kind marlin]

mm what i mean is that collections are kind of intuitive whereas sets do imply more things about them

[warped helm]

absolutely

[rugged nexus]

ffs

[rancid umbra]

this is hilarious

[kind marlin]

im not really gonna argue about this though since i dont really know what the contention is

[rancid umbra]

oh, i didn't mean for it to be an argument lol

my b

[kind marlin]

maybe argument is a loaded term loll

[rancid umbra]

set

[rancid umbra]

so your goal is just to understand the definition of what a sheaf is... for personal satisfaction?

i'm not trying to discourage you or anything, it's uh. just kinda funny like, i have been able to get through an undergraduate degree in math without knowing what a sheaf is

[rugged nexus]

ish?

it just kinda piqued my interest and i wanted to look into it more

potentially good to use in other context where rejecting lem is important and/or helpful

[rugged nexus]

im not even a math major lol

[wide kayak]

you could probably come to some kind of grip with the sheaf of continuous functions on R reasonably quickly

it's the simplest example I know of

[rugged nexus]

yeah im pretty sure i already have one

but i want to know a formal definition

[wide kayak]

gotcha. Yeah, that'll take some weeks of topology work

it's not THAT bad, though

[rugged nexus]

ok

idk how to explain

more of a propositional logic concept

probably better to ask in #proofs-and-logic

[cosmic mirage]

lowkey tho this is very right. even if you speedrun learning definitions you cant do anything with them unless you have seen all the surrounding theory

[quartz horizon]

the best prerequisite for category theory is linear algebra

[cosmic mirage]

smtg smtg why climb a mountain if you're blindfolded at the top

[spare dome]

Hi people. I'm just having a look over Willard's General Topology for chapter 9 on uniform spaces. It starts with the following definition:

but somehow I am stuck at the first remark. And the second too, for that matter...

How would we prove it? Maybe I'm just too dumb but I can't quite figure it out. I'm just starting with uniform spaces, and any tips would be appreciated!

[cosmic mirage]

anyways @rugged nexus idt anyone is trying to discourage you from learning what a sheaf is, its great that youre interested. the point is that taking shortcuts means you miss out on a lot of other really interesting math, and means that you wont be able to fully appreciate sheaf theory. so it is good to eat your munkres vegetables

ime when people recommend not cutting corners its coming from experience haha (myself included)

[pallid lake]

@rugged nexus If you're interested in sheafs for their application to computer science, you might look at "Topology via Logic" for learning topology

https://pages.jh.edu/rrynasi1/FoundationsOFMath/Literature/Topology/Vickers1989TopologyViaLogic.pdf

[worn mortar]

I don’t think this is fruitful for your mathematical foundations whatsoever at this stage

[worn mortar]

Δ=Δ(Χ), which is the diagonal relation on X, if I’m not mistaken

[spare dome]

Yes. I am mostly fine with the definition. But I'm not sure how to use it to prove the first remark (D in script D ==> D inv in script D)

[worn mortar]

Yea…the way the passage is phrased makes it seem like the statement follows immediately from (a), which doesn’t make sense

But actually I think that is just labeling the remarks

[spare dome]

I see

I also was having a look at Kelley and he on the other hand listed remark (a) as part of the definition of a uniformity. 35.2(d) was not in definition, on the other hand. So that's what made me think, maybe those two facts can be used to prove each other

but I can make peace with taking it for granted I guess

But 35.4(b) is no good. How do we arrive at $E\circ E^{-1}\subseteq D$?

[gentle osprey]

Naman

[worn mortar]

Yea i dont quite see how it follows either, and i know you would rather not leave things to the imagination lol

[spare dome]

Yes!!

Maybe I am jumping straight to this section and that's the issue. And in fact, Willard lists some sections as dependencies for this one... but I checked them out, I've already studied majority of them in my coursework last semester

I suppose I'll just keep trying, and hopefully something will click someday

[lament steppe]

If we have set A that is countable, and an onto map f, then the image of f(A) is also countable correct? To me this means that f(A) has at most the same amount of elements as A, so this would be true. However, my book defines a set to be countable if there is a bijection from a countable set to another.

[warped helm]

you can say “at most countable”

[tender halo]

what is your definition of countable?

it might mean that it has to have exactly N elements

countable might mean "countably infinite or finite" or it might just mean "countably infinite"

[lament steppe]

ah sorry. Yeah, the author means countably infinite (or less) here.

And they say that countably infinite means that the set has the same cardinality as N. And having the same cardinality means that there is a bijection between the sets.

[tender halo]

To me this means that f(A) has at most the same amount of elements as A
note that this is a special property of countable (or, more generally, well-orderable) sets, it is not a theorem of ZF for arbitrary cardinalities

its a form of choice (naturally weaker than full choice) called PP (partition principle)

[lament steppe]

So you are saying it DOES result from ZFC tho?

[tender halo]

yes

[lament steppe]

This is all good to know! Thanks for the clarifications 😄

[rugged nexus]

im not trying to take shortcuts i just have no idea what counts as redundant for this scenario
again the first thing i asked for was a textbook or something like that

[rugged nexus]

if they're ever lacking i can make it up in other ways

[worn mortar]

It’ll be much easier to learn after learning both abstract algebra and point-set topology

And there’s a lot of fun stuff in both of those

[rugged nexus]

what exactly do u mean by abstract algebra

cause i already know the groups/rings/fields stuff

[worn mortar]

Oh well if that’s the case, category theory should be relatively understandable

At least the basics

Have you worked through all of the important theorems?

[rugged nexus]

wdym work through?

[worn mortar]

For example the isomorphism theorems

[rugged nexus]

i only did the ones for vector spaces

[worn mortar]

I see

It’s good to get used to that kind of thinking for cat theory

[rugged nexus]

i believe

yeah ok i did a little bit of that

[quartz horizon]

.

[unreal stratus]

Also if you are doing category theory before linear algebra you should ask yourself why

[quartz horizon]

yeah that's fair

[pallid lake]

I hated linear algebra, It is obviously more than critical in retrospect - and that has always moderately annoyed me.

It's math's "practice this simple thing over and over again until you get it"

[quartz horizon]

oh i absolutely loved linear algebra

[pallid lake]

You're lucky!

