point-set-topology

[gaunt linden]

It's a possible definition of "subbasis for tau", assuming you have already defined "basis for tau".

[fading meteor]

Yes

A set $\mathscr{B}\subseteq \tau$ is a basis if every $U \in \tau$ is of the form $\bigcup_{i \in I} B_i$, for $B_i \in \mathscr{B}$

[gentle osprey]

Nico

[gaunt linden]

I'm afraid I'm still not getting your confusion.

[fading meteor]

The elements in the subbasis $\mathscr{S}$, they aren's in the basis described here

[gentle osprey]

Nico

[fading meteor]

Or are they?

[gaunt linden]

Each of them ends up being in the basis too, by setting n=1.

[fading meteor]

Then isn't this an insanely large basis?

If $|\mathscr{S}|$ is big?

[gentle osprey]

Nico

[gaunt linden]

Might be, but that's not a problem.

The entire topology is a basis (as well as a subbasis) for itself too.

(This is not like in linear algebra, where a "basis" can only have one size).

[fading meteor]

That's probably where my confusion comes from

So a basis for a topology is kinda like a generating set for a vector space?

[gaunt linden]

Yes.

A subbasis is even more like a generating set.

[fading meteor]

So a basis is kinda like a generating set with union, and a subbasis is a generating set with union and finite intersection?

[gaunt linden]

Yes.

[fading meteor]

Right

[gaunt linden]

Another way of looking at the subbasis is that you're saying "I have these sets that I want to be open -- which other open sets do I need before it qualifies as a topology?"

[fading meteor]

Yeah

Right, and then he says that every basis is also a subbasis

How does that make sense?

[quartz horizon]

Conversely, if S is a sub basis, then the collection of finite intersections of elements of S is a basis

[fading meteor]

That's kinda this definition, right?

[gaunt linden]

I'm not immediately seeing why it wouldn't make sense.

[fading meteor]

But the finite intersection isn't a defining trait of the basis, right?

[gaunt linden]

No, but if you have a basis, then every element of the basis is still a finite intersection of basis elements -- namely itself!

"Finite intersection" can also mean "intersection of a single set", which means the set itself.

[fading meteor]

Okay but

If we have $U_1 \neq U_2$ both in our basis

[gentle osprey]

Nico

[fading meteor]

Then by definition of a topology $U_1 \cap U_2$ is in our topology

[gentle osprey]

Nico

[fading meteor]

But why does that necessarily mean that it is also in the basis?

[thorny agate]

What is the definition of a basis of a topology I think start there

[fading meteor]

Also, maybe stupid question , but is the empty set in every basis?

[midnight umbra]

it doesn't need to be

[gaunt linden]

U1 cap U2 is not necessarily itself in the basis, but it must then be the union of other basis sets.

[fading meteor]

For every $x \in X$, there must exist an open set $U$ in our basis $\mathscr{B}$ such that $x \in U$

[gentle osprey]

Nico

[polar storm]

no, not enough.

[midnight umbra]

because to achieve the empty set as a union of basis elements, you can take the empty union

[fading meteor]

No that was local basis for x

Ok nevermind

I'm stupid

[polar storm]

local basis is defined differently

[quartz horizon]

I’ve kind of wondered to what extent you can use this trick for measures

[midnight umbra]

how do you mean?

[quartz horizon]

If an empty union gives an empty sum you get mu(emptyset) = 0

But usually I check mu(emptyset) = 0 explicitly

[fading meteor]

We defined basis as: a set of open sets B is a basis if every open set in X is the union of elements in B

[fading meteor]

Yeah I'm just stupid

Sorry

[polar storm]

no, why r u calling urself stupid

u r not

[quartz horizon]

A nice alternative definition is the following

[fading meteor]

Because I don't understand this subject at all

And I keep coming here with stupid questions

[polar storm]

I mean I found point-set topology really hard, too

[quartz horizon]

B is a basis iff for any open set U and any x in U, there exists a B subset U which contains x

[quartz horizon]

Same

[crisp lintel]

You don't actually need to check this if any set has finite measure at allz

[fading meteor]

Do any of you guys know a lot about the Lebesgue integral? If so, do you need a lot of topology for it? For some reason, the second chapter of this course is the Lebesgue integral and I'm kinda tired of topology for now

[gaunt linden]

One problem with learning point-set is that finite examples of topologies are either trivial or weird, so it demands more of the imagination to think of simple examples of many of the concepts.

[quartz horizon]

I’ve learnt a lot about the 1D lebesgue integral recently!

I think, there is a decent amount of topology that’s helpful for it

[fading meteor]

Damn

[quartz horizon]

Compactness especially is vital

[fading meteor]

So I need to finish topology before looking at the Lebesgue integral?

[quartz horizon]

Well finish is a strong word

[fading meteor]

I only have a decent understanding of the absolute basics

[quartz horizon]

Lemme think about what exact topology was needed

[fading meteor]

Still struggling to remember all the definitions, even

[quartz horizon]

Ok, here’s one result you’ll definitely need

[gaunt linden]

You don't need all of abstract topology, but in particular topological notions in R and R^n will be vital.

[quartz horizon]

Let $(C_n)$ be a sequence of nonempty closed bounded subsets of $\mathbb{R}$ that is decreasing, so $C_{n + 1} \subset C_n$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

Then $\bigcap_{n=1}^\infty C_n \neq \emptyset$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[fading meteor]

Yeah

I saw this as early as calculus last year

[quartz horizon]

Ok nice

[fading meteor]

For R, that is tho

[quartz horizon]

This statement is used in showing that you have a premeasure on the half-open intervals

[fading meteor]

Premeasure?

I'm not at all familiar with measure theory, btw

[quartz horizon]

A thing you use to make a measure

In this case the lebesgue measure

[fading meteor]

You know what, I'll just finish all of topology and then start with the Lebesgue integral

[quartz horizon]

another useful statement is that any open subset of R can be written as a countable disjoint union of open intervals

[fading meteor]

Sounds logical

Matter of fact I saw it

[gaunt linden]

Better change that to "finish this particular topology book". There are arbitrarily deep rabbit holes within "all of topology".

[fading meteor]

I saw this theorem that said "V is an open and connected set in R iff V is an open interval"

[fading meteor]

Yeah, I'm not planning on doing a phd in topology anyways

[quartz horizon]

A nice way of saying this is that “a subset of R is connected if and only if it is an interval”

This is used as part of the topological proof of the intermediate value theorem

[fading meteor]

I'm not seeing the intermediate value theorem in this course, I think

Or at least, there's nothing in my syllabus labelled as the intermediate value theorem

There is stuff about path connectedness tho

[quartz horizon]

Do you prove that “continuous image of a connected set is connected”?

[fading meteor]

Yes

[quartz horizon]

So this gives a proof of the IVT

Let $f : I \to \mathbb{R}$ be continuous, where $I$ is an interval

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

Thus $I$ is connected, meaning $f(I)$ is connected, meaning $f(I)$ is an interval

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

So if we choose $y$ between $f(a)$ and $f(b)$ for some $a, b \in I$, then there must exist a $c \in I$ with $f(c) = y$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

This is (essentially) the IVT

[fading meteor]

Let $\mathscr{S}$ be a subbasis of a topological space $(X, \tau)$. Then $x_\lambda \rightarrow x$ iff for every $U \in \mathscr{S}$ with $x \in U$ there exists a $\lambda_0 \in \Lambda$ for which $x_\lambda \in U$ for every $\lambda \succeq \lambda_0$

[quartz horizon]

This is true

[gentle osprey]

Nico

[fading meteor]

Apparently

It is

[quartz horizon]

This is one of the most useful things about bases and subbases

Here’s another

[fading meteor]

Nono

I mean

Help

[quartz horizon]

Oh

[fading meteor]

The proof is left as an exercise to the reader

[quartz horizon]

Right right

[fading meteor]

Literally half of my syllabus btw

[quartz horizon]

So, have you met the definition of net convergence?

