point-set-topology

[gentle girder]

the point is that if you have a path on S^1 v S^1, then choose a starting point on the covering space, we get a unique lift of the entire path to the covering space. (here, going around A and picking the origin corresponds to f~, and going around B and picking the origin corresponds to g~)

[blissful yoke]

yeah i get it now, like the x-axis segments are mapped to A and each of the circles that are tangent to the integer points are mapped to B

i saw another covering map and it clicked

thanks!

[gentle girder]

you have a name and profile pic that are homotopy-equivalent to someone that I know - must mean that the underlying humans are the same

:3

[signal fox]

Hello, does anybody have a hint for the 6) ?

[whole matrix]

for 2-6 in this image while trying to show that g is continuous, i got that g^{-1} (U) has to be open for all open subsets U of Y but the only open sets in Y are Y and empty set, any tips on showing that g^{-1}(Y) is open in Z

also how are constant maps defined because im also struggling with showing that part too

[signal fox]

g^{-1}(Y) = Z

[signal fox]

fix z in Z then a constant map g Y -> Z is defined by g(y) = z for all y in Y

[whole matrix]

wouldnt that require g to be surjective?

[signal fox]

g^{-1}(Y) is defined as y in Y such that g(y) is in Z

[alpine nest]

I'm spitballing here, but maybe fix epsilon and consider the set of A_n of x in [0,1] such that for every m \geq n you've got |f(mx)| \leq epsilon

These should be closed, and their union is all of [0,1], so you can pick finitely many and that should help

But there might be a simpler argument because this feels more complicated than 7, and 7 has a star 😄

[whole matrix]

ahhh i see thanks

[signal fox]

yeah i already tried something like that but i'm gonna do it again

[signal fox]

a star doesn't mean it's complicated it's just mean it's non examinable ah ah

[alpine nest]

I think that should work, once you've got your finitely many sets assume that f doesn't converge to zero and reach a contradiction.

[tiny obsidian]

? I might be missing something but how?

[signal fox]

yeah i don't think this is correct, it would be if the sets were open

[signal fox]

@alpine nest I dit it in this way, i didn't find an easier way

[tiny obsidian]

By definition E δ > 0 tel que

[signal fox]

oups, i'm french

tel que = such that

@tiny obsidian do u think this is correct ?

[tiny obsidian]

it looks right but I'm also not going to look at the algebra near the end too closely

[signal fox]

okay thanks ah ah

[scenic pecan]

Is my answer correct?

$f(x) = 1/x $ when $x >0$ and $1/(1+x)$ when $x=0$. This function when restricted to the closed sets is continuous but the function f is not continuous. The sequence of closed sets are $A_0, A_1, A_2 .....$

[gentle osprey]

NotAntiMatter

[valid zenith]

Lgtm

[scenic pecan]

The map $(x_1,x_2,x_3....x_n) -> ((x_1,x_2...x_{n-1})/(1-x_{n}),|x_n|)$ is the required stereographic projection that maps the closed ball (except the north pole) to $H^n$ ( Upper half plane ) The map is bijective and continous. Is this correct?

[gentle osprey]

NotAntiMatter

[red yoke]

No?

[scenic pecan]

Oh sorry there should be a modulus on x_n

[red yoke]

What is this mapping from and to

Unit ball at origin to upper half plane?

[scenic pecan]

Unit closed ball to upper half plane

[red yoke]

But this is stretching along the x1x2…x(n-1)-plane

[scenic pecan]

This is the surjective map that the book defines

[red yoke]

I think you need to fix your last coordinate (at least)

Cuz the boundary doesn't map to a plane

[scenic pecan]

$((x_1,x2...x_{n-1})/(1-x_{n}),|x_n|)$ Is this not right?

[gentle osprey]

NotAntiMatter

[red yoke]

That looks like a weird bump to me

If you plug in last coordinate = 0 in your formula you should recover x1²+…+xn² = 1 right

[scenic pecan]

yes

oh i see

This should work?

$((x_1,x_2...x_{n-1})/(1-x_{n}),1-|x|^2)$ The boundary of the sphere now maps to the boundary of the half plane.

[gentle osprey]

NotAntiMatter

[red yoke]

Ye

Can you show bijectivity

Actually I don't think it be bijective

[red yoke]

I think a thing you can do is ||translate up by 1, invert about the origin, and translate down by 1/2||

[scenic pecan]

I don't get it.

Oh got it you mean for the last coord

Right?

[red yoke]

Yea

[scenic pecan]

Ok cool now it is bijective

[scenic pecan]

By the way one more terrible mistake was that if the last coord is a large number in the upper half plane then it doesn't have a preimage so it is wrong anyways

[scenic pecan]

But what if the last coordinate in H^n is let's say 100 then |x| is negative, no? so no pre-image

[red yoke]

?

[scenic pecan]

if x_n = 100 then is there a preimage for that in the closed ball? (1/(|x|+1) - 1/2) ? was the formula for last coordinate right?

[red yoke]

I think the coordinates are
(x1, …, x(n-1), xn+1)/(x1²+…+xn²+2xn) - (0, …, 0, 1/2)

[tidal lynx]

How do we know said neighborhood doesn’t intersect E’

To establish that the complement of cl(E) is open, we have to show every point is an interior point, but afaik the proof only finds a neighborhood of each point that doesn’t intersect E, but it may intersect E’ hence cl(E)

[gaunt linden]

If p is not in E, but every neighborhood intersects E, then by definition p would be a limit point of E. Since we're assuming that is not the case, there must be neighborhoods that don't, so pick one of those.

[tidal lynx]

But what if this neighborhood intersects E’

[gaunt linden]

Hmm.

We might need to use that X is a metric space here.

[tidal lynx]

“If N_r(p) n E = {}, then N_r(p) n Ebar = {}. By contradiction: if there was a limit point q of E in N_r(p), then with r' = r - d(p,q) triangle inequality says N_r'(q) subset N_r(p), and since q is a limit point there would have to be a point of E in N_r'(q), so in N_r(p), contradicting N_r(p) n E = {}.”

Copied from reddit

This makes sense

[gaunt linden]

The neighborhood we choose must contains some open ball around p, say one of radius d.
Then the ball of radius d/2 is (by definition) also a neightborhood, and it cannot contain any limit point of E, because the ball of radius d/2 around such a limit point would be disjoint from E, a contradiction.

[gaunt linden]

This works too.

[tidal lynx]

Not fun how rudin decides to omit these things

[tidal lynx]

Hmm not seeing why the justification of “it cannot contain any limit point of E” works

[gaunt linden]

Suppose q is in B(p, d/2). Then by the triangle inequality, B(q, d/2) is a subset of B(p,d) which was assumed disjoint from E. Since q has a neghborhood that is disjoint from E, q is not in E'.

[tidal lynx]

Ahhh I see

[gaunt linden]

The you found on Reddit does essentially the same, just with more footwork so it doesn't need to divide the original d by 2.

[tidal lynx]

Right

Also worth noting that Rudin means “open ball” when he says neighborhood

[white oxide]

Hey so given a tychonoff space of weight k you can embed it in [0,1]^k. So far so good, but how does that alone tell me that [0,1]^k has weight k?

like, transfinite induction on some kind of minimality of k?

I am just wondering how the former statement would be helpful at all in showing the second. I could "brute force" the second statement through the def of the product topology but that isnt exactly intended

[hexed steppe]

what weight does [0,1]^k have?

[white oxide]

k

But thats what I am intended to show using the universality of [0,1]^k

[hexed steppe]

i mean

doesnt it imply [0,1]^k has weight at least k

[white oxide]

Oh right

If it had less weight, then the induced base on the subspace would have weight less than k, which cant happen with a homeomorphic embedding

[hexed steppe]

idk if it gives an uppe rbound

[hexed steppe]

i think if you work along these lines, the upper bound is much easier than the lower bound

so probably that is just the point

you know a countable base for each factor

see eg https://math.stackexchange.com/questions/3593042/weight-of-0-1k-le-k-where-k-is-an-infinite-cardinal

[white oxide]

Yeah, constructing a weight k base is pretty straightforward

And with the universality (and the implicit assumption at least one tychonoff space of weight k exists) I also get a lower bound of k

Thanks!