I wish I loved how it was presented to me

It took me a long time to see the beauty

[quartz horizon]

3b1b's videos helped a lot

[pallid lake]

The "don't underestimate linear algebra" moment for me was when I tried to model rotations without linear algebra. Given the data of two angle axis pairs (θ₁, l₁), (θ₂, l₂) - how do you find the resulting rotation and axis? But wait, how'd we even know a rotation is just a rotation about an axis?

Without linear algebra (and especially without the help of quaternions) this problem is a nuisance.

[queen prism]

I mean it's true

well linear algebra and a little combinatorics

[rancid umbra]

point-set topology doesn't feel very linear

neither does set theory i guess

are you lumping this in with combinatorics?

[queen prism]

it's linear if you squint your eyes enough dw

[cosmic mirage]

@ aluffi

[tender halo]

its very wide

none the parts that are needed for the other areas of math are deep

you can put all the lectures in a bin except like the first three and pull them out in a random order and it will be fine

[quartz horizon]

Compactness is a very cool concept

[cosmic mirage]

Society if compact spaces were precisely the compact objects in spaces

[unreal stratus]

this is why homotopy theory is better

[cosmic mirage]

don't you still need to include retracts if you do this in S

[unreal stratus]

yee

this is why stable homotopy theory is even nicer

[unreal stratus]

ye like retracts of finite CW complexes

but that is already a class that appears naturally so it's ok

[cosmic mirage]

yeah exactly

o right bc fg spectra are the compact ones

[opaque scroll]

X compact space iff X is a compact object in the poset of open sets in X.

[rugged nexus]

i did strang lin alg does that count

reddit seems to hate strang lin alg and im tbh inclined to agree

i dont think it even mentions linear maps

[rancid umbra]

i think strang gave a more computational approach

which is fine

because the target audience probably wasn’t people going into pure math

[rugged nexus]

interesting

good to know

but do you think that it's good enough as a foundation?

[rancid umbra]

depends on what for. but it's never a bad thing to revisit and review concepts

[rugged nexus]

fair enough

[rain nova]

Hello guys, happy weekend. I got a problem concern connectedness in topological space. A subset $S\subseteq X$ is connected provided there are no two non empty disjoint open subsets $A,B$ such that $S=A\cup B$. I wonder ``is there any topological space $O$ such that there exist a subspace $U$ and $V\subseteq U$ connected in $U$ but not connected in $O$? '' All subspace is equipped with the subspace topology. Thanks in advanced!

[gentle osprey]

Tanxs

[rain nova]

I googled it and AI gave a counterexample: ``let $X=(0,1)\cup (2,3)$ and $Y=(0,1)$ so Y is connected in Y, then take $A=(0,1.5)$ and $B=(1.5,3)$ we have $A\cap B=\emptyset$ , both open in $X$ and $Y\subseteq A\cup B$, this implies Y is not connected in X." However I find this explanation not valid since $A, B$ is not both non empty, indeed $B\cap Y=\emptyset$.

[gentle osprey]

Tanxs

[velvet salmon]

If V was not connected in O, we would have opens A, B in V (in the subspace top w.r.t. O) that are disjoint, non-empty and cover V. But then A and B are also open in the subspace topology of U: there exist opens A', B' in O that satisfy A = A' \cap V (equivalently for B) so A = (A' \cap U) \cap V and the term A' \cap U is an open subset of U by definition of the aubapace topology, so A is open in V in the subspace top w.r.t U

So this is impossible

[opaque scroll]

Your definition is not quite right.

A topological space is connected if it is not the disjoint union of two open subsets.

A subset of a topological space is connected if it is a connected subspace. The property is not relative, S is not connected "in X", it's just connected.

Also the definition you wrote would imply any non-open set is connected, which would be weird.

[opaque scroll]

Your definition said that S was connected if it could not be written as a union of two disjoint open sets. If S could be written as such a union S would be open

[rain nova]

How would this implies non open subset is connected? Another equivalent definition of connected is "a space is connected if and only if its only clopen subset is empty set and itself." Consider the statement: ``if a subset $C$ is connected then there are no disjoint proper non empty subsets $A,B$ such that $C=A\cup B$. Suppose the assumption and consider contradiction. That is we take such clopen subset $O\subseteq C$ then we can make $A=C\setminus O$ and $B=O$, this complete the proof. The converse of this statement is obviously true.

[gentle osprey]

Tanxs

[rain nova]

@opaque scroll

[opaque scroll]

Take for example {0, 1} in R. It cannot be written as the union of two open sets, because it's not open.

My point being that connectedness is a property of a space, not a set

You seem to have the right definition in mind. But since the definition only depends on the space, a question like V being connected in U but not in X wouldn't make sense. V is V whether it's in U or X.

[rain nova]

But you can verify that ``a space is connected iff its only clopen subset is empty set and itself.'' With this fact we can in turn show that "a space is connected iff it cannot be union of two disjoint proper non empty subset."

[opaque scroll]

Yes, both the definition you've written here are correct

[rain nova]

I know, but considered R and its subspace Q. Singleton is connected in Q and R simultaneously and I wonder is there a counterexample

[opaque scroll]

What does "connected in Q" mean?

[rain nova]

I mean, when subspace is involved we usually add ``...in (that subspace)'' right? Like open in XX, closure in XX

[opaque scroll]

If "connected in X" just means "connected", then
"connected in Q" implies "connected in R"

Because both just mean the same thing

[unreal stratus]

These are things that depend on a bigger space, but connectedness does not (they are relative whilst connectedness is absolute)

[rain nova]

Let's say $\mathbb{Q}$ with Euclidean topology and $O\subseteq \mathbb{Q}$ with subspace topology. By totally disconnected property of $\mathbb{Q}$ the cardinality of connected subset is the cardinality of $\mathbb{Q}$. What about cardinality of connected subset in $O$? Can this number bigger than cardinality of $\mathbb{Q}$?

[gentle osprey]

Tanxs

[opaque scroll]

A subset of O is also a subset of Q...

Like if you have a subspace of O, that's also a subspace of Q.