[fading meteor]

Yes

[quartz horizon]

Could you spell it out? There are two versions I can think of

[fading meteor]

$x_\lambda \rightarrow x$ if for every $U$, neighbourhood of $x$, there exists a $\lambda_0 \in \Lambda$ such that $x_\lambda \in U$ for every $\lambda \succeq \lambda_0$

[gentle osprey]

Nico

[fading meteor]

That's what's in my syllabus

[quartz horizon]

Ok nice

[fading meteor]

For the forward implication, I'm guessing to just apply the definition of net convergnce

[quartz horizon]

I think, then, this will be a helpful warmup exercise

Suppose $x_\lambda \to x$. Then for every open set $U$ containing $x$, there exists a $\lambda_0 \in \Lambda$ such that $x_\lambda \in U$ for every $\lambda \geq \lambda_0$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

Can you prove this?

[fading meteor]

Just the definition of net convergence?

[quartz horizon]

More explicitly, please

The definition of net convergence references neighbourhoods, but this references open sets

[fading meteor]

Uhm

Well we can just take a neighbourhood of x

And take the intersection with U and that neighbourhood?

[quartz horizon]

Which one?

[fading meteor]

Any one that fully contains U?

[quartz horizon]

Such as…? Like, how do you know one exists

[fading meteor]

Because we have a subbasis, and therefore a basis

?

[quartz horizon]

What I am asking has nothing to do with subbases

[quartz horizon]

It is this statement I am asking you to prove

[fading meteor]

Ah

Well can't we take B(x,r) for a certain r > 0?

This will surely have a nonempty intersection with U

[quartz horizon]

We’re in a topological space, not necessarily a metric space

[fading meteor]

Ah, right

[quartz horizon]

Though it’s not a bad intuition

[fading meteor]

Do we have a local basis for x?

[quartz horizon]

I think you’re getting a little off-course, so let me help

Suppose $U$ is an open set containing $x$. Then $U$ is a neighbourhood of $x$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

Can you prove this?

[fading meteor]

Either U is a neighbourhood of x, or it contains a neighbourhood of x?

[quartz horizon]

I want you to prove specifically that U is a neighbourhood of x

[fading meteor]

That's our definition of neighbourhood

[quartz horizon]

It… sounds like a circular definition? Are you sure?

[fading meteor]

$V \subseteq X$ is a neighbourhood of $x \in X$ and $x$ is interior of $V$ if there exists an open $U$ with $x \in U \subseteq V$

[gentle osprey]

Nico

[quartz horizon]

Ok, that’s the definition I’m familiar with

Using this definition, I want you to show that if U is an open set containing x, then U is a neighbourhood of x

[fading meteor]

Because $x \in U \subseteq U$?

[gentle osprey]

Nico

[quartz horizon]

Yes, that works

[quartz horizon]

Happy with this result now?

[fading meteor]

So now since we have a neighbourhood, the conditions are met for the net to converge?

[quartz horizon]

Well, we’re assuming the net converges to x already, so that’s not what we’re trying to prove

We’re trying to prove that the net eventually lies in U

[fading meteor]

I don't understand what else we need to prove for this statement to be true

[quartz horizon]

So the way I’d do it is

Let U be an open set containing x

Then U is a neighbourhood of x

Thus, by the definition of net convergence, there exists a lambda_0 where for any lambda >= lambda_0, x_lambda in U

[fading meteor]

But we had that already?

[quartz horizon]

We needed this step

[fading meteor]

Isn't this just enough to conclude that this is true?

[quartz horizon]

No, not quite

[fading meteor]

Because if $U \in \mathscr{S}$ and $x \in U$, we just proved that $U$ is a neighbourhood of $x$ and if $\lambda_0$ exists such that $x_lambda \in U$ for every $\lambda \succeq \lambda_0$, then $x$ converges, because every neighbourhood can be written as the union of intersections of elements of $\mathscr{S}$?

[gentle osprey]

Nico

[fading meteor]

You know what, I'll ask about this later

[quartz horizon]

So this shows one direction

[fading meteor]

The library is closing in 5 minutes

[quartz horizon]

If x_lambda -> x then your subbasis statement is true

Now you need to show the other direction

It’s an “iff”

[fading meteor]

So the forward or backward implication?

Didn't we just prove both?

[quartz horizon]

No, we did not

We showed that if x_lambda -> x, then for every element S of your subbasis containing x, x_lambda is eventually in S

We have not shown the converse

I.e. you have not shown that if, for every element S of your subbasis containing x, x_lambda is eventually in S, then x_lambda -> x

[fading meteor]

Sorry, but I really have to go

They're basically kicking me out

[quartz horizon]

That’s ok

[fading meteor]

I might catch you this weekend, if not maybe monday

[frozen valve]

Is the order topology on R^2 using the dictionary order comparable to the standard topology on R^2? My thought was yes, cause the set of rectangular regions forms a basis for the standard topology, and for any point inside a rectangular region we can embed it inside an interval of the type on the right

[urban zinc]

Yea, so any open set in the standard topology is an open set in the dictionary order topology (why?)

But not the other way around (why?)

[gaunt linden]

||Both of the two above dictionary-order opens are counterexamples.||

[rotund prism]

i know its a standard result that if a topological vector space V is hausdorff, then V is locally compact if and only if V is finite dimensional

Is there a version for if V is not hausdorff? Are there any non-hausdorff, infinite dimensional topological vector spaces that are locally compact?

[crisp lintel]

dumb example: take any infinite dimensional V with the indiscrete topology. The vector space operations are both continuous because the codomain is V in both cases, and the indiscrete topology is even compact

[rotund prism]

...fair enough. Anyone able to say anything about stronger topologies?

[iron bolt]

afaik any TVS where {0} is closed is already Hausdorff, so all counterexamples are of the boring form where the topology is just indiscrete on some nontrivial subspace

the quotient by this subspace cl({0}) is a Hausdorff TVS and hence finite-dimensional, so the only thing that can go wrong is really that cl({0}) is infinite-dimensional

[limber wyvern]

Hi, the other day I asked for help in one of this channels and I've been struggling to understand the structure of the set, I know a base set is a familiy of neighborhoods, and so, I don't understand why shouldn't enclosure all the base inside another brackets

To leave it like this

[quartz horizon]

This is to do with set builder notation

If I write ${2n \mid n \in \mathbb{N}}$, what set do you think I’m describing?

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[limber wyvern]

the even numbers

[quartz horizon]

Mhm

[limber wyvern]

wait

"a even number"

[quartz horizon]

We have a function $\mathbb{N} \to \mathbb{N}$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

Doing $n \mapsto 2n$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

No, it’s all the even numbers

[quartz horizon]

This set is the image of that function

Similarly, we have a function $\mathbb{N} \to P(\mathbb{R})$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

The codomain is the powerset of the reals

This sends $n$ to the interval $(x - \frac 1n, x + \frac 1n) \subset \mathbb{R}$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

So, what set is ${(x - \frac 1n, x + \frac 1n) \mid n \in \mathbb{N}}$?

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[limber wyvern]

that should be the powerset right?

[quartz horizon]

Of what? This set isn’t the powerset of anything

[limber wyvern]

This set should be (x-1,x+1)

[quartz horizon]

No, that’s not correct

Maybe this alternative way of writing the set will help

${S \subset \mathbb{R} \mid \exists n \in \mathbb{N}, S = (x - \frac 1n, x + \frac 1n)}$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[limber wyvern]

mmm

I would probably say the same

[quartz horizon]

Can you explain your logic?

[limber wyvern]

If I know that S = (x - 1/n, x + 1/n) and start going through the naturals I start with the set: (x-1, x + 1) and ending with (x,x), because x is inside the set (x-1,x+1) I would say that the set is (x-1, x+1)

[quartz horizon]

Ok but what if S = (x - 0.5, x + 0.5)

Then clearly S doesn’t equal (x - 1, x + 1), right?

[limber wyvern]

well, but (x-0.5,x+0.5) is inside (x-1, x+1)

isn't it?

[quartz horizon]

This is true, but I don’t see how that’s relevant

Yes, $(x - 0.5, x + 0.5) \subset (x - 1, x + 1)$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

However it is still true that $(x - 0.5, x + 0.5) \neq (x - 1, x + 1)$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[limber wyvern]

mmm, I think of this as being an open set

[quartz horizon]

It is an open set, yes

Again, I don’t see how that fact is relevant

[quartz horizon]

This is still true, after all

[limber wyvern]

You asked me about what set was that

[quartz horizon]

I am asking you what the set ${S \subset \mathbb{R} \mid \exists n \in \mathbb{N}, S = (x - \frac 1n, x + \frac 1n)}$ is

[limber wyvern]

and so I think it's the open set {(x-1,x+1)} even though now I know is bad😅

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

Ok here’s a concrete question

[quartz horizon]

Is $(x - 0.5, x + 0.5)$ an element of this set?