[shadow charm]

what are some nice sufficient conditions for a metric to be "continuous" in the sense that d(x,y)<a implies the existence of a z with d(x,z)<a/2, d(y,z)<a/2

or even better yet for every lambda a z with d(x,z)<lambda*a and d(y,z)<(1-lambda)*a

[gleaming warren]

I do not know what you are interested in but this property feels like beeing true for normed spaces, indeed taking z=(x+y)/2 works

[shadow charm]

yeah it's definitely true there, i just wanted slightly more generality

without requiring absurd conditions on the space

if that can be done

(like is compactness + a few other conditions enough for example)

[gleaming warren]

@shadow charm First thing i thought of was completeness but after a few seconds you notice that this ain't it since you can come up quite easy with couterexamples

i think normed spaces are not that absurd, but my main interest is functional analysis so my opinion might be a bit biased

[fast viper]

Hey, I have a question: if we have a base B of top. space X and consider L some finite set of elements of B, can we say that B\L is still a base? is it true in T1,T2,T3 spaces? why? i'm sure it's extremely easy, but that's the only party in the proof i'm studying rn that i don't understand 😦
it seems true for T1(so others too), but I have problems with strict proof
thanks for answers

[ocean narwhal]

No. Take a finite topological space.

[fast viper]

fair enough, maybe in regular space this holds?

[gleaming warren]

Something like [0,1] union {2} with the subspace topology should cause problems i think

[tiny obsidian]

or discrete topology (metrizable hence regular) with basis given by each point

[gleaming warren]

You're right this is even easier whilst it is the same idea

[opaque scroll]

So for each point, the intersection of all basis elements containing said point is the same as the intersection of all neighborhoods of said point.

If this intersection is open, then there is a minimal open set containing your point, and removing that will get you in trouble.

If there is no minimal, you can always find a smaller basis set, so you can freely remove finitely many.

So T1 + no isolated points should be enough for example.

[fast viper]

Seems more than logical, also it's clearly now why T1 is so important(since in T1 {x} is closed), but where in your proof there are problems with isolated points?

[opaque scroll]

If {x} is open, then removing {x} will change the topology

[fast viper]

But isn't every one-point set {x} in T1 closed?

[opaque scroll]

Sure

But we want it to be not open

Wheter it's closed or not is irrelevant

[fast viper]

but it can't be open if it's closed can it?

[opaque scroll]

Consider the discrete topology for example

[fast viper]

oh, right, there are not only real numbers😅
thank you, now it's all clear

[opaque scroll]

https://youtu.be/SyD4p8_y8Kw?si=57pXXzOBn-QfY5vh

[opaque scroll]

I mean something like {0, 1} as a subset of the real numbers have {0} as a clopen set. So you don't have to look to very weird spaces.

[fast viper]

oh, true😂

[fast viper]

saved it lol

[white oxide]

the space itself and the empty set are always clopen

[gleaming warren]

Our professor recommended to us to watch this video lol

[gleaming warren]

Yes, by definiton of a topology they are open and complement of each other hence closed

[fast viper]

well at least this i know, but i forgot there are non trivial examples

[fast viper]

im gonna recommend this to our proffesor then lol

[gleaming warren]

Just a nice theorem characterising this: a top. space is connected iff these are the only clopen sets

[white oxide]

Right, lower or upper limit topology work too and are a bit more geometric flavored where clopen sets are everywhere

[regal flint]

Is every subspace of the Sorgenfrey line separable?

[shadow charm]

Turns out my question is pretty related to the idea of a convex metric space which I hadn’t heard of before so I’ll look into that

[gleaming warren]

interesting

[feral copper]

Why had I never seen this before? 😮

This is pure gold!!

[brittle rapids]

second countable spaces are* separable

subspaces inherit cardinality properties of the base

[empty grove]

The Sorgenfrey line isn't separable though is it

[plain raven]

yeah it's pretty good

[merry geode]

I think it is fair to say that something without boundary at all is both closed and open

Like if you only think about the surface, entire Earth is closed in some sense, and open in another sense

Just my 2c

Non-closed open points are cursed tho

[white oxide]

Do you define the sorgenfrey line as being the topology generated by sets [a,b)? If so each of those contains at least one rational

The sorgenfrey plane isn't separable tho

[empty grove]

Ah sorry I meant second countable

It not second countable

[gentle girder]

there was a professor at my school that sent this video to everyone in the department lol

[heady skiff]

the converse of a is false right

[hexed steppe]

justify your answer

[heady skiff]

lmao

well im trying to come up with a counterexample

if it really is false, which i'm guessing it is

[hexed steppe]

don’t guess

well guess

[heady skiff]

lol

[hexed steppe]

and then think why you guessed it

[heady skiff]

well

[knotty vine]

Guessing is one of Polya's strategies

[hexed steppe]

what mechanism do you think is responsible for it being false

[alpine nest]

It's not called guessing, it's called heuristics

[hexed steppe]

even blind guessing is useful insofar as it makes you get started

[heady skiff]

well i'm supposing that O is open in X x Y, so that implies that it's open in X' x Y'. thus, for all (a, b) in A x B = O, we can find open sets U of X and V of Y such that (a, b) in U x V subset A x B. but then we can also find open sets U', V' in X', Y' such that (a, b) in U' x V' subset A x B

however this doesn't carry the implication that U is an open set in curly U

and similarly with V

so essentially, i'm trying to find an open set in X x Y and X' x Y' so that we can find open sets of the respective topologies for each members of A x B, but that they're not contained in each other

and this is where i'm stuck lmfao

[hexed steppe]

what if you used O = U x V

[white oxide]

How do the open sets of X' x Y' look like?

[hexed steppe]

(recall that the product topology has basis given by sets of the form UxV)

[heady skiff]

hmm how would this provide a counterexample? cuz O = U x V open in X x Y => open in X' x Y' => for all (u, v) in U x V we can find open sets U' of curly U and V' of curly V such that (u, v) \subseteq U' x V' \subseteq U x V, but isn't that exactly the statement that curly U is contained in U' (and similarly for curly T)

oh wait so that's a proof for the converse

because we showed that for all basis elements of the product topology on X x Y, we can find a basis element for each member of a fixed basis element that's contained in that basis element no?

[hexed steppe]

that was my thought

[heady skiff]

yeah makes sense

okay thanks

is this theorem still true if the criteria is for (x, y) subset S?

oh wait no that doesn't make sense

wait no yea

because x and y are already in S

so if X is an ordered set and Y is a convex proper subset, then Y is an interval

[alpine nest]

I'm not sure how this works for X other than the reals

Well, or a subset of the reals.

[heady skiff]

true

i guess the proof of this result does rely on AOC

never mind

i guess it's not true

[hexed steppe]

really?

[heady skiff]

yea i looked it up online

[hexed steppe]

like if it’s R

[heady skiff]

oh

well if it's R then it's true

[hexed steppe]

just your original question

[heady skiff]

Let X be an ordered set. If Y is a proper subset of X that is convex in X, does it
follow that Y is an interval or a ray in X?

[hexed steppe]

about (x,y)

sure, that i have no intuition for

i dont see how it could go wrong with (x,y)

because if you have that property

and you assume x,y in S

oh you said this already

[white oxide]

R is an interval under that def?

[heady skiff]

wdym

like

i was just seeing if convexity => being an interval in R

[white oxide]

if I pick two points out of R then [x,y] lies in R. So by the above R is an interval

[heady skiff]

no yea i know that

sorry i'm probably being confusing

[heady skiff]

this was the original question i'm working on

at first i thought it was true because of my intuition/this theorem regarding R

but it doesn't apply to arbitrary topological spaces

[white oxide]

Yeah I just wanna understand what the def is here

[heady skiff]

of convexity?

[white oxide]

interval

[hexed steppe]

its wrt an order

so just take th usual definition

[heady skiff]

oh well that was how they characterized intervals in my real analysis text.

here's how munkres defines it

[white oxide]

"the usual definition" I have in mind is that in the pic above 🙈

[tulip bluff]

If X is a Hausdorff space and A a subspace such that there exists a retraction r : X -> A how can I show that A is closed? I thought about considering x in X \ A and r(x) in A and using the Hausdorff condition to find distinct neighborhoods of these points, but I can't construct an open neighborhood for x that is inside X \ A to conclude that X \ A is open.