And if it happens to be connected, then it's a connected subspace of both

[unreal stratus]

A subspace cannot have larger cardinality

[opaque scroll]

I think they mean the number of subsets

[unreal stratus]

Oh lol the set of connected subsets

[rain nova]

This question encountered when I try to verify that subspace of totally disconnected space is totally disconnected. So, yeah I knew the answer is it will not bigger than cardinality of $\mathbb{Q}$ but how can we shows this is the case. My attempt is ``if there is a component $C_O(q)$ in $O$ of $q$ which is not singleton, says $C_O(q)={x,y}$ for some $y\in O$ then it will contradict the fact that $\mathbb{Q}$ is totally disconnected since $C_O(q)={x,y}$ is a connected set in $\mathbb{Q}$ ''. But I could not ensure this.

[gentle osprey]

Tanxs

[rain nova]

Oh crap I think I got it... connected subset means those subset as a subspace is connected, so yeah if it is connected in $O$ (which mean it is a subspace of $O$) then it will automatically connected in $\mathbb{Q}$ (since subspace of a subspace is also a subspace of the original space)...

[gentle osprey]

Tanxs

[frank path]

Hello i am looking for a graphing software that can graph relation graphs and cant find one. I just need one that can plot points and connect them using differing colored line segments.

[kind marlin]

this might not be the perfect tool but quiver lets you plot points and color the segments

https://q.uiver.app/

[worn mortar]

Quiver is the perfect name for such a thing

[quartz horizon]

Quiver meaning “bundle of arrows”, of course

[raw ledge]

If one has a dynamical process that does not have a preferred or intrinsic underlying geometric structure, is there a way to encode something about this system using a set of points and have whatever it encodes about the dynamical process natural in a categorical sense?

[gaunt laurel]

I don't think this belongs here

[woven trail]

can you recover a topological space by what each sequence in it converges to?

[alpine nest]

As in "can you characterize a topology by describing which sequences are convergent, ant to what"?

[woven trail]

yes

[alpine nest]

In general no, for example if you consider R with the discrete and with the cocountable topology, then in both cases the only convergent sequences are eventually constant ones.

[woven trail]

aw

[alpine nest]

You can characterize topology in this way if you generalize to nets (or filters)

But sequences as such will not suffice

[woven trail]

ohh

[alpine nest]

(this is Willard)

I think sequences will be enough to describe a metrizable topology, but I'm not 100% on that

[woven trail]

I was thinking of that specific case too

[alpine nest]

Apparently yes: https://mathoverflow.net/a/36382

[woven trail]

:000

this question was basically by convenience cus I see a lot of topologies being kinda described by what mode of convergence they imply

like for function spaces yk

[alpine nest]

Yep

Convergence in measure and in L^p is in fact metrizable (or semimetrizable if you don't identify functions that only disagree or a null set)

Weak topology isn't metrizable, so I'm not sure to what extent weak convergence characterizes the weak topology

[woven trail]

this is very convenient to know then indeed yey

[woven trail]

:(

makes sense

[alpine nest]

I mean, the statement "a sequence converges in the weak topology if and only if <it satisfies the standard definition of weak convergence>" is still true.

But just stating the definition of weak convergence might not be enough to satisfyingly describe what the open/basic sets of the weak topology look like.

[woven trail]

I'm reading some old ass french paper on hamornic analysis in locally compact abelian groups and I'm seeing a lot of topologies

not too lot but some

[woven trail]

makes sense

yeah you still need nets apparently

[crimson terrace]

For a metric space such that it is not necessarily true that every infinite set has a limit point, would the values that the metric take not be "smooth"? I know this is very informal but I am not sure how I would explain it... Like, the closest points aren't arbitrarily close and the space has alot of air between it

Also, is there a different notion of "compactness" (compactness as in packed together) such that we can have this fake "compactness" while allowing "some air" in between?

[opaque scroll]

I'm not sure what you mean by smooth, but R seems pretty smooth and has an infinite subset without limit points.

This should just correspond to whether your metric space is compact

[crimson terrace]

for some context, my question is motivated by this

[crimson terrace]

aw okay I see

Are there some other measure of how much a set is "packed together"?

[opaque scroll]

Well, I'm not sure what you mean exactly

[crimson terrace]

hmm okay i guess sequential compactness and these other formulations

yeah sorry I guess it isn't really clear what I mean (and I don't have a clear idea of what I am asking)
Thank you!!

[warped helm]

condensation points perhaps?

these are points where every neighborhood intersects the set uncountably many times

[crimson terrace]

yes!!! something like that
but I guess this shows that the set is "highly packed"
I guess something more relaxed is useless which explains why it doesn'te exist

[lament steppe]

If we have a map f: X -> Y, an open set in Y named U, an open set in X named V such that the restriction of f to V ( which ill call f|_V ) is continuous, this implies that (f|_V)^(-1)(U) is open in V, right? Is it also true this inverse image is open in X?

V here is not defined to a a subspace of X, so the only topology to go off of are the open sets in X. So saying that (f|_V)^(-1)(U) is open in V means that this inverse image is just one of the sets that is an element of the topology of X. Is this correct? Maybe there is another way to look at this?

I say that V is not defined to be a subspace of X because this comes from a proof in a book Im looking at, and this proof is BEFORE they introduce subspaces... so hoping to piece this together without subspaces in mind.

[somber gorge]

do you mean that U is a subspace of X? or do you mean the domain restriction of f to f^{-1}(U)?

[lament steppe]

Neither U or V are defined to be subspaces. Im saying that "(f|_V)^(-1)(U)" is the inverse image of the f restricted to V acting upon U.

[warped helm]

er

i missed where you said that V was open in X

if V is open in X then sets that are open relative to V are also open in X

[lament steppe]

There is a Proposition in the book that says exactly what you said @warped helm , but with the added condition that V is a subspace of X. 🤔

Im wondering if restricting f to V ends up working the same as calling V a subspace of X

Since (f|_V)^(-1)(U) = f^(-1)(U) n V

[warped helm]

that works

hm well its not granted that f^(-1)(U) will be open

[lament steppe]

right

[warped helm]

weird

are these just general topological spaces?

or metric

[lament steppe]

Yes. Let me post an image to makes things more clear from what the book is saying.