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[limber wyvern]

That is a subset as you said

[quartz horizon]

Incorrect

I think you really need to review set-builder notation

[limber wyvern]

I always confuse in what is an element and a subset when I start going with families and so on

[quartz horizon]

If I let $A = {S \subset \mathbb{R} \mid \exists n \in \mathbb{N}, S = (x - \frac 1n, x + \frac 1n)}$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

It is completely false that $(x - 0.5, x + 0.5) \subset A$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

So I don’t know what you mean here

[limber wyvern]

i was refering to this

[quartz horizon]

This is true, but $A \neq (x - 1, x + 1)$

[limber wyvern]

Could you elaborate why it is an element?

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

It follows from the definition of set-builder notation

$S \in A$ if and only if $S \subset \mathbb{R}$ and $\exists n \in \mathbb{N}, S = (x - \frac 1n, x + \frac 1n)$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

Now, do you agree that $(x - 0.5, x + 0.5) \subset \mathbb{R}$?

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[limber wyvern]

is S a set?, because of it being a capital letter

[quartz horizon]

Please answer my Q first

[quartz horizon]

It being capitalised has nothing to do with it

[limber wyvern]

yes, for n = 2

[quartz horizon]

???

the statement i put does not reference $n$ at all, so i don't know what you mean

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

i am just asking whether $(x - 0.5, x + 0.5)$ is a subset of $\mathbb{R}$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[limber wyvern]

yes it is

[quartz horizon]

ok, good

now, do you agree that $\exists n \in \mathbb{N}, (x - 0.5, x + 0.5) = (x - \frac 1n, x + \frac 1n)$?

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[limber wyvern]

yes

n = 2

[quartz horizon]

good

so then, you are forced to conclude that $(x - 0.5, x + 0.5) \in A$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

where i have defined A here

right?

[limber wyvern]

yes

[quartz horizon]

you would also be forced to conclude that $(x - 1, x + 1) \in A$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

in particular, $A$ has at least two elements

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

because $(x - 0.5, x + 0.5) \neq (x - 1, x + 1)$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[limber wyvern]

yeah

[quartz horizon]

do you see why this is false, now?

[limber wyvern]

that set would look like: {(x-1,x+1),(x-0.5,x+0.5),(x-0.25,x+0.25), ... , (x,x)} right?

[limber wyvern]

yes

[quartz horizon]

there's no (x, x), but otherwise correct

i think what you were doing is the following

A is a collection of subsets of R

so in particular, A itself is not a subset of R

but i think, for some reason, you'd convinced yourself that $A \subset \mathbb{R}$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

and so implicitly what you were doing is $\bigcup_{a \in A} a$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

because it is true that $\bigcup_{n=1}^\infty (x - \frac 1n, x + \frac 1n) = (x - 1, x + 1)$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

but this is not what A is

[limber wyvern]

well, I was thinking the set aboug being just the first case: (x-1,x+1) and just considering A like (x-1,x+1)

[quartz horizon]

that's a type error

A is not a subset of R

A is a collection of subsets of R

[limber wyvern]

yeah

I always confuse that

[quartz horizon]

said differently, $A \not \in P(\mathbb{R})$, but $A \subset P(\mathbb{R})$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

in fact $A \in P(P(\mathbb{R}))$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[limber wyvern]

when I start seeing collections, families and so on I start getting a mess in my head

[quartz horizon]

it's helpful to remember that in set theory, everything is a set

it's perfectly reasonable for sets to contain other sets

so long as you're careful with set-builder notation, you'll be fine

[quartz horizon]

the main thing you were doing is, if $\mathcal{F}$ is a family of sets, for some reason you were replacing this with $\bigcup_{a \in \mathcal{F}} a$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

but $\bigcup_{a \in \mathcal{F}} a \neq \mathcal{F}$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

what is true is a slightly different statement

$\bigcup_{a \in \mathcal{F} } {a} = \mathcal{F}$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

this works for any set, and just says that "any set is a union of singletons"

but $\bigcup_{a \in \mathcal{F} } {a} \neq \bigcup_{a \in \mathcal{F} } a$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[limber wyvern]

okay

so

Now I understand that this represents set of neighborhoods and so it doesn't require another {}

[quartz horizon]

yes, so this would be ${(x - 1, x + 1), (x - \frac 12, x + \frac 12), \dots, {x} }$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[limber wyvern]

because {(x-1/n,x+1/n)} end up being a collection of sets that has an union with a set {x}

[quartz horizon]

in this case, $\bigcup_{a \in B_x} a = (x - 1, x + 1)$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

but of course, this is not what $B_x$ itself is

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

B_x is this set, which in particular has countably infinitely many elements

[quartz horizon]

whereas this set has an uncountable number of elements

it's helpful to remember the identities $x \in P(a) \iff x \subset a$, and $x \in {a} \iff x = a$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[limber wyvern]

I have to fight with the confussion between elements and subsets when talking about collections or families😅

[quartz horizon]

yeah you very much just need to trust what the set builder notation tells you

[limber wyvern]

Ill work on that, anything you recommend?

[quartz horizon]

Maybe ask in #proofs-and-logic ?

[limber wyvern]

thanks for the help and time 🙂

[heady bloom]

Hey there I was wondering if any1 could help me prove this result:

Let $Q \coloneq \bigcap_{k\ge 1}k^{\mathbb{N}}$. Show that $Q$ is not a $G_{\delta}$ subset of $\mathbb{N}^{\mathbb{N}}$.

[gentle osprey]

dan

[heady bloom]

The question right before it was this, which I think might be helpful to this problem:

Let $X$ be a complete metric space. Prove that a subset $Y\subseteq X$ is comeagre if and only if $Y$ contains a dense $G_{\delta}$ set.

[gentle osprey]

dan

[quaint spruce]

to me Q just seems like its equal to {(1,1,1,...)} and so is the intersection of the open sets {1}xNxN..., {1}x{1}xNxNx..., ...

did you mean something like union instead of intersection?

[heady bloom]

Oh god yeah my bad

Lmao

Union

I think it can be solved by remarking that Q is dense in the Baire space and thus if it’s a G_delta set it would be both meagre and comeagre since each k^{\mathbb{N}} is nd but the Baire space is Baire so it’s necessarily nonmeagre which implies that it has no meagre and comeagre sets

But we didn’t see that nonmeagre => no meagre and comeagre sets in class, i just got that from wikipedia

[quaint spruce]

You could probaby use just the baire category theorem. Q is meagre and dense as you noted. If it was G_delta set, then by baire category theorem, Q is comeagre. But this means that N^N = Q U (N^N\ Q), both of which are meager which is not possible

[heady bloom]

Ohh right I kinda forgot that meagre sets form a sigma ideal haha

Thanks for your help

[rancid umbra]

is there an example of a non-hausdorff space where the intersection of any two compact subspaces is compact?

oh

i guess there should be a small example with a weird topology on the integers

i guess the integers with two origins works

[gaunt linden]

Some set with the trivial topology?

[rancid umbra]

ah yea. i guess you just need any set with more than one element and that works

this was inspired by the question of whether or not, in a lattice, the meet of two compact objects is compact

my first instinct was yes because in a hausdorff space this is true

the intersection of compacts is compact

[polar storm]

try cocountable topology on R.

any compact subspace must be finite. So their intersection is finite, and thus compact.

[rancid umbra]

would be interesting if there is some sort of "hausdorff" condition for a lattice, or some characterization of those lattices where the intersection of any two compact objects is compact.

just for completeness, here is an example where the intersection of compact opens is not compact https://math.stackexchange.com/a/229821/803927

[unreal stratus]

Btw for schemes this is called quasiseparatedness and this is a common condition to care about

(Because this condition is equivalently saying that you can write any intersection of open affines as a finite union of open affines)

[rancid umbra]

which condition? that the intersection of compact opens is compact?