[heady skiff]

i don't really understand how this is a counterexample. isn't by definition Y = (0, infty)?

moreover the definition of convexity of a proper subset Y of an ordered set X talks about points of Y

Given an ordered set X, let us say that a subset Y of X is convex if for each pair of points a < b of Y, the entire interval (a, b) of points of X lies in Y

[hexed steppe]

what

what pic

[white oxide]

above*

[heady skiff]

i think he was talking about this one

[hexed steppe]

that is what i meant

by usual definition

[hexed steppe]

“interval or ray in X”

[white oxide]

Yeah and what I meant is that by the theorem that was posted R would be an interval which seems incompatible to me

[hexed steppe]

oh sorry

i see what you meant

a,b are allowed to be -infty, +infty

[heady skiff]

i'm confused, isn't Y = (0, \infty) an interval of X = (-infty, 0) U (0, infty)?

[hexed steppe]

no as 0 is not in X

[heady skiff]

okay now i'm confused about the definitions of intervals 💀

[heady skiff]

going by this definition

[hexed steppe]

dont look at your analysis book

just look at munkres

[heady skiff]

(0, infty) is of the form (a, b)

no?

[hexed steppe]

i mean

this is about deducing consequences from definitions right

[white oxide]

oh, so the analysis book conflates intervals and rays into one concept ig?

[hexed steppe]

so you need to use the precise definitions that the stackexchange poster likely has in mind

[hexed steppe]

i think so yes

[white oxide]

a bit awkward ngl

[heady skiff]

oh makes sense

i didn't read the hypotheses of the interval definitoin

or rather i did read them before but didn't bother to look at them this time

[white oxide]

intervals are either open, half open, or closed and they are sandwiched between two points that actually lie in your set. Rays just specify "all points left/right of this one"

(either strictly or not)

[heady skiff]

let's see if i got my quantifiers correct - a space X is not Hausdorff if there exists a distinct pair of points x1, x2, such that every neighborhood U1 of x1 and U2 of x2 have nonempty intersection

[unreal stratus]

Yes sure

(Though I'd say pair of distinct points rather than distinct pair)

[gentle girder]

how the freak do i compute homology

[unreal stratus]

what is the context lol

[gentle girder]

hatcher exercises getting me down, so not much more context than literally being asked to compute homology

[unreal stratus]

One hint is in your nickname ig

[gentle girder]

so true

[unreal stratus]

No I mean it depends how hard the computation is

if it is hatcher probably mayer vietoris lol

[gentle girder]

oh lol for sure not with a spectral sequence

I am just complaining

because I'm lazy and need to actually read the chapter

[tribal palm]

waw relateable

[knotty vine]

hatcher chapter 5

[unreal stratus]

Nice pfp btw Semer, recently replayed both games lol

[knotty vine]

Ye classic turrets with their funny lil moustaches

[red yoke]

Cellular, MV, relative sequence, excision, quotient, homotopy equivalence, π1, etc

Chapter 2 TLDR

[fervent root]

could i get a sanity check on this

[brittle rapids]

abcd

[fervent root]

good catch. is that all 12?

[red yoke]

Ye that's all

[hidden crag]

Read your nickname my sister in christ

Nvm potato already said that

[unreal stratus]

To be fair somebody said cellular homology which is from a spectral sequence :)

[red yoke]

[knotty vine]

huh? Do you mean the Serre sequence from the CW skeleton filtration? Do you really need all that?

[red yoke]

For reference this be Hatcher

[unreal stratus]

Hm idk if it's due to Serre but if you just filter homology with the skeleton you get a spectral sequence which immediately degenerates and gives you cellular homology

Like the E1 page is the cellular complex and converges on E_2 to the homology

i think

[fervent root]

does this look okay?

[feral copper]

Heyo! So the space of real plane curves of degree d is a real projective space of dimension d(d+3)/2.
It should have a natural stratification, where the codimension zero stratum is the (open dense) subset of non-singular curves, and the codimension one stratum is the subset of curves with a single singularity which is a double transverse singularity. What are strata of greater codimension? How does one define this neatly?

In the codimension two stratum, there should be curves with two transverse singularities, or one unique singularity which is a parabola/line tangency point?

[heady skiff]

are the open sets of R under the cofinite topology of the form R - F where F is a finite set + the empty set + R itself? is that a complete characterization of them?

i'm trying to think of open sets in R under the cofinite topology which are not the empty set or R

and that are not equal to my claimed characterization

i can't think of any lol

so i'm guessin git's true?

[brittle rapids]

the intersection of finite sets is finite

and the finite union of finite sets is finite

[heady skiff]

sorry what do you mean by that

[brittle rapids]

i actually don't completely understand what you mean lol

[heady skiff]

like if O is an open set of R under the cofinite topology

[brittle rapids]

the cofinite topology is defined as taking open sets to have finite complement + empty set

[heady skiff]

and O is not R

or the empty set

then must O = R - F for a finite subset F of R

certainly O is open in the cofinite topology

Cuz R - (R - F) = F is finite

i'm wondering if the converse holds

[white oxide]

Check if the cofinite sets plus the empty set form a topology

If they do, then surely the cofinite topology is the one where every non empty open set is cofinite

[heady skiff]

what do you mean by cofinite sets?

[white oxide]

Finite complement

[heady skiff]

oh okay never mind

i was overthinking it hella

i was trying to show that R under the cofinite topology is not Hausdorff

[white oxide]

It definitely isn't but why

[heady skiff]

but I can just use this: let x and y in R, consider any neighborhood U of x and V of y, we know that R - U and R - V are finite; if R - (V u U) is empty, then V u U is all of R which is a contradiction since U and V are arbitrary; hence it must be nonempty. But R - (V u U) = (R - U) n (R - V) which is nonempty and must be finite; since U and V are infinite, we conclude that U n V must be nonempty

idk if that works

because this would show that it's true for every points x, y of R

but the hausdorff condition only needs a counterexample

so i'm pretty sure there's some faulty logic

[unreal stratus]

Uhh why does U and V being arbitrary mean you get a contradiction

Isn't that sort of assuming what you want

[heady skiff]

oh yeah oops

there it is

[unreal stratus]

This seems a bit convoluted to me

Also you can have two infinite subsets of R with nonempty intersection

Try to involve U n V more explicitly

[white oxide]

Try to make it constructive, that should help clear things up

[unreal stratus]

What is R \ (U cap V) equal to

[white oxide]

You mean cup there?

[unreal stratus]

No

[white oxide]

Right, U and V were the complements not the finite sets themselves

[unreal stratus]

Yeah U and V are opens

[brittle rapids]

ugh work with the complements period

[unreal stratus]

Or at least neighbourhoods, may as well assume open

[brittle rapids]

translate everything implicitly

[white oxide]

Yeah, let U = R - A and V = R - B. Can you phrase U n V in terms of A and B?

[brittle rapids]

even though R cofinite isn't hausdorff, you can do this related exercise: show that the intersection of all hausdorff topologies on R gives the cofinite topology

where we take a topology to be a subset of the powerset of R

[unreal stratus]

oh interesting

i know that the cofinite top is the smallest t1 topology but didn't realise also intersection of hausdorff ones

[white oxide]

Huh?

[unreal stratus]

just the normal definition

[white oxide]

I guess

[heady skiff]

yeah that was i originally tried, their intersection is equal to R - (A U B) which is nonempty since A and B are finite

but i was unsure if that worked

[white oxide]

It does depend on the cardinality of your space

If it is finite then you just have the discrete top

[brittle rapids]

i love using filter-induced topologies in constructions

instead of patching together R^n in different pathological ways

[unreal stratus]

Yes this is the right thing

[heady skiff]

huh okay cool

thanks topologists

[unreal stratus]

In short: U, V have finite complements. So the union of their complements is also finite, and hence not all of R, so U and V must intersect somewhere

This works for any non-finite space

In fact there is a fun game where basically if $\kappa$ is an infinite cardinal (read: "set size") and $X$ a set, then you can put a topology on $X$ where all sets of size $< \kappa$ or $\le \kappa$ are closed e.g. for $\kappa = # \mathbb N$ and taking $<$ you get the cofinite topology

[gentle osprey]

potato

[unreal stratus]

Then you can see that basically the behaviour depends on how |X| relates to kappa

e.g. you can do cocountable topology on a set, and you'll get smth discrete or smth non-hausdorff

[white oxide]

Oh filters... I am still recovering from the whiplash of ultrafilter schenanigans on ßN

[brittle rapids]

i actually wanted to read up on βN and how it relates to stone-cech and dynamics

since they seem to use semigroups a lot

[white oxide]

We only used it to show certain properties of ultrafilters over N, like them having no countable filter base and stuff like that

Might need to get back to it since I heard some cool stuff, but learning and doing the assignment in one day with an approaching deadline wasn't exactly motivating

[brittle rapids]

un four chew net

[white oxide]

What

[brittle rapids]

the free ones?