That statement Im talking about is right at the end there
"and is therefore also an open subset of X"

[warped helm]

yeah so it seems they messed up the ordering of things if they havent talked about what “open subset of V_x” means

[lament steppe]

FWIW, I have gotten way further in my book and have had some introduction to subspaces, and I just happened to come back to this proof because it was referenced later on and Im stumbling at this part (tho im wondering if I ever understood this when I first read this now)

So yeah, if we assume that V_x is a subspace of X then f^(-1)(U) has to be open in X (based on the equation at the bottom of my image)
Wasnt sure if there was another way to understand this, but yeah, maybe its just an oversight in the book.

I guess since f^(-1)(U) is open in V and V is open in X, then f^(-1)(U) is open in X if V is a subspace topology (doesnt seem like f^(-1)(U) n V = (f|_V)^(-1)(U) actually proves what we want here.)

[warped helm]

how does the book continue the proof after the last equation

since i notice there's a comma

[warped helm]

and what is proposition 2.8(g)

[warped helm]

is it that U is open if every point has a neighborhood contained in U

yes

then that's how they reconcile the set being open

[warped helm]

where is this stated in the book?

[lament steppe]

The following chapter, page 49

[lament steppe]

Im still trying to process thoughts here... not totally convinced of this fact yet using the idea that all points in V_x have a neighborhood

Oh excuse me, thats where subspaces are introduced... that proposition is a little after page 51 (if that helps at all)

[warped helm]

so the order is just wrong

[lament steppe]

So I stumbled upon an errata of the book Im looking at and there is some extra clarification for this problem:

Not sure if this topology is standard for function restrictions... but this topology isnt even quite the subspace topology (tho kinda close)

[warped helm]

it is the subspace topology

[lament steppe]

Wait... it is? It says all open subsets of X that are contained in Y. The subspace topology is about the intersection of the subspace with open subsets of the parent space, right?

[warped helm]

U cap Y is contained in Y

[lament steppe]

ohhhh.. right. Exercise 2.5 doesnt say Y has to be open... but the proposition does say this.

So yeah, I think you are correct.

[warped helm]

exercise 2.5 does say Y is open

[lament steppe]

😮‍💨

Its been a long night... thanks 😄

[calm olive]

Yw

[young stone]

<@&268886789983436800>

[velvet grove]

Hi,what book is this from? Thanks

[wide kayak]

If I'm not mistaken that's John M Lee, Introduction to Topological Manifolds

[viscid blade]

thoughts on gamelin greene ?

how it differe to munkres

[quick delta]

What’s the contents page?

[viscid blade]

i think its similar to smth standard like munkres im windering more about exposition

[copper quiver]

From Hatcher's VBKT. Note that paracompact is defined as paracompact Hausdorff.

I don't understand the last part of the proof, namely the one that claimed that every varphi_gamma is supported in some V_k.

Also in the last line gamma and k are switched lmao.

[copper quiver]

Ig you could add a V_infty, and it would still be a disjoint union of opens contained in some U_alpha, right?

No, V_infty would be empty.

Wait, we're overcomplicating things, just take a new partition of unity wrt V_k 💀.

[lament steppe]

Yes it is John Lee's introduction to topological manifolds (as others pointed out) 2nd edition.

[little vine]

This is from munkres calculus on manifolds, I have no clue how this holds. I do have something in mind but if what I say doesn't make sense then you may ignore it.

Can I do something like taking an open cover omega of {x} X B,
then for each of the open sets consider the R^m part of the set or something which forms an open cover of B in R^m,
which gives us an finite subcover of B and then we consider the R^(m+n) versions of those sets that we took from omega,
then adjoin it with an open set in omega that contains x to finally get a finite subcover of {x} X B

then for each of the open sets consider the R^m part of the set or something

by this I mean if pi_i is a projection function then considering pi_m(W) for W in omega or smth

[queen prism]

nvm what I had in mind is not gonna help for compactness

[warped helm]

compactness is a topological invariant

what should x X B be homeomorphic to

[little vine]

the set of points (x1,...,x_m+n) where (x1,..,xn) = x and (x_n+1,....x_n+m) in B?

[warped helm]

x is fixed here

the answer is B

[little vine]

oh..

Yeah

Alright makes sense, ty

Also while there is practically no difference between them, R^(n+m) and (R^n) X (R^m) are different sets right

[queen prism]

they're different sets but there's an obvious bijection between them

[little vine]

alright, thanks

[quartz horizon]

product of compact sets is compact

[little vine]

Here they were basically building up some theory to prove that only

[quartz horizon]

i see

[copper quiver]

Ah yes, using Tychonoff to prove that {*}xK is compact.

.

[warped helm]

You dont necessarily need tychonoff since its just a finite product

[copper quiver]

Still, the meme applies.

[queen prism]

is a cinema compact

[copper quiver]

Depends, is it complete?

[unreal stratus]

Maybe it's not metrisable though

[thorny agate]

who is that 💀

[wide kayak]

Serge Lang

[lament steppe]

So something Ive been grappling with is when Im asked to show that 2 topologies are equal. In my head there are 2 routes for this. Let my 2 topologies be T1 and T2

1.) The sort of "set theory inclusion approach." Show T1 includes T2 and vice versa to show that T1 = T2 (so they have all the same elements)
2.) The homeomorphism approach. Show there exists a homeomorphism between T1 and T2

To me, #1 means these 2 topologies are literally the same thing (same elements). #2 feels like a lesser condition where the elements might be different, but they "behavior" the same way (I know Im packing a lot of details into the word behavior here...)

Anyone have thoughts that might help with these perspectives?

My motivation right now is that Ive been asked to show that Product Topologies are "associative." In other words, I need to show that X1 x X2 x X3 is "equal" to X1 x (X2 x X3) etc etc...

At the moment, it feels like I can't actually use the #1 approach because that would mean Im trying to say elements of the form (a,b,c) are exactly the same as elements of the form (a, (b,c))... which are not structurally the same (even though I just want to say they are the same and it seems kinda trivial)

So I guess Im sorta "stuck" with using method #2 where I make a homeomorphism using canonical projections to map between these different things.