[unreal stratus]

Yes (given that you have a scheme)

[rancid umbra]

too bad i don't know what a scheme is

[cosmic mirage]

its an object in the category of schemes

uh morally schemes are to Spec(some commutative ring) as manifolds are to R^n

except that your ring can vary over the cover

[rancid umbra]

im going to take that to mean that it locally looks like a ring

don't know what Spec is either

[cosmic mirage]

fair

[rancid umbra]

alg geo scares me

[maiden junco]

c scared

[cosmic mirage]

me too, I'm currently learning this stuff

[rancid umbra]

my condolences

[cosmic mirage]

Spec of a ring, as a set, is the set of prime ideals

and there's some canonical topology

and a sheaf you can put on top of that which remembers all the data of the commutative ring

[unreal stratus]

Spec R is the functor Hom(R,-)

[unreal stratus]

Sorry lol

[rancid umbra]

woah what? ring homomorphisms out of R are the same as prime ideals of R?

[covert nexus]

a scheme is a space that locally looks like the set of prime ideals of a commutative ring endowed with the zariski topology

[cloud kindle]

Like if a manifold was terrible

[cosmic mirage]

wdym "if"

[rancid umbra]

i think they mean life if a manifold was terrible

[cosmic mirage]

a manifold is already terrible

[cloud kindle]

All topological spaces are terrible

Should never have been invented

[rancid umbra]

oh i misread it whoops

[unreal stratus]

Tat was a bit of a joke message but the point is like there is a "functorial" perspective on schemes which is that you can study functors {commutative rings} -> {sets} and some of these can be thought of as representable by geometric objects. One pleasant thing about this is that it is easy to come up with examples. Like take the equation { x^3 = 5 } and taking solutions to this equation give you a functor as above. It turns out that this functor is actually corepresentable, in this case by the ring Z[x]/(x^3 - 5).

[unreal stratus]

Other way round

[cosmic mirage]

If you hand me a random topological space I will generally be happy, but if you tell me it can be given a manifold structure then I will think lesser of the space

[unreal stratus]

Fibre product of smooth manifolds hmmm

[cloud kindle]

Like if terrible was a manifold?

[unreal stratus]

But also like we often needn't rly think of schemes as topological spaces

[cosmic mirage]

is there some reason to say fiber product instead of pullback lol

[covert nexus]

colife if coterrible was a comanifold

[unreal stratus]

Habit lol

[cloud kindle]

To flex their ability to spell correctly

[covert nexus]

beacuse that's what AG people say and we're cool and everyone should listen to us

[rancid umbra]

oh that is kind of cool actually

[unreal stratus]

Yeah I didn't even say fiber [sic.] product

[cosmic mirage]

Im too woke for this. I will call these pullbacks and also directed limits are just colimits to me

[cosmic mirage]

my question was deliberately worded

[cloud kindle]

Directed and inverse limits are a mistake and we should get rid of the terminology

(Unironically)

[unreal stratus]

What stuff do u do Dirichlet

I feel silly asking Dirichlet

[covert nexus]

WOKE GEN Z GENUS STUDIES PHD IS DESTROYING AMERICA \silly

[unreal stratus]

Now I feel even sillier

[cosmic mirage]

there is a reason to call it a fibre product iff a) there is a reason to call it a fiber product and b) you hate democracy and freedom and oil

[rancid umbra]

ohhhh so if you are american

[unreal stratus]

Saw smth like this recently where someone was talking about not being able to get a job despite a degree from Cambridge and someone commented about how when they hear cambridge they think top engineer or medic, not humanities

Ragebaited me lol

[covert nexus]

AG mostly with a sideline in thinking tropical geometry and matroids cool. As for what in AG, well i've spent the last year thinking very hard about some moduli spaces of ideals in matrix orders that show up in deformation theory stuff -- more generally i don't know exactly what i wanna do for phd

and thankfully i don't have to decide yet because lsgnt

[unreal stratus]

Ah very cool

Ah yes you are who I thought you were from another server lmao

[cloud kindle]

I could’ve joined you but I forgot what a vector space is in my interview 👍

[covert nexus]

hehe

[unreal stratus]

This is v cool tho!

I also do deformation theory lol

[covert nexus]

honestly no clue how i got in lmao

[cloud kindle]

I got waitlisted which I honestly take as a win after how badly I bombed my interview

[covert nexus]

tbf the thing i flubbed was constructing a genus 2 curve which i had seen before and should have been able to do but couldn't recall lol

[cloud kindle]

Nah you deserve it

[covert nexus]

but that isn't something that's entirely trivial so i guess they didn't mind too much lmao

[novel acorn]

Me when the uh
Me when uh something something hyper elliptic

[covert nexus]

also it helps being already based in london because everyone at lsgnt kinda already knows me

or well has at least seen me

[unreal stratus]

Do you know who ou're gonna work with? Without doxing lol

[covert nexus]

no

[novel acorn]

Doesn't y^2=f(x) where f(x) is a degree 4 polynomial work hm

[unreal stratus]

Oh or are you just about to start at lsgnt

[cloud kindle]

Yeah I forgot that like normal subgroups exist in my Manchester interview and they let me in lol, I forgot groups are shit

[covert nexus]

ye currently in masters

[cloud kindle]

Yes

[unreal stratus]

Ah nice

[unreal stratus]

Draw two donuts

[covert nexus]

i drew so many pictures

[unreal stratus]

Ngl I am lucky I managed to do AG stuff without my AG being assessed

[covert nexus]

that's the secret to geometry interviews lol

[cloud kindle]

I spoke about semi simple algebras in mine lol

[unreal stratus]

Cause I am homotopy theory

[covert nexus]

tbf i got into 2/4 places i got interviews and the latter two i didn't get were both for "no funding" reasons

[novel acorn]

I feel like age wise I'm the oldest but career wise the youngest lmfao

[covert nexus]

i am apparently decent at interviewing

[unreal stratus]

Lol realising I did some geometric stuff and never drew a picture I think rip

[covert nexus]

you did not give the people what they wanted

[unreal stratus]

Lol

[cloud kindle]

Who did you interview with at LSGNT out of curiosity?

[unreal stratus]

67

[cloud kindle]

(If you feel comfortable sharing)

[covert nexus]

shepherd-baron and sivek

[unreal stratus]

Ngl I forgot S-B was there lol

Very cool

[covert nexus]

he does very neat stuff

he also told me to read sga 4.5 in that interview tho so idk how to feel

[cloud kindle]

Ah nice

[novel acorn]

AHAHAHAAHAH

what a legend holy shit

[unreal stratus]

Why is that bad

Based

Irony you know what I find funny

Hilbert thm 90 seemed a bit pointless when I first saw it cause it is often stated in a slightly random way

[covert nexus]

fr*nch and also it's the one where deligne etale cohomologies all over everything

[unreal stratus]

But it is a goat

[covert nexus]

thm 90 my beloved

[unreal stratus]

Yes those are both good

[covert nexus]

it came up in the interview actually lol

[novel acorn]

I like it but also hate it just because of how sneaky it can be

[unreal stratus]

nice lol

[novel acorn]

like

[covert nexus]

he asked me brauer group stuff

[novel acorn]

I was reading Milne's algebraic groups book

[unreal stratus]

Cause of like.applying to function fields of curves ?

Yeah sure

[unreal stratus]

Pain.

[novel acorn]

and he used some very nontrivial corollary of hilbert 90

He now has an errata on his website from me and a friend because of that LMFAO

[unreal stratus]

Yeah it feels like a thing where people just say by hilbert 90 to mean a family of theorems lol

[covert nexus]

bro 'd themself

[novel acorn]

I said I wanna read SGA3 but my friend didn't wanna

tho what I don't like is how strew all about the place the theory of algebraic groups is

there truly is no good source

especially for finite flat stuff

[unborn hawk]

read it bro

[unreal stratus]

Yeah Idk lol

[novel acorn]

ur a bad influence but I love you so this'll pass

[unreal stratus]

Idk if it is about target audience idk

[unborn hawk]

luv u too buddy

[unreal stratus]

I have friends doing alg groups who don't rly know schemes / don't like them but then as an AG-y person it feels sad for there not to be a good accessible book in modern language (imo)

[covert nexus]

something something be the change \silly

[unreal stratus]

Ig you can just take Milne's notes and do the somewhat easy adaptations and it's decent

[covert nexus]

tho tbf my main opinion about algebraic groups is that "algebraic groups" is actually the worst named thing in math

[unreal stratus]

I'm just glad Milne doesn't just work over C

[novel acorn]

finite commutative stuff also seems so incredibly rich but nobody really talks about it

[unborn hawk]

Milne is a shit writer lol

[novel acorn]

like Deligne has so many interesting results

[unreal stratus]

Lol yeah and p-divisible group

[novel acorn]

god yeah

[novel acorn]

somebody told me to go read like Bhatt's work to understand p-divisible groups LMFAO

[unreal stratus]

This is similar but not even as clear it is AG

[unreal stratus]

What lmao

[covert nexus]

i liked his fields and galois thery book but yea a lot of his stuff is painful

[unreal stratus]

Learn p-divisible groups by viewing them as a special case of prismatic F-gauges

[novel acorn]

yeah I asked about Dieudonne theory and p-divisible groups and someone told me like "do u know prismatic stuff" lmfao

[unreal stratus]

fppf.