[brittle rapids]

malapropism

[white oxide]

Yeah

[heady skiff]

oh cool

how hard would y'all say this exercise is

i meant attempt it later but i don't want it to take up too much time

[white oxide]

You should manage a) in finite time

[heady skiff]

good to know

[brittle rapids]

god this exercise was so fun to do

[white oxide]

I'd just recommend starting to play around with it

[dusty talon]

disagree lol

[white oxide]

Think of ways how the formal operations might simplify and when you only get one-way inclusions (so you might assume they are proper). Then try some examples out for b and try to figure out why they fail

[brittle rapids]

why

[dusty talon]

idk it just didn't feel that interesting to me

and it took longer than it should have

[brittle rapids]

are you the type of person who gets frustrated when the solution takes so long to come up with, but it's actually just two or three lines

[heady skiff]

okay fuck it i'll try i tout

is this proof invalid because it doesn't imply that every neighborhood U of x intersects some A_alpha?

or idk how to put it into words

but yeah, like this neighborhood happens to intersect one A_alpha_i

but the next neighborhood might not intersect A_alpha_i

we just know it intersects at least one of the A_alpha

[white oxide]

I mean, I can give you a counterexample to that statement so no, I don't think it's valid

[brittle rapids]

yes exactly

[white oxide]

Let A_n = {1/n} and consider 0, does any A_n manage to hold 0 long enough?

[white oxide]

My blind ass thought that was your proof 🙈

[heady skiff]

LOL all lgood

[brittle rapids]

and the way you put this makes it (hopefully) clear that under certain finiteness conditions, the statement holds

[heady skiff]

this is true tho with intersections right

[heady skiff]

i think it's clear

[brittle rapids]

yes

[heady skiff]

when are things not true with intersections

[unreal stratus]

Wdym

[heady skiff]

nah i was kinda half joking

but to me, it always seem that if A and B have a given property, then A n B automatically has this property

but i'm sure that's not true in general

i.e. if H and K are subgroups then H n K is a subgroup, blah blah blah

anyways how exactly is this true? (-1, 0] is open in (-1, 1) under the subspace top but is not open under the order topology on (-1, 1), right? (since we can't find a basis element of (-1, 1) that contains 0 which is contained in (-1, 0])

[unreal stratus]

No, (-1,0] isn't open in subspace top

In fact if U is an open subset of a space X then a subset of U is open iff it is open in X

(The nontrivial direction: if V is open in U which is open in X, then V = W n U for some W open in X, so V is an intersection of two open subsets of X)

[unreal stratus]

One easy way to see it is true is to use the fact open intervals form a basis for the topology on R (and hence on (-1,1))

[heady skiff]

Nvm I don’t know what i was thinking 😂😂😂

[balmy field]

Ur learning every subject

Thats dope

[heady skiff]

lol wdym

[balmy field]

Like i see you in every channel

Whats your favorite subject? Algebra? Analysis? Or top?

[heady skiff]

oh yeah rn i'm just reviewing point-set, doing/reviewing some galois theory and learning measure theory

negl at this moment i'm leaning towards analysis

[balmy field]

Crazy

[heady skiff]

probably because thsi sem i was traumatized by my algebra class

[balmy field]

...?

[heady skiff]

na not rlly lol i'm learning it hella slow

[heady skiff]

like

i'm kinda tired of algebra atm negl

it used to be my favorite

[balmy field]

How is measure theory?

[heady skiff]

i just started on axler, and just got to the section on outer measures

so i can't really say that much lol since the first two sections were reviews on integration

seems like an interesting subject tho

[brave geode]

if you like nonsense, then the intersection is the limit of the diagram

[brave geode]

and the forgetful functor Grp -> Set is right adjoint, so it preserves limits

(this is the wrong explanation though)

[hidden crag]

This can be formalized somehow with logic iirc?

Someone told me about this a long time ago I think

[naive trench]

This is not the channel but its a pretty abstract and so much formalism

[tulip bluff]

If i : A -> X is a cofibration, then there exists a retraction r : X x I -> X x {0} ∪ A x I. Is there a way to picture what this retraction is doing?

[red yoke]

the retraction is not canonical, as you can imagine from say [1, 2] -> [0, 3]

But composition with r provides a way to extend a homotopy of A and a map of X to a homotopy of X

[tulip bluff]

Can I get a retraction from X -> A from this r : X x I -> X x {0} ∪ A x I somehow?

[red yoke]

No

E.g. global stuff like induced maps on pi1 and (co)homology can obstruct existence of retractions

But cofibrations are more of being local point-set niceness at A

E.g. disk does not retract onto its boundary, but it is a cofibration

[tulip bluff]

But if i : A -> X is a cofibration and X is say Hausdorff the space A should be closed right?

[red yoke]

Yes iirc

[heady skiff]

That is indeed nonsense to me

[empty grove]

The innocence of a child

[white oxide]

One day you too will be blessed by the light of universal properties

[unreal stratus]

yes

often one restricts to convenient categories in which this is always true

[gentle girder]

got blasted on the AT exam

[hidden crag]

what were the exercises

[red yoke]

Wait were you cramming few days ago

[gentle girder]

literally basic stuff, computing homology of not-too-complicated spaces, proof of exactness of Mayer-Vietoris, another diagram chase that I somehow messed up

I just sort of understood almost nothing this semester ngl, I should not have taken this class, I'll take a bad grade on my transcript and move on because what else is there to do?

[woeful coral]

On the 1 dimensional sphere one can traverse the entire thing using a “pencil”

Can you do the same thing for a 2 dimensional sphere? I.e. draw a line and cover the entire thing?

[opaque scroll]

Space filling curves is a thing.

[woeful coral]

Ah perfect thanks

Thats what I needed

[white oxide]

Might stretch your definition a bit since usually the space filling curves are the limit of a sequences of curves that become increasingly dense

[opaque scroll]

Hey, A curve is a curve. They don't all need to be smooth

[red yoke]

Space filling be continuous

But not smooth

[tulip bluff]

If I let X = [0,1] x [0,1] and consider A the subspace to be the union of line segments from the origin to the points (0,1) and (1/n,1) for positive integers n I want to show that there is no retraction from X onto A. My idea was to first note that both spaces are contractible and connected so if such a retraction would exist then the incuded map r_* : H_0(X) -> H_0(A) should be surjective so we have a surjective homomorphism Z -> Z. Would this imply that r would need to be the identity map or something which would not be possible?

[red yoke]

Unfortunately r* is the identity

It just tells you the (only) component is sent to the (only) component

[red yoke]

Hint: ||apply continuity at (0, 1) and get contradiction||

[knotty vine]

I suppose this space is ||the comb space||

[tulip bluff]

Continuity of r?

[red yoke]

Yup

[tulip bluff]

Will r have some sort of an jump at (0,1) or why it cannot be continuous?

[red yoke]

More generally, locally connected spaces can only retract onto locally connected subspaces

[tulip bluff]

A neighborhood of (0,1) in A can't be locally connected as it contains infinitely many disjoint line segments?

[red yoke]

ye

[fervent root]

this question was on my exam today, but the topology was lower limit

i wasn’t sure what the open sets would be beside Q, empty, a and b, and a and b and c

i guess also c?

[heady skiff]

how did it go?

[fervent root]

i mean other than this question i think it went okay

i think i missed c being an open set on this question. i kinda came wondering if i missed anything else haha

[fervent root]

since R with lower limit is finer than R with standard, would it have been right to say that the open sets in R standard would’ve also been open in R with lower limit?

[tiny obsidian]

yes

[winged viper]

Why is the map \tilde{H} here the identity map on the fibers?

i think i'm missing something because it looks like the only condition on \tilde{H} is that the first diagram commutes? which shouldn't imply that it's a bundle isomorphism

[pearl holly]

it does tho

[winged viper]

why?