Am I approaching this correctly, or am I being dense?

[quartz horizon]

You’ve identified a good distinction here

My preferred approach would be to use the universal property here

Approach 1 can’t work

Approach 2 can, but often you want something a little stronger than “they are homeomorphic”

You want them to be homeomorphic by some canonical map

Are you familiar with the universal property of the product?

[lament steppe]

Maybe that is the same thing as the "Characteristic Property of the Product Topology?"

[quartz horizon]

It’s likely similar, but let me state the version I like

Given two topological spaces A and B, what the product space A x B “does” is the following

[quartz horizon]

If you have a continuous map $f : Z \to A \times B$, then you can unpackage it to a pair of continuous maps $g : Z \to A$ and $h : Z \to B$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

By writing $f(z) = (g(z), h(z))$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

And the reverse is also true - if you have a pair of continuous maps $g : Z \to A$ and $h : Z \to B$, you can package them into a continuous map $f : Z \to A \times B$ defined by $f(z) = (g(z), h(z))$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

So, what the product topology “does” is let you package and unpackage continuous maps

In a natural way

[lament steppe]

Yeah, that seems to agree with the definition I have on my end. So composing a homeomorphism using this idea is probably the best I can do, right?

[quartz horizon]

Well, you can approach it in a neater way

There’s a general result from category theory which says that if two objects “do” the same thing, then they are isomorphic

So what you can do is the following

Take an arbitrary topological spaces $Z$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

Then continuous maps $Z \to (A \times B) \times C$ naturally correspond to pairs of continuous maps $Z \to A \times B, Z \to C$, right?

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

By the universal property

But those naturally correspond to triples of continuous maps $Z \to A, Z \to B, Z \to C$, again by the universal property

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

I’ve just “unpackaged” the map Z -> A x B

Now we can start doing some repackaging

These naturally correspond to pairs of continuous maps $Z \to A, Z \to B \times C$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

Which naturally correspond to continuous maps $Z \to A \times (B \times C)$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

Did you follow all that?

[lament steppe]

Yeah, I think I do follow the high level thoughts here, I suppose Im not confident in the rigor here (though I believe you)

[quartz horizon]

I don’t blame you for being a little concerned about the rigor

The missing piece is defining what “natural” means here

Have you come across this notion beforehand?

[quartz horizon]

Once you do, this argument is fully rigorous

[lament steppe]

Natural being similar to the word Canonical?

I have zero experience with Category theory, but it sounds interesting 😄

[quartz horizon]

Kind of, but in this case it has a formal meaning!

You should intuitively think of “naturality” as a categorification of “orthogonality/independence”

You have two different operations that are “independent of/orthogonal to” each other, in the sense that the order in which you apply them doesn’t matter and gives you the same result

For example, algebraic structures have “internal” operations like the group operation, and “external” transformations like homomorphisms

[crisp lintel]

They aren't the same, for method 2 what you want to do is show that specifically the identity map is a homeomorphism

[quartz horizon]

These are independent of/orthogonal to each other, because f(xy) = f(x) f(y)

[crisp lintel]

Of course if two topological spaces are homeomorphic that's it's own thing, but for showing two topologies on the same set are equal it's not sufficient

[quartz horizon]

In fact, homomorphisms are defined to be orthogonal to the internal structure of your algebraic objects

Another good example is precomposition and postcomposition

Associativity tells you that $(f \circ g) \circ h = f \circ (g \circ h)$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

You can view this as saying that postcomposition (by f) and precomposition (by h) are orthogonal/independent operations

You get the same result regardless of the order in which you apply them

Do those examples make sense?

[lament steppe]

I definitely need to take a second to digest these thoughts! I gotta run to work but I will be back 😄

[quartz horizon]

Ok

[lament steppe]

Ok, yeah what you are saying definitely makes sense, now that I had a bit to think about it. Thanks for shedding some light on Category Theory related fun.

[lament steppe]

For the specific example I highlighted above, I dont believe I can use an identity map here. I think the best I can do is have some canonical projections to construct a homeomorphism.

To me, an identity map is one that outputs w.e input is given to the map. So if Id is our identity map, and x is an elements of (X1 x X2) x X3, then Id(x) is another element of (X1 x X2) x X3 and NOT an element of X1 x X2 x X3.

Maybe canonical projections fall under the umbrella of what you mean by identity? (so a homeomorphism that is composed of canonical projections really DOES mean 2 topologies are equal?)

[crisp lintel]

yeah for product topologies the topology would not be equal

so you would need to exhibit a homeomorphism

[lament steppe]

Thank you!

[quartz horizon]

So we can relate this to the product topology now

The key idea is that packaging/unpackaging is orthogonal to/independent of precomposition

So if I have $g : Z \to A$ and $h : Z \to B$ and $\alpha : W \to Z$, then $(g, h) \circ \alpha = (g \circ \alpha, h \circ \alpha)$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

Can you see why this is true?

[lament steppe]

Are there any steps to seeing why those 2 concepts are independent?

[quartz horizon]

you just need to check the equality of functions

by which i mean, demonstrate that $\forall w \in W, (g, h) \circ \alpha = (g \circ \alpha, h \circ \alpha)(w)$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[lament steppe]

I haven't thought through exactly how to accomplish that, but that seems reasonable.

[quartz horizon]

it's mostly just unfolding definitions

you know that the definition of function composition is $(\beta \circ \gamma)(x) := \beta(\gamma(x))$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

and the definition of packaging is $(g, h)(x) := (g(x), h(x))$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

so, try applying those

and lmk when you’ve done that

[lament steppe]

Ah yeah, I think I see this coming together. Question for you, I like to really dig into a subject when I'm learning, so do you have any recs for intro category theory books? I'm a self learner if that gives you some perspective.