[unreal stratus]

Lol

[novel acorn]

cuz I was reading Moonen's book on abelian varieties

and they don't talk about Dieudonne theory since that's incomplete

so I asked where can I learn about this

[unreal stratus]

Yeah you should start with basic stuff like Scholze–Weinstein first

And then move on to prismatic dieudonné theory

[novel acorn]

I'm actually reading about rigid analytic stuff rn for my thesis :)
Because p-adic heights bring me pain 😔

[unreal stratus]

Nice

I like p-adic geometry

Wait maybe we should be in alg chill atp

[covert nexus]

i keep alternating rapidly between feeling like all that stuff is cursed as fuck and is proof that math is haunted, and equally feeling that its really cool and neat and interesting

[unreal stratus]

I do wonder what p-adic geo will look like in 20 years time lol

[novel acorn]

I feel like the stuff I'm doing rn is pretty cursed because genuinely it seems like magic

but Mihyong Kim somehow came up with it

non abelian chabauty scares me

even like classical chabauty is magic

[unreal stratus]

Nice

[novel acorn]

"this method doesn't work for R, but we made it work for Q_p and it's the most efficient way we have today of finding rational points on curves"

[unreal stratus]

Minhyong left my uni just before I swapped to maths

[covert nexus]

undergrads will be thinking they need to look like this in order to be good enough to phd \silly

[novel acorn]

he's at edinburgh now right?

[novel acorn]

lol what did u do before

[unreal stratus]

Ye I think though he may have moved again

[unreal stratus]

Did physics for a year

[novel acorn]

LMFAO same

[covert nexus]

my condolences

[unreal stratus]

Yeah I vaguely thought about swapping before I started

[quick delta]

I did some p adic dynamics stuff at one point and err

[novel acorn]

I really wanted to do like

mathematical physics

but I kinda started hating physics lol

[unreal stratus]

Yeah lol

[quick delta]

It’s a very funny thought that probably the most applied/physics person in my college at the start of our degree is now the most abstract nonsense person lol

[covert nexus]

just AG so hard that you can start talking about applications in physics and mean string theory \silly

[unreal stratus]

I was thinking like everything I was doing was so far from math that it was a bit weird to do that before math physics lol

[covert nexus]

many such cases

though personally i have been consistent throughout

im cool like that

[quick delta]

I’ve become slightly less pure maths inclined, but not by much

[covert nexus]

i nearly took statistical mechanics because it seemed neat

went to two lectures

bailed super hard lol

was taught by a physicist tho idk what else i was expecting

[quick delta]

I took QM2 for supos for this reason
I did not take it to exams

[unreal stratus]

Statistical mech ah lol

[covert nexus]

here people take qm for free points because it is aggressively Just Linear Algebra™

well qm1 is

[quick delta]

I considered taking it, but it was taught by someone well known to be transphobic so errr

[covert nexus]

qm2 is a bit more interesting

[unreal stratus]

One day I need to go back to thermo cause I could never understand it fully and then realised it's cause nothing was defined properly lol

[quick delta]

Yeah same here
That’s why I was suckered into QM2

[unreal stratus]

UK moment

[novel acorn]

I've gotten dragged into some pretty interesting AI work cuz it pays well so now I'm learning a shitton of like dynamical systems stuff on the fly LMFAO

[unreal stratus]

Irony AI arc...

https://tenor.com/view/it's-over-over-it's-gif-3551756699416095564

[novel acorn]

and it's not even like LLM crap

[covert nexus]

blocked banned filed for restraining order \silly

[novel acorn]

I liked this gif and my computer started lagging for like 30s straight

[covert nexus]

my computer is lagging super hard but this is because i have had sagemath enumerating invariant lattices in Q_(19)^4 for two days now and it still hasn't finsihed

and at this point it's too late to stop

[novel acorn]

lmfao

that's such insane slop haha

[unreal stratus]

Lol

[novel acorn]

I had to do shit like that for my topic before this but my advisor saw me come to the meeting and said "you hate this topic don't you" and I said yeah and he said "well then let's find you something more interesting"

[quick delta]

I wonder what it would’ve been like if Cambridge had a more typical master’s thesis

[covert nexus]

i mean this is not a calculation i have to run, it's a calculation im running because i have nothing better to do with the cpu time and my room is kind of cold \silly

[quick delta]

Can you calculate how much sanity I’ll have come October?

[covert nexus]

-3

[quick delta]

Generous :3

[covert nexus]

actual reason is because im letting the computer run while i try to solve the problem with my brain and whoever finishes first gets a cookie

[quick delta]

Hmmmmm
Do computers like cookies?

[covert nexus]

yes

or else why would my browser have so many of them

[quick delta]

I solved it can I have the cookie?

[covert nexus]

only if u can gib proof that wahl orders are not closed

[quick delta]

The solution is ||3||

[zealous berry]

<---- topologist

[urban zinc]

Nooo stat mech is awesome

David Tong has great notes on it online

[grave solstice]

am I the only one who doesn't like Tong's notes

[urban zinc]

😔

Do you like Pathria's textbook? It's what we used for my stat mech class

[grave solstice]

I haven't used it

So far I have mostly read from Bowley-Sánchez (very physics oriented) and for lattice models I think Friedli-Velenik is pretty neat. But I'm not the right person to ask about this, and I don't like Tong's notes generally, not just the statistical physics ones

[quartz horizon]

tong's notes are absolutely goated

(mostly)

[cloud kindle]

This was like the one physics course I did well in and i think it’s largely because I was taking stats at the same time lol

[unreal stratus]

Gg lol

[urban zinc]

One day I will understand cumulants better

[copper quiver]

Bro, Tong is the literal goat.

[limber wyvern]

Hi, would this be an appropiate drawing for the following problem:

Let A be a open set in a metric space (X,d). Suppose that we eliminate a numerable amount of points from it. Would A still be open?

Red x are the points that are eliminated

[polar storm]

yeah

[limber wyvern]

My principal idea is that in this case it wouldn't be open because you could eliminate all the points from the bigger open ball of x, therefore there will be an isolated point with no open ball

am I right?

[polar storm]

wait numerable means finite? or just means countable?

[limber wyvern]

numerable means infinite but countable

[polar storm]

oh icic

in that case A needs not be open

[urban zinc]

Do you get to choose which points you remove?

Can you post the original wording of the question

[polar storm]

mmm, so u mean something like the open interval (-1,1), and u exclude the points 1/n for every integer n?

so that u cannot form an open ball around zero?

(cuz u excluded arbitrarily close points near 0, at least the picture u drew looks like this situation for 2D case, instead of 1D)

[midnight umbra]

surely it depends on the metric? for example in the discrete metric A \ {your points} would still be open since literally every subset is open

[limber wyvern]

mmm

[limber wyvern]

thats the original wording

The problem is in spanish

but well, Ill post it anyways

I already proved the first part

[limber wyvern]

yes

well

more likely

[polar storm]

yeah, ur intuition works for standard euclidean metric.

[limber wyvern]

suppose you have an open ball, remove all the points of the open ball except the center point

This exercise is for generic metrics

there is no metric given

[quasi forum]

you can likely just pick your favourite metric space

[limber wyvern]

wdym

[polar storm]

oh sorry

[quasi forum]

the question just presents some metric space

so if the claim fails for one of them i think youre fine

[polar storm]

^^^ but this can change if u use different metrics

[polar storm]

I meant ^^^. Replied to wrong message

[limber wyvern]

but

in the discrete metric there are only points at 0 and at 1

right?