[pearl holly]

or wait a sec, maybe you need principal ones?

[winged viper]

yeah idk, i thought it was something like that but there seems to be no indication of that

https://math.stanford.edu/~ralph/fiber.pdf

the beginning defines a map of fiber bundles to just be a map where there's a commutative triangle

so it doesnt need to be an iso on fibers

[pearl holly]

ah I see

[winged viper]

tbf i didnt read anything else so idk if halfway through he switches

[pearl holly]

I think inspecting the proof of the lifting/covering homotopy theorem might do it, but like I said, with principal stuff it's a lot easier if you're working with those

[winged viper]

yeah i was trying to do that, but the proof is pretty hairy and i didnt want to get into it :')

[pearl holly]

I can't say much else, sorry. I wish I had something better to say

[winged viper]

haha no worries

ill probably try to read the covering homotopy theorem later

[pearl holly]

hopefully someone else will give a satisfactory answer

[pearl holly]

I don't think it's that bad honestly

and I do believe it should follow from it

[winged viper]

yeah i dont think its bad it just has a lot of notation :(

yeah it seems to just be from the proof of the covering homotopy theorem

i guess bundle map here means it induces iso's on fibers

[pearl holly]

yeah I think that's standard but idk, I'm no expert

but yeah I believe the basic idea should do it

[winged viper]

yeah, the covering homotopy theorem requires that the maps have the same fiber, so like it would make sense if thats what bundle map meant

[pearl holly]

ye it's how Steenrod does it so, so I do too lol

[winged viper]

ohh okay

wikipedia doesn't tho https://en.wikipedia.org/wiki/Bundle_map

[winged viper]

does this lemma have a name?

[warm quiver]

If we’re restricting to spaces with the homotopy type of a CW complex. This is just the homotopy lifting/extension property. You would typically see X be A \times I in which case i is the homotopy equivalence.

[winged viper]

Right, I was wondering if this general version had a name

I guess its just like an axiom

[coarse night]

that's an axiom

nvm wrong context.

[knotty vine]

This lemma, together with the lemma that says any map can be factorized with (co)fibrations/htpy equivs makes a weak factorization system

Or two weak factorization systems ig

Which is again part of what makes a model category.

Specifically, the lemma says that (co)fibrations have the right(left) lifting property against trivial cofibrations(fibrations)

A trivial (aka acyclic) fibration is just a fibration which is also a weak homotopy equivalence.

As Flyon said, if eveythings a CW complex, then weak homotopy equivalences are just homotopy equivalences

I suppose we dont need cw complexes if we're talking about the Hurewicz model structure, but then fibrations are slightly different. Depends on how the book defined them

[ebon galleon]

That's not quite the factorization system, as you'd need Hurewicz cofib followed by Hurewicz fib with one of them being a homotopy equivalence (in fact, you technically need a slightly stronger version of a Hurewicz cofibration: a "closed cofibration" as it's sometimes called).

The mapping path space/mapping cylinder factorizations give you these weaker facorizations, but getting the proper factorizations for the model structure are very nontrivial

You can use something along the lines of Cole's cofibration hypothesis (which IIRC is slightly faulty, use Riehl and Bartel's monomorphism hypothesis instead) + some categorical construction to get it, or you can use the Moore mapping path space for this; either case has quite a few technical lemmas

[knotty vine]

I think I will stay in CW land

wtf is this emote

This server really has a bias in the emoji department...

[opaque cloud]

riehl shit

[opaque cloud]

they're just math department emojis

How do people deal with manifolds with boundary?

[hidden crag]

Stokes

[opaque cloud]

Because if you consider for example the closed interval [a, b] it isn't a manifold because there is no open neighborhood around a and b which is homeormorphic to an open set in R

i.e a manifold with boundary...isn't a manifold?

but surely that cannot be the case as I've heard people talking about manifolds with and without boundary

like stokes theorem for example

[hidden crag]

manifolds with boundary have their own definition

[opaque cloud]

oh I see, how are they defined?

[hidden crag]

look it up

[opaque cloud]

oh "half space"

https://ncatlab.org/nlab/show/manifold+with+boundary

I see this makes sense

[opaque cloud]

so now an open neighborhood around a or b can be homeomorphic to a subset of R^+ U {0}

[gaunt linden]

That is a bit too liberal -- it would make an arbitrary subset of R a "manifold with boundary", which isn't the intention.

There should be a neighborhood homeomorphic to R, or a neighborhood homeomorphic to [0,infty) -- taking arbitrary subsets of the latter will not do.

[red yoke]

Open subsets work though

Just like for manifolds without boundary

[knotty vine]

If the math department consisted of only categorical homotopy theorists... There's Lurie, Riehl, Quillen, Schreiber. The O.G. category theorist Noether. (and Grothendieck for good measure)

[empty grove]

https://ncatlab.org/nlab/show/red+herring+principle

The mathematical red herring principle is the principle that in mathematics, a “red herring” need not, in general, be either red or a herring.

nlab humour

[ebon galleon]

top tier

[feral copper]

Top tier, Grp tier, or even [Δ^op,SSet^J] tier

[knotty vine]

J-indexed bisimplicial set tier

[feral copper]

Sure I honestly don't know what kind of category this is, I just copied random shit from nlab

Anyways; is anybody familiar with the stratification of the space of real algebraic curves?

I'm scared to ask in #algebraic-geometry because I'm only working over R or C, and I care about actual topology and actual geometry

[ebon galleon]

Simplicial objects in simplicial co-presheaves on J....

[heady skiff]

does this exercise just imply that the closure of (a, b) under the order topology is just (a, b) itself? because it was [a, b], then we would obviously have equality, and if contained some point other than [a, b], then we wouldn't have inclusion

i guess that would make sense

because if x is not in (a, b), and say x >= a, then [x, b_0) would be an open set containing x that does not intersect (a, b) where b_0 is a largest element of X

hmm i guess this only holds if X has a largest element tho

and similarly for smallest element

[tiny obsidian]

no it doesn't imply the closure of (a,b) is (a,b) in all cases

non-example: reals

example where that is the case: integers

[heady skiff]

oh no i meant that it would imply it if X had no largest or smallest element i think

so we would have equality if X didn't have a largest or smallest element?

[tiny obsidian]

ok X = [0,1] is a non-example then

[heady skiff]

wdym

oh

but like, say we're considering (1/4, 1/2) in X = [0, 1]; then (1/4, 1/2) has no limit points not contained in (1/4, 1/2) , because any such x must be greater than or equal to 1/2 or less than or equal to 1/4; say wlog x is greater than or equal to 1/2

wait

[tiny obsidian]

1/4 is a limit point

[heady skiff]

wait nvm

i've confused myself

[merry geode]

Lmao

[heady skiff]

can i get a hint on this please? i'm not really sure how to show that if x is not in (a, b) and is a limit point of (a, b), then x = a or x = b. I don't have access to the standard methods of considering neighborhoods like (x - e, x + e) for example since we're not working in R

[tiny obsidian]

do you have any information about what the closure is without using limit points?

[heady skiff]

well munkres originally defined it as the intersection of all closed sets containing A

he also gave a theorem saying that x is in the closure of A iff every basis element B containing x intersects A

[tiny obsidian]

this is the way

so in particular it's a subset of every closed set containing A

[merry geode]

Why can't you use an analogous argument?

Oh, maybe you are missing characterization of open intervals in topology

[heady skiff]

oh so this is pretty easy then right

[a, b] is a closed set containing (a, b)

because its complement is open

if X has a largest or smallest element it's open

[merry geode]

(Ah welp that is easier)

[heady skiff]

and etc for all the cases

yea damn i'm not used to using that definition lmao

[tiny obsidian]

?

[heady skiff]

sorry, like for example if X has a largest element b_0 but no smallest element, then the complement of [a, b] in X is just (-infty, a) U (b, b_0] which is open by the def of the order topology

right?

and then there are a bunch of cases

[tiny obsidian]

oh

yes

that definition is the ✨ nice one ✨ because it says the closure is the smallest closed set containing A

and limit points are ugly because you need to actually consider points

[heady skiff]

yeah i guess so

[unreal stratus]

Limit points definition seems only to have disadvantages lol

[merry geode]

Wdym, points are easy to analyze

[heady skiff]

but for the converse, would you have to look at the limit points?