[quartz horizon]

Awodey's "category theory" is great

anyway, the formal proof goes as follows

$(g, h) \circ \alpha = (g, h)(\alpha(w)) = (g(\alpha(w)), h(\alpha(w))) = ((g \circ \alpha)(w), (h \circ \alpha)(w)) = (g \circ \alpha, h \circ \alpha)(w)$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

since this holds for all $w \in W$, we obtain $(g, h) \circ \alpha = (g \circ \alpha, h \circ \alpha)$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

so packaging commutes with/is orthogonal to/is independent of precomposition

intuitively, packaging happens on the codomain side, while precomposition happens on the precomposition side

unpackaging also commutes with precomposition, but this is a little easier to show

if you have $f : Z \to A \times B$, then the unpackaged functions are just $\pi_A \circ f : Z \to A, \pi_B \circ f : Z \to B$, for $\pi_A : A \times B \to A$ and $\pi_B : A \times B \to B$ the projection functions

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

so this boils down to associativity

$\pi_A \circ (f \circ \alpha) = (\pi_A \circ f) \circ \alpha$, and similarly $\pi_B \circ (f \circ \alpha) = (\pi_B \circ f) \circ \alpha$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

does that all make sense?

[sick elbow]

Hello everyone,

I have been trying to construct a countable topological space (base space is countable) that is not first countable, but I have not made much progress so far. I can see that this isn't possible in metric spaces and I also tried to exploit some necessary consequences of first countability, but they seem harder to use constructively than the definition itself.

A small hint or nudge in the right direction would be very much appreciated.

[crisp lintel]

I'd probably pick a point and try to construct an uncountable set of subsets containing the point such that no subset contains another

idk if that's possible but

that's probably where I'd start

and then if you can do that, maybe try to build a topology from that

[sick elbow]

Ok Ok, i was also thinking something similar to declare all sets around a point to be open..( i mean not singletons)

[crisp lintel]

yeah the problem with that is that then every set around the point contains the singleton, so you have a finite base

[quartz horizon]

@lament steppe so we can go through this argument again

When i say “naturally”, you can read that more formally as “commutes with precomposition”

And all of those statements hold rigorously because we’ve shown packaging and unpackaging commute with precomposition, hence are “natural” operations

The upshot is that we have a natural bijective correspondence between continuous maps $Z \to (A \times B) \times C$ and $Z \to A \times (B \times C)$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

Not only are these sets in bijection, but the bijections we use commute with precomposition

Do you agree with all of that @lament steppe ?

[sick elbow]

This means you want to construct unctbl sets of disjoint sets in a ctbl space which isn't possible right? (Actually finite intersections is possible but not disjoint).

The construction that I was looking for has one example as Arens-Fort space (as given in Steen and Seebach)

[crisp lintel]

They don't need to be disjoint

I think this is kinda what happens for the fort space

[sick elbow]

oh. I thought when you said no subset contains another, you meant almost disjoint, except the point in context, but that would be same as saying it is disjoint because this would imply the we get an unctbl family of disjoint things.

[sick elbow]

Here's the example....

[crisp lintel]

a family of sets can each not contain each other while still having a lot of overlap, for example the sets $\mathbb{N}\setminus{k}$ for $k\in\mathbb{N}$ almost completely overlap but none of them contain another

[gentle osprey]

Blake

[crisp lintel]

anyway I think the fort space construction is a bit better because it proves non-first countability using a concrete sequence example

[rancid umbra]

a simpler example, imo, is \omega_1 + 1 with the order topology, where \omega_1 is the first uncountable ordinal

[opaque scroll]

That's not countable though

[civic verge]

Guys, that existence of Z_{i} confuses me; it says it's because of the topology of the subspace.

[warped helm]

the subspace topology on a subset S of a topological space X is given by {U cap S : U open in X}

[lament steppe]

Yeah pretty much. I feel a tad out of my element because there is a lot here and category theory seems a bit intimidating, but I think I grasp what you are saying!

[quartz horizon]

The final step is using what’s called the yoneda lemma

You can think of it as saying precomposition and postcomposition are “orthogonal complements”

We have a map taking $f : Z \to (A \times B) \times C$ to $f^T : Z \to A \times (B \times C)$ which commutes with precomposition, meaning $(f \circ g)^T = f^T \circ g$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

But that means $f^T = (\text{id}{(A \times B) \times C} \circ f)^T = (\text{id}{(A \times B) \times C})^T \circ f$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

In other words, “being orthogonal to precomposition” is equivalent to “being postcomposition by a fixed map”

Similarly, we have our map $g : Z \to A \times (B \times C)$ to $g^T : Z \to (A \times B) \times C$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

Which has to take the form $g^T = (\text{id}_{A \times (B \times C)}^T) \circ g$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

Then, we know that $(f^T)^T = f$, since these correspondences are inverse to each other

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

That means $(\text{id}{A \times (B \times C)})^T \circ (\text{id}{(A \times B) \times C})^T \circ f = f$ for every $f : Z \to (A \times B) \times C$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

Applying this to $f = \text{id}{(A \times B) \times C}$, we obtain $(\text{id}{A \times (B \times C)})^T \circ (\text{id}{(A \times B) \times C})^T = \text{id}{(A \times B) \times C}$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

Similarly, using $(g^T)^T = g$, we obtain $(\text{id}{(A \times B) \times C})^T \circ (\text{id}{A \times (B \times C)})^T = \text{id}_{(A \times B) \times C}$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

So this gives us the isomorphism between the two spaces! And completes the proof

Lmk if you have any Qs @lament steppe

[rancid umbra]

i like Pseudo's approach, in fact, this is my preferred method too, but it might feel like it has a high barrier to entry, just because of all the categorical jargon. i know that is how i felt when i first started learning category theory, at least.

your method 1.) only works with two topologies defined on the same set. but, as you have noted, (X1 x X2) x X3 and X1 x (X2 x X3) are not the same sets, so approach 1 isn't even applicable!
method 2 is really the only way to go, as far as topology is concerned. "equal" in this context really means homeomorphic. you could potentially have this same discrepancy when working with, say, groups. two groups are literally equal if they have the same set of elements and the same group multiplication, but two groups can be isomorphic without having the same underlying sets, e.g., Z and 2Z are isomorphic as groups, but 2Z contains only the even integers.

here is a more elementary approach:
there is a function a : X1 x (X2 x X3) -> (X1 x X2) x X3 given by a(x1,(x2,x3)) = ((x1,x2),x3). we want to show that this map a is continuous. my advice would be to write a as a composition of continuous functions.

there are a few functions here that we know are continuous:

• the projections X x Y -> X and X x Y -> Y are always continuous,
• if f : X -> Y and g : W -> Z are continuous, then the product function f x g : X x W -> Y x Z is continuous
• if f : Z -> X and g : Z -> Y are continuous, then the function (f,g) : Z -> X x Y is continuous (this is the characteristic property)
  you can use these facts to decompose a as a composition of continuous functions.

a has an obvious inverse, and you can decompose this inverse into continuous functions in a similar manner

[lament steppe]

Thanks for clarifying! What you have outlined here is essentially what I have worked out on my end (tho I was definitely taking for granted the 3rd bullet pt when constructing my homeomorphism so its nice to be able to point that out in my proof.)