[quasi forum]

no? you can give any set you want the discrete metric

[midnight umbra]

the discrete metric takes only values 0 and 1

you can put the discrete metric on any set you like

[limber wyvern]

but, for the discrete metric, it only gives values 0 and 1, therefore if I remove all the values <= 1, except for the point in the center, that wouldn't be considered an open right?

[midnight umbra]

so i can understand correctly, you're saying that for example the set {5} isn't open in the discrete metric?

[limber wyvern]

for example, yes

[midnight umbra]

it is open

for example, the ball of radius 1/2 around 5 in the discrete metric is {5}, because every other point is distance 1 away

[limber wyvern]

I see

my professor just told me that I have to search for one metric that doesn't satisfy this, and therefore I would have proven

[quasi forum]

yeah thats what i wouldve expected

[midnight umbra]

cool

[midnight umbra]

in fact, you can say more, in the discrete metric every subset is open and closed

[limber wyvern]

yeah, he told me that proving that for every topology it's not impossible but almost impossible xD

[midnight umbra]

so it's perhaps the most useless metric lol

[quasi forum]

an explicit counterexample can be written using say R with euclidean metric or something

[limber wyvern]

im now in class but ill try to do it afterwards

[quasi forum]

yeah id suggest doing it, its not too hard

[limber wyvern]

not too hard😔

[urban zinc]

Ah okay I guess the wording is just kinda ambiguous then. Try finding a case where it remains open and a case where it doesn't.

[limber wyvern]

something like that yeah

Could you give a hint?

[urban zinc]

For which one, it remaining open or it not being open anymore?

[cosmic zodiac]

I'm confused about the last sentence. I understand how X would be T1 (as a subset of a Hausdorff space) but what allows for it to be completely regular?

[tender halo]

compact hausdorff spaces are normal

and therefore their subspaces are tychonoff

[fading meteor]

I dont' really understand the topological definition of a quotient space

[ruby delta]

what dont you get?

[gaunt linden]

(And what is the wording of the definition you don't understand?)

[fading meteor]

ust idk how to interpret it

[tender halo]

we want to define the operation of gluing points together and what the resulting space looks like

[limber wyvern]

Hi, I have the following example for proving that function is continous between metric spaces, such spaces are: (X,du) and (X,d*) where d* = min{|x-y|, 1-|x-y|}
It says that X = [0,1) and the function is f(x) = x, and that is not considered continous for f: (X,d*) -> (R, du). It chooses epsilon = 0.1 and wants to prove that there is no delta such that: f(b(0,delta)) is a subset of b(f(0),0.1).

My question is the following, why does the proof take this interval of x: (1-delta, 1) intersection [0.1,1)?

this is the example I'm talking about

[polar storm]

cuz we want to make the error bigger than or equal to the epsilon, which is 0.1 (so contradiction) .

[limber wyvern]

and so, why intersection and not just say, I dont know, (epsilon, 2epsilon)?

[polar storm]

Can you compute the metrics d*(x,0) and d(x,0) for every x in (1-delta, 1) \cap [1/10, 1)?

To be more specific, can u compute the upper bound of d*(x,0) and the lower bound of d(x,0) for such x?

[polar storm]

The example in the image u uploaded is just saying that every x in (1-delta, 1) \cap [1/10, 1) gives the contradiction for continuity. Can u see that?

[limber wyvern]

well, the upper bound nope, but the lower yes

why not pick, lets say (0.1,1)

[polar storm]

yes u can, as long as u take the intersection with (1-delta, 1)

[limber wyvern]

why the intersection?

[polar storm]

okay, so we first want d*(x,0) < delta. Do u agree?

[limber wyvern]

yes

[polar storm]

now delta is arbitrary, so suppose delta = 0.1. Now what happens if u pick x = 0.5? It is still in the interval (0.1,1).

u don't get d*(x,0) < delta

[limber wyvern]

I see

[polar storm]

that is why we r taking the intersection with (1-delta, 1) to ensure the upper bound delta. Can you see why?

[limber wyvern]

Yeah, but I don't get why it has to be the intersection xD

[polar storm]

now X = [0,1) consists of nonnegative numbers. Using this fact, can u simplify d*(x,0)?

[limber wyvern]

(1-delta, 1) \cap [1/10, 1)

I mean, for delta = 0.1:

(0.9, 1) \cap [0.1, 1) -> This is equal to (0.1,1)

Why the intersection?

why not just say that set

[polar storm]

cuz the intersection forces the upper bound of d*(x,0) to be delta.

Can u derive that?

[limber wyvern]

I'm sorry im not seeing it

I even have a picture on my book

But i don't get it

[polar storm]

d*(x,0) = min(|x|, 1 - |x|) = this equals to ?

use the fact that x is nonnegative and x lies in (1 - delta, 1) to derive the upper bound.

[limber wyvern]

|x|

well

it will depend on x

xD

[polar storm]

yes, u can remove | |

now use the inequality 1 - delta < x < 1

[limber wyvern]

for x > 0.5, 1-|x| would be picked

mmm

0.9 < x < 1

[limber wyvern]

right?

[polar storm]

actually, I don't need to do that.

Since delta is arbitrary, do u agree that delta can be made arbitrary small?

[limber wyvern]

yes sir

[polar storm]

in that case, we get x \geq 1 -x. Do u agree?

cuz x lies between 1 - delta and 1 (since delta is very small, u will get 1 - delta to be very close to 1)

[limber wyvern]

mmm

I kind of see it but it's being hard😅

[polar storm]

now use this fact to derive that d*(x,0) = 1 - x

[limber wyvern]

So you are saying that, because 1 - delta < x < 1, then x >= 1-x

[polar storm]

yeah

[limber wyvern]

mmmm

every time I look at it I get more confused

but, translating it to stupid facts:

1 - delta < x < 1, Means that x is arbitrarily small
How then if x is arbitrarily small it can be that x >= 1-x, as smaller the x, the bigger 1-x is

[polar storm]

Since we can make $\delta > 0$ arbitrary small, suppose $\delta \leq 1/2$. Now we are given that
$$
x \in (1- \delta, 1),
$$
so $1 - \delta <x <1$. This gives $0 <1 - x < \delta$ and $1 - x < 1/2$. Moreover, $1 - x < 1/2$ implies $x > 1/2 > 1- x$. This means
$$
d_*(x,0) = \min{ |x|, 1- |x| } = \min{ x, 1 - x } = 1 - x.
$$

[gentle osprey]

Euiseok (Class of 2100)

[polar storm]

from this, can u conclude that d*(x,0) < \delta?

[limber wyvern]

let me think about this😅

mmm

but that is not true?

I mean

1 - 1/2 < 1/2 < 1 -> 1/2 < 1/2 < 1?

but well, i get the point of that part

okay

I got it

yeah

[polar storm]

so this is why we r taking the intersection with (1 - \delta, 1)

[polar storm]

Btw, the same thing happens even if \delta \geq 1/2, but we don't need to consider this case, since \delta can be made arbitrarily small.

[limber wyvern]

Yeah I do get this but I still think, let for example x in (0.1,0.4), then the min{x,1-x} = x

[polar storm]

in this case, u instead get d*(x,0) = x

the same reasoning applies here as well

[limber wyvern]

and that is not a problem?

[polar storm]

yeah

[limber wyvern]

so you take the intersection (1 - delta, 1) because you want to use 1-x?

[polar storm]

yes

(1 - delta, 1) is used to force the upper bound d*(x,0) < \delta, and [1/10, 1) is used to force the lower bound |f(x) - f(0)| \geq \varepsilon

in order to get these two simultaneously, u just take their intersection

and u r done

[limber wyvern]

https://klipy.com/gifs/head-explode-mind-blown-1

I'm sorry to be redundant but this concept is being hard for me: why picking x is a problem?

[polar storm]

um which x?

[limber wyvern]

why to force d* to be 1-x, why not let it be x?

[limber wyvern]

shouldn't you take the union?, if delta is something small, the intersection between 1-delta and 0.1 will never exist

[polar storm]

if delta is small enough, then 1-delta becomes very close to 1. So the intersection 1 - \delta is bigger than 0.1, so the intersection can't be empty

[limber wyvern]

mmm

lets say

as before, (0.9,1) and [0.1, 1), the intersection of this would be (0.9,1), right?