[unreal stratus]

I don't think I've really used limit points since real analysis

[merry geode]

But yea would rather use limit of sequence

Converse?

[heady skiff]

like for what conditions are a and b contained in the closure of (a, b), which seems like it boils down to looking at limit points

like sorry

the converse inclusion, under what conditions do we have equality

[merry geode]

Ahh

[tiny obsidian]

yeah here you need to very specifically consider a and b

[heady skiff]

i.e. [a, b] contained in the closure of (a, b)

ye

[merry geode]

Checking closure of (a, b) could be a bit laborous

Like, e.g. need to check if there is a closed set containing (a, b) not including a

Or just check if a is in closure in another way - using the point

[heady skiff]

wait why isn't a automatically contained in the closure of (a, b)? we only need to check if every basis element containing a intersects (a, b); so we have three cases, either the basis element = (x, y) which contains a, or equals [a_0, y) for the smallest element a_0 of X, or equals (x, b_0] for the largest element b_0 of X. but i'm pretty sure in all cases they intersect (a, b)

[tiny obsidian]

again: think about integers for why your 'pretty sure' is wrong

because in integers you in fact have (a,b) closed and hence its own closure

[merry geode]

Hmm it is a bit hard to see at a glance yea

Might be easier to find closed set E where E \cup (-inf, a] \cup [b, inf) is the entire space.

[heady skiff]

sorry i'm confused, why exactly would (0, 2) in the integers be considered closed? because any basis element in the order topology containing 2 for example would have to be of the form (x, y) right, with x < 2 < y

[tiny obsidian]

yes

(0, 2) = {1} is closed as the complement of open (-infty, 1) u (1, infty)

And any basis element is indeed of that form, e.g. (1, 3)

[heady skiff]

ohhh right

okay my brain is not working today

thank you for your help

[opaque cloud]

Lol so the definition of a manifold with boundary includes all manifolds without boundary as well?

[opaque cloud]

ah right okay

[gaunt linden]

Yeah. The boundary might be Ø.

[unreal stratus]

Yes

[cedar pebble]

manifold with corners

[empty grove]

The piss manifold

[opaque cloud]

this is cursed, this is a disgrace to manifolds

[half vortex]

I get that we split it into subsets calculate them then join them again, but stuck at the part where i actually have to decide what the subsets are suposed to be
in turn i struggle with the rest of the proceedings...

[steel glen]

with anything square related it's usually X - {x} and then some open nbhd around x, where x is inside the square

maybe i said that too fast but im sure (hope) it'll work out nicely

[drowsy moth]

hello anyone

i don't known about this theorem may anyone explains

[hidden crag]

?

[red yoke]

Alternatively you could try inner region of square ∪ outer region of square

And the outer region deformation retracts to just the "4" edges

[tribal palm]

how would one go about defining a metric on S1?

i know the subspace metric from R2 would work

but that feels rather unnatural

does arc length work?

[alpine nest]

It does.

Or normalized arclength (divided by 2pi)

Which lets you isometrically identify S1 with [0,1] with ends glued together

[tribal palm]

lovely

[red yoke]

This is equivalent to subspace metric I think

[alpine nest]

Yep

But less unnatural, possibly

[brave geode]

it's also equivalent to the quotient metric on R/Z if you like those

[tribal palm]

when posing this question i mistakenly imagined the behaviour of the arc distance around opposing poles to be necessarily noncontinuous, but yeah that was wrong

[red yoke]

It's still nonsmooth

[opaque cloud]

subspace metric from R^2? Like just using the R^2 metric by considering S^1 embedded in R^2?

[tribal palm]

yes

[red yoke]

Does anyone have a hint for this problem?

Prove that a nontrivial torus knot is prime, by considering how a sphere S intersecting the knot K transversely at 2 points intersects the torus T.
This is what I have so far: For any decomposition of K into a sum of two knots, there is a sphere S intersecting T and K transversely, such that the two knots are the interior and exterior segments of K with arcs on the sphere. S∩T is a disjoint union of circles. Assume all these circles separate T, and therefore bound a disk in T. Suppose K∩T is contained in two different circles, then K crosses the boundaries of the bounded disks once, contradiction. Therefore K∩T is contained in the same circle, and the segment of K in the bounded disk is trivial.

I'm guessing the circles are indeed separating whenever K is nontrivial

[wind bronze]

Hello! I cannot see (intuitively) how the product of topology of (R, e) and (R, e) (where e is the standard euclidean topology) forms the standard topological space of (R^2, e).

Using the definition of a product topology, I obtain the fact that the base of the topology in R^2 isn't made of open balls, but rather "open rectangles" (cartesian products of open intervals in R).

What does the topology induced by the euclidean distance have to do with any of this ?

[tulip bluff]

How can I show that a pair (X, A) (A is closed in X) has the homotopy extension property if and only if X × {0} ∪ A × I is a retract of X × I? I got a hint that for the first implication the identity map X × {0} ∪ A × I -> X × {0} ∪ A × I extends to a map X x I -> X × {0} ∪ A × I, but I don't know how to show this.

[brittle rapids]

a retract r: Y -> B ⊆ Y is a continuous map such that r|_B = id

so retractions can be viewed as extensions of the identity map B -> B

[red yoke]

If B, B' are bases, then the topology of B' is finer than B iff for any p ∈ b ∈ B there exists b' ∈ B' s.t. p ∈ b' ⊂ b

You can show each is finer than the other

[wind bronze]

Hmmm, I will think about this. Thanks!

[umbral panther]

What do you mean by equivalent?
They produce the same topologies. They produce the same geodesics. But one of them is a geodesic metric and one of them isn’t. It has distances that are shorter than paths

[red yoke]

Bilipschitz equivalence

Oop I forgot there were multiple equivalences

[umbral panther]

I hate bilipschitz equivalence because it mixes together local bilipschitz equivalence with coarse equivalence. In particular, almost everyone who uses the term seems to really mean one of them. Of course in the compact setting, it’s just the local property, so this isn’t a problem

[red yoke]

Okie, will take note

[red yoke]

I think this should work: For any nontrivial torus knot (p,q) we have p, q ≥ 2. By Poincaré duality, a circle intersects the knot at most twice only if it is a meridian, longitude, or null-homotopic. For the first two cases, since the sphere divides the torus into interior and exterior regions, there is another circle which by duality considerations have the same homology class, and thus we have at least 4 intersections, contradiction.

Though I wonder if there is a better method (assuming this is even correct) since this is only the first exercise

[heady skiff]

it's true that $\overline{A \cap B} = \overline{A} \cap \overline{B}$ right?

[gentle osprey]

okeyokay

[hexed steppe]

what do you think

[heady skiff]

yes

[hexed steppe]

why

[heady skiff]

well i was just writing up a proof of why but i wanted to get verified first lol

[novel acorn]

no
Take (0,1) and (1,2)

[heady skiff]

ah

[novel acorn]

one of the inclusions always holds tho I believe

LHS to RHS

[hexed steppe]

you might learn more by trying to fully prove it on your own and then carefully whether what you did makes sense, and finally asking if youre still confused

though that takes more effort

[heady skiff]

yeah maybe

i'm a bit confused tho, say x is a limit point of A and is contained in B (i.e. i'm viewing it as a particular element of cl(A) n cl(B)). then every neighborhood U of x intersects A and B, so why shouldn't it be in cl(A n B)?

also, if x is a limit point of A and a limit point of B, then every neighborhood U of x intersects A and B, so x in cl(A n B)

[hexed steppe]

i think this is how you develop intuition

[heady skiff]

if x is in A n B then this is trivial (I'm looking at RHS to LHS) so i'm assuming otherwise

[heady skiff]

because every point in a set is a limit point of that set right?

[hexed steppe]

no

eg there can be isolated points

[heady skiff]

oh wait

yeah

[hexed steppe]

as in the counterexample just given

[heady skiff]

(0, 1) U {2}

[hexed steppe]

sure

another way to phrase it is to say cl(X) := smallest closed set containing X

cl(A) n cl(B) is closed hence contains cl(AnB)

it feels like there should be a simple characterization of when equality holds

https://math.stackexchange.com/questions/356758/when-is-the-closure-of-an-intersection-equal-to-the-intersection-of-closures

see the answer by ni zhiyang

[alpine nest]

Interesting!