[quartz horizon]

This might be a nicer approach actually

The main thing for my approach is understanding naturality

Would you say that’s the main piece of “categorical jargon” i introduced?

[runic storm]

Are connected hausdorff spaces necessarily path connected? Context, I'm trying to construct a path between any two points in a (connected component) of a smooth manifold

[quartz horizon]

I don’t think so

[quick delta]

No
The standard topologist’s sine curve should work

[runic storm]

Yeah that would do it

[quick delta]

That is possible to do
But it does need “locally path connected” (which is implied by manifold)

[runic storm]

Okay great, thanks

I'll look into it

[rancid umbra]

yea, i also didn’t mean this as a jab either ha

[quartz horizon]

Oh dw I didn’t think it was

[rancid umbra]

but yea, this was how i showed that a few categories were monoidal,

i.e., that the associator is natural

[rancid umbra]

connected + locally path connected => path connected. my fav proof of this is via chain connectedness

[copper quiver]

I like the proof that goes let U be the set of points reachable by x. U is clopen. We win.

[quartz horizon]

(Clopen and nonempty)

[copper quiver]

I mean yeah.

[somber gorge]

does this proof work? im not very used to working with nets so id appreciate any feedback

[desert vortex]

hello how do you get that x U contains at most finitely many x_i ?

and what do you define as a chain? you mean x_a, x_b, x_c, x_d form a chain if a <= b <= c <= d in your directed set I ?

[somber gorge]

yeah that

we know such a U exists by assumption that x_i has no convergent subnet

[desert vortex]

thats the part I dont understand sorry

[somber gorge]

by definition, a convergent subnet is one where every neighborhood of some limit contains infinitely long chains of (x_i)

i think?

[desert vortex]

no that doesnt sound correct to me

[somber gorge]

hmm i mean the definition im working with is that x_i converges to x if the net is eventually in every neighborhood of x

[desert vortex]

yeah

[somber gorge]

so if the net is eventually in U, then it contains an infinitely long chain? since there exists some a such that for every b>=a, x_b is in U

[desert vortex]

do you know what this word "cofinal" means in directed sets?

J is cofinal subset in I iff for every i in I there is some j in J with i <= j

so for example the odd numbers is cofinal in N

[somber gorge]

yeah im familiar

[desert vortex]

but N is not cofinal in N U {+infty}

Im just thinking because I dont see how this relates to what you are saying, subnets have to be cofinal like if you wanna restrict x_i to some set J in I then J has to be cofinal to call x_j a subnet

anyway I found it easier to do a direct proof of this exercise, with nets that is

but I wanna try without nets awell just pure open set

[somber gorge]

i guess just saying that G = K^{-1}K is compact works

[quartz horizon]

oh that's neat

$K \times K$ is compact, and then you use continuous images of compact sets are compact?

[desert vortex]

smart

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[somber gorge]

yes

[quartz horizon]

since the map $(a, b) \mapsto a^{-1} b$ is continuous

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[somber gorge]

indeed

[quartz horizon]

this honestly feels more illuminating than the net proof

[desert vortex]

Im happy I had the courage to open this book

yeah

[somber gorge]

yeah idk why this didnt cross my mind in the first attempt

[quartz horizon]

you might be interested in the concept of "filter quantifiers"

they help formalise this notion of eventually/frequently

Let $D$ be a directed set, and $p : D \to {0, 1}$ be a predicate on this set

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

we say that $p$ holds $\textit{eventually}$ if and only if $\exists d \in D, \forall x \geq d, p(x) = 1$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

we say that $p$ holds $\textit{frequently}$ if and only if $\forall d \in D, \exists x \geq d, p(x) = 1$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

you can think of these as generalisations of $\forall$ and $\exists$ in the context of the directed set

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[somber gorge]

i see, interesting. thanks

[quartz horizon]

for example, the convergent sequence definition may be stated as follows

using $\mathbb{N}$ as your directed set

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

$x_n \to L \iff \forall \epsilon > 0, |x_n - L| < \epsilon \text{ eventually }$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

where you can notice that the statement "|x_n - L| < epsilon" is a predicate on N

some useful general properties of these quantifiers are the following

• if p holds eventually and p implies q, then q holds eventually
• if p holds eventually and q holds eventually, the predicate "p and q" holds eventually
• if p holds frequently and p implies q, then q holds frequently
• if the predicate "p or q" holds frequently, then either p holds frequently or q holds frequently

a "cluster point" of a net $(x_\bullet)$ defined on a directed set $D$ is a point $c$ such that, for every neighbourhood $N$ of $c$, $x_\alpha \in N$ frequently

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

this is the same as a limit of a convergent subnet

beware that, for sequences, being a limit of a convergent subsequence is strictly stronger than being a cluster point, in general

[quartz horizon]

the slogan is that limits capture the eventual behaviour of a sequence up to an $\epsilon$ of room

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

one thing i'm curious about

do you think there's a way to communicate the idea of naturality without needing to introduce new jargon?