[polar storm]

yes

[limber wyvern]

mmm

I still lost :/

[limber wyvern]

so the lower bound will be 0.9 and the upper 1?

[polar storm]

d*(x,0) < 0.1 and d(x,0) \geq 1/10

in that case.

[limber wyvern]

$d*(x,0) < 0.1 and d(x,0) \geq 1/10$

[gentle osprey]

S0S4 - Feel free to ping

[polar storm]

yeah.

[limber wyvern]

I still not get it, let me do write what I've got in my head (apart from cobwebs)

I don't even know how to explain it😅

So the distance is 1-x for a delta <= 1/2. Therefore d*(x,0) < delta

so lost

more lost than a octopus in a garage

[limber wyvern]

right?

and what does the range of x has to do with that

[limber wyvern]

nice caligraphy

[polar storm]

Do u understand now? Or is there anything not clear?

[limber wyvern]

what the hell, how you manage to do that so clear

I mean

I get that clear seeing all the rest

but for an start, I wouldn't even imagine I had to pick that intersection

[polar storm]

yeah, u just practice, and learn some pattern.

do a lot of exercises.

[limber wyvern]

but again, what is the need of the intersection if 1-delta,1 will be always bigger than epsilon,1?

[polar storm]

okay, this is not a good answer, but yeah

[polar storm]

u also have the situation when \delta \geq 1/2, which we just ignored (since the delta can be made arbitrarily small).

In that case, it's the opposite.

[limber wyvern]

I see

so thats why is the intersection there

to cover both cases, and choose one that fits for every delta

[polar storm]

yeah

[limber wyvern]

I see

yeah

I have a clear image now

[polar storm]

but this can be ignored by picking some smaller delta' < 1/2, and pick the x in (1 - delta', 1), which gives d*(x,0) < delta' < delta.

[limber wyvern]

this is crazy man

who invented this

well, thank you very much I have now a clear image

@limber wyvern -> Explicación Ejemplo 1 Tema 3

That is a message for me so I can search this up later xd

Thanks for the effort and time 🙂

[cosmic zodiac]

Oh that makes sense LOL

[steep wedge]

A question was
"Find a set of real numbers which has infinitely many derived sets, distinct from each other"
My soln was
let K = {0,1/2,1/3,...,1/n,...}
the answer is
K U (1+ K + K) U (3 + K + K + K) U ... U (n(n-1)/2 + K + K...(n times)) U ...

I am almost certain my answer is correct, but is there a nicer answer?

[low stone]

Your answer is definitely correct

As for a "nicer" answer, you could replace the n(n-1)/2 by just n or n^2

or any increasing integer function at all really

Though, I still think the triangle numbers make sense to be used; it's elegant enough

[steep wedge]

I meant it as in my answer looks very contrived.

[astral mango]

It's been a while since I've even seen the topic of derived sets, but is
n(1+K)
not doable?

I'm assuming the question is supposed to be "Find a set of reals whose derived set is discrete"?

[steep wedge]

It's supposed to be that if you take the first derived set, second derived set, and so on and so forth, these are distinct

[astral mango]

Ah n-derived, gotcha.

[low stone]

I think you can achieve it without sumsets, but it would require a recursive definition

[steep wedge]

True, I was thinking maybe there was something along the lines of "the set of reals who is base 3 rep only consists of finitely many 2s" tho this doesn't work

[low stone]

I was thinking more along the lines of $A_{0} = { 0 }, K={ \frac{1}{n} \mid n \in N }, A_{n+1}=(A_{n}+K) \cup A_{n}$ and then let $S= \bigcup_{i=1}^{\infty}(i + A_{i})$

Ofc it still uses sumsets

and also even more contrived, but this is the form I think it will take

[steep wedge]

You don't think there is any nice soln?

[low stone]

Might be

[steep wedge]

Hm

[low stone]

Yeah, I can only really think of more ways to right this type of solution

[astral mango]

I like this because it also explicitly makes it clear that you're essentially constructing the set by "undoing" the "derivation" by using the properties of K.

[low stone]

There might be a more closed form for this

[astral mango]

I wouldn't even call it contrived because of that.

[gentle osprey]

Yours truly, Almond

[low stone]

I just realized I had made a mistake lol. but nothing too much, just wrote K as X in one instance

What about this: $\ A_{n} = { \sum_{i=1}^{n} \frac{1}{k_{i}} \mid k_{i} \in N } \ S = \bigcup_{i=1}^{\infty} (i + A_{i})$

[gentle osprey]

Yours truly, Almond

[low stone]

A little bit cleaner, but still the same idea

[gaunt linden]

What does an infinite union of numbers mean?

[low stone]

it's a union of sumsets

[gaunt linden]

Ah, I can't read.

[low stone]

issok, happens

[low stone]

@steep wedge I really can't think of simplifying this further

Sorry to disappoint blud 😔

[gaunt linden]

Is (k_i) a fixed sequence of integers?

[low stone]

No, its any naturals

sorry, i couldnt think of a better way to write it

basically I want n arbitrary naturals for each A_n

[gaunt linden]

Ah, so A_n is the set of all possible sums of n unit fractions.

[low stone]

Precisely

[gaunt linden]

Perhaps we can define something recursively. Set A0 = {0} subset [0,1].

Then let A(n+1) be the union of countably many copies of An squeezed and translated so they stretch from 1/2 to 1/1, from 1/3 to 1/2, from 1/4 to 1/3, and so forth.

[low stone]

ooh, this is nice

I like it a lot more than the unbounded ones we have done so far

[steep wedge]

This wouldn't work, its derived set would be itself (I think)

[low stone]

It would not be it's own derived set

[gaunt linden]

Pretty sure the derived set of A1 is A0.

And A2' = A0 cup A1.

We can simplify it a bit by including An itself in A(n+1); then we should -- by induction -- have A(n+1)' = An for all n.

[low stone]

That's a really clever solution

[steep wedge]

Exactly, so their union would have its derived set be itself

And any given An would only have finitely many distinct derived sets

[low stone]

No, each derived set would have one less layer of copies

[gaunt linden]

We're still following Almond's plan of translating the Ai's away from each other before unioning them together.

[steep wedge]

Oh

Yeh, than thatd work

[gaunt linden]

Because each Ai is a subset of [0,1) and has a maximum, they won't interfere with each other after translation.

[urban zinc]

Never thought I'd see Tropo posting about AI translation...

[gaunt linden]

Boo

[fierce mesa]

Chat, can anyone explain this question?

Here is my reasoning:
A) is not true because x_n may not necessarily converge.
B) is not true via counterexamples (i.e x_n = (-1)^n diverges, but consecutive differences converge)
C) is irrelevant to the definition of completeness
D) is not true, because we don't know this result holds for all Cauchy Sequences?

That is, if we know that any sequence (x_n) with d(x_n, x_(n - 1)) < n^(-3/2) converges, that's great. However, not every Cauchy Sequence has this sort of setup - not every Cauchy Sequence satisfies this inequality. So, it can't be used to conclude anything general about Cauchy Sequences in this space?

Since for a metric space to be complete, we require all Cauchy Sequences to be convergent, hence D) is not enough to say X is complete?

Is my logic for each option correct, or am I missing something fundamental? Thanks.

[warped helm]

are you sure about B?

just a forewarning, its late and im tired so i might trip up slightly but the problem seems interesting

[fierce mesa]

I mean yeah I did provide a counterexample so unless there's something wrong with it, then 🤷

Hmmm

I did initially think B as well

[opaque scroll]

The question is about the sum of the consecutive differences

[fierce mesa]

But one of my friends pointed it out

[cosmic mirage]

its also late for me and im tired but i believe B is false but your counterexample isnt quite right

[fierce mesa]

oh

Oh wait

Yeah whoops

I added them instead of subtracting them

Yeah thanks for the heads up

[warped helm]

thats what i was thinking too

[fierce mesa]

I swear though that this implication only goes one way though

The sum being finite implies that the sequence converges

[opaque scroll]

Your reasoning for D is also very thin.

[cosmic mirage]

and why is that?