[hexed steppe]

i think it suggests that the counterexample (0,1),(1,2) is the only thing that can go wrong

[heady skiff]

is this reasoning accurate to show that $\overline{\cap A_\alpha} \subset \cap \overline{A_\alpha}$? Suppose $x$ is a limit point of $\cap A_\alpha$. Then every neighborhood $U$ of $x$ intersects $\cap A_\alpha$, and hence $A_{\alpha}$ for all $\alpha$. Hence, $x \in \cap \overline{A_\alpha}$.

[gentle osprey]

okeyokay

[rocky kraken]

seems correct

[heady skiff]

also we don't necessarily have $\overline{A} - \overline{B} = \overline{A - B}$ right? because if we take $A = (0, 1)$ and $B = (0, 2)$ then we don't have equality. however, we always have $\overline{A} - \overline{B} \subset \overline{A - B}$ I believe

[gentle osprey]

okeyokay

[white oxide]

i am getting a dejavu

or was that about unions

[heady skiff]

[heady skiff]

wha

[white oxide]

you asked this or something related before, not that deep

[white oxide]

thats right

Maybe as an alternative way. How else do you define the closure of a set?

[heady skiff]

Yeah I think I asked about the closure of union is not equal to the union of the closures or rather if my disproof was correct

[heady skiff]

to show that the order topology is Hausdorff, would this proof work? let any $x_1 < x_2$ in $X$ and suppose that there exists $t \in X$ with $x_1 < t < x_2$. Suppose without loss of generality that $X$ has no smallest or largest element. then $U_{x_1} \coloneqq (-\infty, t)$, $U_{x_2} \coloneqq (t, \infty)$ are open sets containing $x_1$ and $x_2$ respectively, satisfying $U_{x_1} \cap U_{x_2} = \varnothing$. If there exists no $t$ with $x_1 < t < x_2$, set $U_{x_1} \coloneqq (-\infty, x_2)$ and $U_{x_2} \coloneqq (x_1, \infty)$. then these sets satisfy the desired properties.

[gentle osprey]

okeyokay

[white oxide]

You defined rays as open?

Anyways there is a mistake or rather a missing case. There being no element between x1 and x2 (think N or Z)

[heady skiff]

oh that was my second case

that there's no t with x_1 < x_2

[heady skiff]

ya

[white oxide]

oh, yeah

Yeah I'd just say : "take the left ray and right ray of t" or whatever your def. of rays was

[heady skiff]

ye makes sense

[white oxide]

the no smallest or largest element thing makes it more awkward than it needs to. And imagine you'd be doing proofs pages long, a bunch of text with little to nothing happening yknow?

[heady skiff]

yea that's a lot easier

i thought you would have to consider four cases or something

[white oxide]

the definition of the topology doesnt care too much about it

[heady skiff]

wdym

[tiny obsidian]

the rays are open and defined regardless of whether there is a largest/smallest element or not, so there's no need to look at those cases either way

[white oxide]

what edward II just said

i shouldnt mix my casual speaking habits with math talk

The topology doesnt care about min and max at all

[unreal stratus]

Hi joe

[heady skiff]

why is it that every basis element can be expressed as a finite intersection of subbasis elements in a topology generated by a subbasis S? how do we know that it's not a union of finite intersections of elements of S?

[tiny obsidian]

you're going to have to explain which basis you're talking about

because if it's an arbitrary basis for the same topology as that generated by subbasis S, you can't

dumb example: Some X with at least two points, S = { {x} : x in X} a subbasis (and basis) for the discrete topology, and the basis consisting of all open sets is obviously not the collection of finite intersections of singletons

there is an 'obvious' basis for a topology generated by a subbasis, but that's defined as the collection of finite intersections of elements of S and hence what you want is by definition

[heady skiff]

oh i see

[red yoke]

Are these exercises supposed to be doable

[naive trench]

Quick question, every vectorial space V have the second axiom of countability? Assuming that V is a metric space with a distance function defined from a norm etc etc

Finite dimensional ofc

[brittle rapids]

finite dimensional normed spaces?

isn't that R^n or C^n

[alpine nest]

Pretty much, yes

[naive trench]

Ugh well yeah there exist an isomorphism between V and R^n or C^n

I literally forgot about it

[brittle rapids]

more than that

an isometry

[naive trench]

Ideed

I feel dumb

[unreal stratus]

eh i mean

multiple different inner products on a vector space

pls don't feel dumb alex

[brittle rapids]

please don't feel dumb i agree

[brittle rapids]

we care that it's an isometry because it needs to be at least homeomorphic ig

[opaque scroll]

No matter what inner product you put on it, you'll get something isometric. But there are norms that are not given by inner products.

[unreal stratus]

Hm really

I thought you can get different inner products w different norms

[opaque scroll]

Yeah, just pick an orthonormal basis, there's your isometry

[unreal stratus]

Oh lol

Yeah sure lol

Now i feel silly

I guess I was more thinking like sure the identity map needn't be an isometry

But ofc you can just change the map lol

[brittle rapids]

oh i see

[ebon galleon]

Yeah they won't be exactly the same, but there's still an isometry

[opaque scroll]

But for example the Lp norms give spaces that are not isometric for different ps. They will be homeomorphic though

[unreal stratus]

Indeed an (ordered) orthormal basis is just a choice of isometry with R^n (with dot product) lol fair

[brittle rapids]

oh ya you need an ip

now i feel silly

[unreal stratus]

internet protocol

intellectual property

[brittle rapids]

inner product

ya they aren't isometric to R^n or C^n in general

[naive trench]

Is just after thinking a bit makes sense that V is 2AN. I just forgot the fact and didnt try to thinking for my own

@unreal stratus @brittle rapids thank you

[tribal palm]

there are people who study mathematics and don’t feel dumb?

[alpine nest]

Yes, the bad ones

[naive trench]

There are some moments when I think omg Im so dumb then realize omg its trivial and took me like secs or mins to understand what I did not it

Related to basic stuff

Like my question, totally forgot that a finite real or complex vector space is isomorphic to R^n or C^n

[safe torrent]

Average math experience

[naive trench]

Fr fr

[radiant cedar]

I know that we can't classify all the homotopy groups π_k(S^n) of spheres, but if we restrict to n ≥ k ≥ 1 do we know these?

[hidden crag]

Yes they’re trivial for k smaller than n and Z for k =n

Proving this is a good exercise

[fading fern]

hi i gotta go soon but can someone tell me a little hint on how to compute homology groups? specifically H(R,Q)

[unreal stratus]

not me

[unreal stratus]

ye

[unreal stratus]

hm isn't it either easy or super hard depending on which theorems you have

like if you have something like cellular or simplicial approximation (or similarly powerful) it is easy, and otherwise i don't know how to do the k < n case

[hidden crag]

It’s not hard yes

Im assuming they have the standard machinery you have when dealing with higher homotopy groups

[unreal stratus]

Idk I would just add that as a warning like if you don't have those technology tings dw lol

ye

[umbral panther]

First do the smooth case

For continuous maps, this is necessarily harder than proving that R^n is not homeomorphic to R^m

[radiant cedar]

Is there an explicit formula for a retraction from the unit square [0,1]^2 to the downwards shaped T space? That is [0,1]^2 -> [0,1] x {0} ∪ {1/2} x [0,1].

[knotty vine]

Do it first from the unit square to the L shape [0,1] x {0} u {0} x [0,1]. Then flip that along the y-axis and glue the two together (use continuity of a piecewise-defined function) to get a retraction [-1,1]x[0,1] to the upside down T centred at 0. Then translate and scale to size

[unreal stratus]

Lol I was trying to do it without reading yours properly and then wound up with the same thing Semer

[radiant cedar]

Is there an explicit map from the unit square to the L shape? I kinda see that I want to push the top right corner to 0, but I don't know how to define this.

[unreal stratus]

I think you can just join each point by a line to (-1,-1)

and see where it intersects the L

(You can easily write an explicit description, and it's obvious that this is continuous)

[knotty vine]

You could do (x-y,0) in the bottom right triangle and (0,y-x) in the top left triangle

That avoids having to do "raytracing" but requires glueing again

ig thats kinda raytracing too but then orthographic...