[fickle tendon]

its the idea of there being a "canonical" way to go from F(X) to G(Y) if there is a "transformation" η : F ⟹ G and a morphism f : F → G
though fill in all this in with the context youre trying to explain it in

[rancid umbra]

hmm. im not satisfied with cannonicity

[fickle tendon]

what i mean is: given this information there seems like there would be two obvious ways to get from one to the other

naturality says that these two obvious ways are, in fact, the same

[rancid umbra]

think you mean f : X -> Y here

[fickle tendon]

lmao whoops

[rancid umbra]

all good

[quartz horizon]

Me neither

[fickle tendon]

it is a deceptively hard question lmao

[rancid umbra]

naturality and products are really closely related

i almost want to say that if you understand product categories and maps out of product categories, you understand naturality

the homotopy perspective might be a good way to communicate it without introducing new jargon.
feels like a lot of people are familiar with that idea

[quartz horizon]

(I’m biased but I agree with this)

[rancid umbra]

like, yea, at face value, this is what a natural transformation does

but i don't think it gets across the idea of naturality?

it doesn't give me the impression that there is some mental model that i can create/latch onto with this

[quartz horizon]

Just speaking from personal experience, I think this can work with a little more depth to the idea

Saying it’s a homotopy alone didn’t work for me because it felt too “discrete” to be one

[fickle tendon]

yeah i was thinking about this but i felt it would be too much jargon lol

[quartz horizon]

Since to me homotopy = continuous transformation

[rancid umbra]

i felt this way when i learned about chain homotopies

[quartz horizon]

But if you talk a little about viewing categories as geometric objects, it makes more sense

[rancid umbra]

yea

[quartz horizon]

In that sense, commutative squares are the “filled in” squares

[rancid umbra]

was just about to say this

[quartz horizon]

How would that work in this example though

[fickle tendon]

we had a convo about this a while back

i believe

[quartz horizon]

One important flavour of category is “morphisms generalise functions”

But I’d say an equally important and vastly underemphasised flavour is “morphisms generalise paths”

Viewing categories as a kind of combinatorial model for a directed space

[fickle tendon]

and looking at finer and finer homotopies of paths (higher homotopies) corresponds to going upwards in the hierarchy of n-categories

[quartz horizon]

Mhm

[quartz horizon]

It’s to the extent that I’m tempted to introduce categories this way rather than the standard approach

At least, it feels a bit more compelling to me

[rancid umbra]

just to be clear, you are talking about the natural isomorphism Hom(Z,X x Y) <-> Hom(Z,X) x Hom(Z,Y)?

[quartz horizon]

Yes

I feel like using “homotopy” here might only lead to further confusion

Because it’s in a topological context so there is a genuine topological notion of homotopy

[rancid umbra]

yea, that is definitely a concern

this one has the added difficulty of being a presheaf

we can work in the opposite category, of course, but still

[rancid umbra]

im not sure tbh

[ruby delta]

then the concerns would become ncerns

[hollow pawn]

would it be a better idea to learn topology (point set) first or measure theory?

[unborn hawk]

point set topology

[hollow pawn]

ty 👍

[trail charm]

do open covers need to be given by proper subsets?

or if X is a topological space can I just say that {X} is an open cover of X

[alpine nest]

Yep

[trail charm]

yay

[alpine nest]

It usually won't be a particularly useful open cover, but it is a valid open cover

[trail charm]

yeah

[alpine nest]

In point-set topology and analysis in general; "subset" typically doesn't mean "proper subset" unless specified

The default inclusion relation is usually weak.

[trail charm]

yeah I was looking at the wiki page and it used \subset instead of \subseteq or \subsetneq so just wanted to make sure

[alpine nest]

Yep \subset in topology almost always means weak inclusion

[trail charm]

got it ty

[prime elbow]

How do I show this defined metric is finite? I don't want use the completeness of R here

[alpine nest]

Cauchy sequences are bounded

[quartz horizon]

this is an important part of the usual proof of cauchy completeness of R

cauchy sequence => bounded

bounded sequence => convergent subsequence

and then cauchy sequence + convergent subsequence => convergent sequence

[torn dome]

If I have a basis for my topology can I write every open set as an arbitrary union of finite intersections of those basis elements?

[tender halo]

yes

[torn dome]

Cool

[tender halo]

but also the intersection part is superfluous

you can do with just unions

[torn dome]

Ah nice, I was gonna use that to abstract out the intersections

Didn't realize that worked in general rather than in my specific case

[tender halo]

being able to write every open set as a union of basis sets is what being a basis is

or at least its equivalent

[torn dome]

Coolio, ty

[alpine nest]

As bussy beaver points out, if it's a basis then you can express any open set as just union of basis sets; the description you're providing characterizes a subbasis

Also conversely, if you start with any family of sets, then their finite intersections form a basis of a topology (which is the smallest topology containing your starting sets)

[torn dome]

Good to know. I'm working on Alg Geo right now and right now the fact I haven't done a proper topology review in like a year is really working against me

[prime elbow]

but bounded sequence implies convergent subsequence doesn't need completness of R?

[prime elbow]

sorry, i don't get it

how boundness helps me here

i have to show d(x_n, y_n) has limit

[torn dome]

Both the sequences x_n and y_n and bounded on account of being Cauchy

[prime elbow]

yes

[torn dome]

So what can you say about the distance between two bounded things?

[prime elbow]

i know d(x_n, y_n) is cauchy

[torn dome]

Then what's the problem?

[prime elbow]

but i don't want to use the completenss of R

[warped helm]

why not…

you have to in order to show the metric’s outputs live in R

[quartz horizon]

Hm the proof i know of does

[warped helm]

this actually is equivalent to completness

[quartz horizon]

It finds a monotonic subsequence, and then uses the monotone convergence theorem

And that requires completeness cause you use a sup or an inf

[warped helm]

you can take that as an axiom but its equal to completness

[quartz horizon]

Yeah there are lots of equivalent forms of completeness

One of the weirder variants i know is the intermediate value theorem

[warped helm]

why ‘weird’? i think its perfectly natural

well

i suppose as a starting point it is a bit strange

i suppose in a roundabout way you can prove that whatever the convergent subsequence converges, the original sequence must converge to as well

[prime elbow]

So how do I show that metric is well defined?

[opaque scroll]

Take two equivalent sequences and write out the definition of the metric I guess

[prime elbow]

Because if I want to construct R from Q so then how can I use that metric if it is not finite?

[prime elbow]

I want to show the metric is finite

I mean lim d(x_n, y_n) exists

[opaque scroll]

Then use that the sequences are Cauchy

I.e. write out what that means to get a bound

[prime elbow]

okay then?