[opaque scroll]

A Cauchy sequence converges if any of its subsequences converges, so you might try to construct a nice subsequence from a given Cauchy sequence

[fierce mesa]

I don't have the proof in my head at the moment, but this is the theorem that I was introduced to

[cosmic mirage]

ah i dont think i knew/remembered this fact lol

[fierce mesa]

Oh okay

So

It's possible that

Ohh okay I see so,

From option D), I can take an arbitrary Cauchy sequence, then find a particular subsequence such that x_(nk) satisfies the conditions in option D?

If so, I guess I need to work on proving that then.

[opaque scroll]

If you did that it would prove D

[fierce mesa]

Noted, thank you!

[fierce mesa]

Okay, better counter example:
x_n = sqrt(n + 1) I think.

sum of sqrt(n + 1) - sqrt(n) = 1

wait

no

[opaque scroll]

It might be worth noticing that
Sum xn - x[n-1] = xn - x0
And that
Sum d(xn, x[n-1]) = Sum |xn - x[n-1]|
For X the real numbers

[fierce mesa]

Okay I think I got it. Just take x_n to be the partial sum of the harmonic series. The sums obviously diverge, but the difference between partials converge.

That's a better counterexample.

[warped helm]

its asking about the sums though

sums of the differences

[opaque scroll]

You're closer to something, but the differences between the partial sums just gets you back the series in this case

[warped helm]

the consecutive differences of s_n is 1/n and you recover the harmonic if you add these up

yep precisely

[fierce mesa]

Ahhh

Okay I'm going to stop myself from rushing

[cosmic mirage]

oh i guess if youre complete then its clear enough that such a sequence will be cauchy by iterating the triangle inequality, so then you converge. the reverse implication is actually not so clear to me

the naive thing would be to say that if youre cauchy then that series is bounded and then x_n converges, thus completeness. but there are counterexamples to that assertion that the series is bounded

[opaque scroll]

The reverse implication is essentially the same idea as for doing D in the original problem.

(find a nice subsequence)

[cosmic mirage]

ahh makes sense

and then nice subsequence which is forced to converge will drag the whole cauchy sequence along with it

oh okay then yeah that's fine, use cauchy-ness to find a bunch of points that are 1/n^2 apart from the next one at most and then that converges

neat

or chef's choice of series with positive and decreasing terms which converges

[opaque scroll]

Like n^-3/2 for example 😉

[cosmic mirage]

do we even have a closed form value for zeta(3/2)

i like n^{-2} because we do actually know zeta(2)

[fierce mesa]

Option B)
Counterexample
x_n = (-1)^(n + 1)/(n + 1), this sequence converges.
d(x_n, x_(n - 1)) = |-(-1)^n/(n + 1) - (-1)^n/n| = 1/(n + 1) + 1/n = (2n + 1)/[n(n + 1)]
Because
(2n + 1)/[n(n + 1)] > 2n/[n(n + 1)] = 2/(n + 1), the sum of differences diverges. This means that finite sum of differences are not required for convergent sequences.

Option D)
Let x_n be cauchy. We know that for all eps_1 > 0, we have d(x_m, x_n) < eps_1 for some n1 such that m > n > n1.
Let eps_x = x^(-3/2) for x = 1, x = 2, etc.
By the above definition, there exists n1, n2, n3 ... etc such that the distance between any two terms in the tail of the seqeunce are less than eps_x.
So, the construction of the subsequence is to just choose the n0 + 1th term, n1 + 1th term, etc.

By the conditions in option D), we have constructed a convergent subsequence of x_n. Hence, every Cauchy Sequence is convergent.

I think this does the trick, thank you everyone for your help!

[cosmic mirage]

nice, yeah that is a good example for b

also the alternating partial sums of (-1)^i 1/i work, that should converge to ln 2 or smtg

[fierce mesa]

Ahhh yeah that's a nice one too.

[vagrant sapphire]

Does anyone have some kind of intuition or way to visualize balls in ultrametric spaces?

[gaunt linden]

I tend to visualize an ultrametric space as a tree, and balls as subtrees.

(Really, I suppose, the elements of the space are infinite branches through the tree).

[vagrant sapphire]

So with that visualization, for example, if two balls intersect is because they are in the same branch, so one is a subset of the other?
And if an element is on a ball, that element is also the center of that ball, you could see this as if the ball collapses all its elements into the "node" it represents on the tree?

[gaunt linden]

Yes: ultrametric balls are either disjoint or one contained in the other.

And every element of an open ball is a center.

[vagrant sapphire]

Thank you!

[urban zinc]

(For p-adic integers, there's a pretty clear correspondence here between p-adic numbers and paths through a p-branching tree)

@vagrant sapphire See if you can spell out the isomorphism there

[vagrant sapphire]

I remember seeing the p-adic integers as an infinite product Z/pZ x Z/p^2Z x ... . I suppose we can think each node of the tree as picking a number in 0,...,p-1 and for each subsequent level we pick again those for each node, so at level n of the tree we have p^n nodes. And a number Will be represented by a path on the tree, something like that?

[urban zinc]

Yeah but you need some compatibility relations for the product (e.g. if you're 8 mod 9 then you can't be 1 mod 3), so it's an inverse limit rather than a product—the way that the tree is connected together directly enforces this compatibility condition (you can't jump from a node on one subtree to a node on another subtree at the next level)

[vagrant sapphire]

Oh yeah i also saw the inverse limit notation but i dont know what it is.
And how can that compatibility constraint be expressed with the product notation?

[ruby delta]

you can't, not really

you're going to need notation to denote the inverse limit

[rancid umbra]

you can describe the inverse limit as a subset of the product in the case of the p-adic integers.

because all limits are made out of products and equalizers/kernels

[opaque cloud]

You mean limit

Me when direct limit = colimit and inverse limit = limit

Chat, why is any invertible linear map a proper map (i.e inverse images of compact sets are compact)?

[opaque cloud]

Say, from R^n to R^n

nvm it's obvious

[crisp lintel]

yeah any homeomorphism is proper lol

[hidden helm]

hi im taking a grad point-set top course next semester for my topology requirement, and i just wanna know

what is the study of point-set topolgoy

[kind egret]

I think it is relevant because in biology they use ultrametric spaces for the classification of species, and the space itself is a tree. So "natural" ultrametric spaces are trees, like Baire's or Cantor's spaces, and I know that there also are representations of the p-adic spaces as fractal subspaces of R^2 (the plane)

You can find nice ones on wikipedia

[novel acorn]

<@&268886789983436800>

[urban zinc]

It's the study of properties of shapes that remain constant if you're allowed to stretch and bend them

[rancid umbra]

hmmmm

[urban zinc]

And the formalization of these ideas when you define a shape as a set of points

It's like geometry but stretchy

[polar storm]

I thought the purpose and motivation for point-set topology is more general than just to cover geometric notions. Idk lol I haven't studied enough amount of math yet.

[warped helm]

that's true but it's not going to mean anything to someone who hasn't studied the subject

since like, yes, the point-set topology you're doing in analysis is not necessarily the "rubber-sheet" geometry of coffis cup = donut

but even then it's the the building blocks for eventually talking about those things

[polar storm]

I see.

[urban zinc]

I think you're underselling geometric notions. We as humans are fundamentally visuospatial creatures in many ways and therefore interpret the world of abstract relationships through a geometric language. Part of the beauty of topology is the ability to use geometric notions to capture the essence of abstract concepts like the prime ideals of a ring, as well as to fundamentally question what we mean by notions of space and nearness, and to classify all their possibilities.

[polar storm]

I see. I have never studied algebraic geometry (which I guess u r talking about (?)). Seems like I just need to study more math. So far, I only know very basic analysis lol.

[urban zinc]

topology is ubiquitous, not just in algebraic geometry, but in logic, manifold theory, etc

[crisp lintel]

I would say topology is used for formalize the idea of approximations

which is very general but also topology is used everywhere so I think it's kinda accurate

[median sand]

Let X be a space, do I have it right that the following are equivalent
i) X is initial wrt C(X,[0,1])
ii) X is initial wrt C_b(X,R)
iii) X is initial wrt C(X,R)
iv) X is completely regular

[crisp lintel]

what do you mean initial here

[median sand]

The initial topologies induced by these families coincide with the original one

[crisp lintel]

ah ok yeah