[merry geode]

Hmm, Raytracing?

[violet summit]

What goes wrong with defining locally connected to just be that every point has an open neighborhood that is connected?

As opposed to every neighborhood containing an open connected set containing the point

[red yoke]

Then you get every connected set is locally connected

Which is not actually what you want

"Local" usually means "this property holds in any neighbourhood"

In the case the property is existence of some open connected region containing the point

"Local" is useful when you don't get to control how small your neighbourhoods get

[fading fern]

sorry for asking twice but how would you do it, im stuck at this

[lime sable]

what do you know about homology

[fading fern]

most i know is the basic definition, quotient of singular chain

[fathom steeple]

am i correct that in the uniform topology on R^N, given x=(x1,x2,x3.....) then for eps<1, the ball of radius eps centered at x is \prod_i=1^n [xi-eps , xi+eps] ?

[lime sable]

do you have a guess as to what the homology might be

[fading fern]

my best guess is countable many direct sums of Z

[lime sable]

the idea is that H_n counts n-dimensional holes by taking the quotient of cycles by boundaries

the intuition is that if you have an n-cycle that isn't a boundary, then it goes around a hole. if it was a boundary, then there would be an (n+1)-chain which the cycle is the boundary of

so saying that H_n(X,Z) is Z is intuitively saying that there's one n-dimensional hole generated by the homology class of some n-cycle, meaning that every nontrivial n-dimensional hole comes from going around this n-cycle ..., -2, -1, 0, 1, 2, ... times

[lime sable]

with that in mind, do you have another guess

[cedar pebble]

H^k(R,Q) should be trivial unless k=0 where it's just Q

is that what you're asking

[opaque scroll]

Do you know that it's the same for homotopy equivalent spaces?

[fading fern]

wait what would H_n(X,Q) represent in this case?

[fading fern]

yes i think so

[opaque scroll]

Well, then you just have to show that R is homotopy equivalent to a point

[fading fern]

yeah but the extra Q is complicating stuff

[opaque scroll]

Not really, no

Or what do you mean is the complications?

[fading fern]

like

[opaque scroll]

Wait, is this relative homology, not homology with coefficients in Q?

I see, then it's a little more complicated yes

[fading fern]

i mean

[opaque scroll]

You'll have to whip out the long exact sequence in relative homology

[fading fern]

homology as in standard definition with singular simplices group chain

also sorry for not replying earlier i got distracted

the complication for me is that i have to take quotient by the groups of Q

and then compute homology

[opaque scroll]

Yeah, so instead of trying to do all that computation, you should use the long exact sequence in relative homology. If you're familiar with that

[fading fern]

nope, book doesnt mention it
but what is it?

[opaque scroll]

It relates the homology if R, the homology of Q and the relative homology of (R, Q)

[fading fern]

also before you continue

can i know if the homology on pairs is related to the quotient topology?

[opaque scroll]

It is for nice enough pairs, using excision, but (R, Q) is not very nice.

[fading fern]

yeah that makes sense

[opaque scroll]

Anyway I have to go, but you should look up the LES

[fading fern]

i would ask what excision is but i dont wanna branch off

?

whats the LES

long exact sequence?

alright thanks for the help

[lime sable]

oh i thought it was with coefficients in Q lmao

[fading fern]

yeah the LES is what i was talking about sorry i didnt know the term

[tidal cedar]

excision more like

excession

[fading fern]

i just realized i took homology as S_n(X)/im(δ_n+1) while calculating

ill try again now

[tribal palm]

what is the idea behind this notation?

it's used as an example throughout the book (viro's elementary topology problem textbook)

(which is an interesting book leaving all proofs to the reader haha... with plenty of hints thankfully)

[knotty vine]

It is the specialization poset of the topology

[tribal palm]

u rel v if u is contained in the closure of v ?

[knotty vine]

Ye aka all opens that contain u also contains v

Did I get that the right way around?

From bottom to top: d c b a

[tribal palm]

thank you very much

[knotty vine]

I hope that wasnt one of the exercises

[tribal palm]

not at this point

[coarse mist]

I'm self studying munkres topology, realistically how many of the problems in each section should i be able to solve

should clarify the book lol

[hexed steppe]

all of them, on your first try

[coarse mist]

its over for me then

time to be a farmer in minnesota

[white oxide]

no

more

Prove every statement you see

[hexed steppe]

yes

first try as well

[coarse mist]

oh

oh

[hexed steppe]

in fact dont read the statements

[coarse mist]

oh

[hexed steppe]

just discover them on your own

[white oxide]

idk, id do it on first try like 5 times

otherwise just become possesed for a sec and write down the proof

actually thinking about the steps is cheating

ok i think i filled my bullying quota

[coarse mist]

ok whatr about your giving helpful advice quota

[hexed steppe]

what book

[coarse mist]

muynrkes

munkres

[hexed steppe]

idk

there are a lot right

[white oxide]

As many as possible ideally.

The better question is when to move on for any given exercise

[coarse mist]

i dont know how long to bang my head against the wall for for some problems

yup

[coarse mist]

there are around 10-15 per section

[hexed steppe]

you can take breaks from problems and come back to them later

oh i thought like 20+

[white oxide]

Hm I guess munkres fits into a semester?

[hexed steppe]

i mean it depends on you and the timeframe you want

[white oxide]

See if you split it up to weeks and then do like at least 5 harder looking ones that week and give the rest a look

Just waiting after reading an exercise can make a lot of difference

[coarse mist]

Thats fair

[white oxide]

Nut in water analogy type of situation.

Also like, just look some of them up

If you insist on doing everything on your own you never really see how other people approach these problems and what strategies they use

Which is what you have to a degree in an actual course

[coarse mist]

Hm ok

Thats solid

[hexed steppe]

yeah

i think generally you should allow yourself a few hours per problem

if not more

like before googling it

[coarse mist]

Dang

So a section could take me potentially 30 hrs?

[hexed steppe]

sure

the most important part of munkres is like 2 chapters

[coarse mist]

Would that be normal for a course? There are 50 sections

[white oxide]

i kinda suck at self control when self studying, its like an itch to get those list of like 20 problems done in one weekend day

[hexed steppe]

munkres is not supposed to be done in a semester

[white oxide]

like 1-2 problems from each

[coarse mist]

I think the first half is, the second half is algebraic topology but the point set stuff is 50 sections

[coarse mist]

That makes sense

[hexed steppe]

chapter 2/3 are the most important

[white oxide]

if you cut off at the covering space part it fits

but beyond? yeah not really

[hexed steppe]

wym

i think munkres algebraic topology treatment is pointless

[white oxide]

thats what I meant

[hexed steppe]

and should just read hatcher or something

oh ok

[white oxide]

why did he add watered down algebraic topology at the ned

[hexed steppe]

idk

a lot of courses do it

[white oxide]

when stretching munrkes over 2 semesters?

[hexed steppe]

no 1 semester

like chapter 2,3, then the alg top

it is dumb

[coarse mist]

Hm. Maybe that'll be my goal for break then, to cover that + ch4 if possible. I just am about to finish chapter 2 but skipped a number of problems in the product topology and the metric topology section is giving me difficulty

[white oxide]

what happened to 1

[hexed steppe]

1 is just sets

[white oxide]

oh

[hexed steppe]

like de morgan's law or something

[coarse mist]

Or is ch4 even that important? Given I can only do another chapter after chapter 3 what should I do

[white oxide]

lemme check my perfectly legal munkres pdf

[hexed steppe]

you should do the product and metric

chapter 4 is important

i guess

[white oxide]

oh jesus

[hexed steppe]

its important to learn urysohn's lemma

[coarse mist]

But the nagata smirnov metrization theorem in chapter 6 sounds lit

Oh ok

[white oxide]

wait are you for real?

[hexed steppe]

yes what

like 2,3,9 say

[white oxide]

chapters 1 and 4-8 will be missed

and stop at 9??

[hexed steppe]

i mean

are you looking at the table of contents

[white oxide]

yeah

[hexed steppe]

are 4-8 really necessary

[white oxide]

to be called point set topology, yeah

[hexed steppe]

6 is not

[white oxide]

rest goes into alg top

[hexed steppe]

5 is not

i mean i agree it is bad pedagogy