point-set-topology

[hoary kayak]

Suppose that I have a 2 cosimplicial diagrams $[n] \mapsto C_n$ and $[n] \mapsto D_n$ in $Pr^R_{st}$ (the category of presentable stable $\infty$-categories with right adjoints as the morphisms). Suppose that I am given a morphism of diagrams such that each $ F_n: C_n \to D_n$ preserves colimits. Is it true that the induced functor $colim_n F_n : colim_n C_n \to colim_n D_n$ preserves colimits?

[gentle osprey]

Arun

[ebon galleon]

Hmm is this even true at the 1-categorical level?

[prime elbow]

Yes

So for the question, which topological structure has exactly one base, it is indiscrete topology, right?

What does that mean?

In Topological space X, open subset of X, it is the same as the open set of X ?

[hoary kayak]

Oh, I forgot to add that all the transition functors $C_n \to C_m$ and $D_n \to D_m$ induced by $[n] \to [m]$ are colimit preserving

[gentle osprey]

Arun

[viral atlas]

Yeah

[viral atlas]

The definition of a base (the one I'm familiar with) doesn't exclude the possibility of having an empty set in your basis

So you end up with two elements in your basis in general, the emptyset and the entire set

[tender halo]

does anyone know a nice example of a prime filter in the lattice of open sets that is not an ultrafilter?

one can take a pair of points in a non-hausdorff space that cant be separated and then take the filter of open sets containing them and thats an example, but in some sense its a principal filter and i would like an example of a non-prinicpal filter

[formal tide]

Here Pi(K) is a homotopy from Pi(f) to Pi(g). I'm not seeing how Pi(K*L) implies that Pi(f) is an equivalence. It seems to me that the argument is:

• f is a homotopy equivalence, hence gf ~ id and fg ~ id.
• Hence Pi(gf)=Pi(g) . Pi(f) ~ id and Pi(fg)= Pi(f) . Pi(g) ~ id.
  and that's it?

not seeing how Pi(K * L) = Pi(L) . Pi(K) is used

[prime elbow]

Thank you

[ebon galleon]

Yeah I don't know nearly enough about ∞cats to answer this, so I'll just say you'll get better responses if you post it in #category-theory instead.

[proper fiber]

The answer to this was this Cantor set like construction

Is any other type of construction possible

Also is there a way one could motivate this construct

[knotty vine]

In the plane at least these are called Lakes of Wada.

[proper fiber]

I see

The construction is lit

But it's like very nontrivial to come up with this

[umbral panther]

An example that is easy to write down is the basins of attraction of using Newton’s method to find a root of a cubic polynomial
I suspect that if you take a cubic polynomial with 3 real roots, this gives you an example in the line, but I imagine it’s harder to check

[proper fiber]

I agree it's hard to check without a computer

But it's cool

[tidal cedar]

There's a large algebraic topology server which is probably better able to answer these sorts of questions, I can DM an invite if you'd like

[bright acorn]

Do we have the following equivalence of categories?

Is category of O(n)-principal bundles over a manifold equivalent to the category of vector bundles over it endowed with a inner product?

Is the category of SO(n)-principal bundles also equivalent to the category of orientable vector bundles endowed with an inner product + an orientation?

[umbral panther]

Do you have a functor? If it’s well defined and true over a point, it’s probably globally true

[bright acorn]

hmmm ye, I will check the details but

I was just trying to play around and see if thinking about a principal G-bundle for G some matrix lie group as being just a vector bundle endowed with some more structure reflecting the nature of G makes sense in general.

[grave solstice]

Is topology a part of geometry or the other way around, or none of the two?

[umbral panther]

As we learn in high school, math is divided into three parts, algebra, geometry, and calculus. Under this classification system, topology is clearly part of geometry

[quartz horizon]

Where’s number theory lol

[hidden crag]

Algebra

All of number theory is solving for x

[quartz horizon]

I never knew..

[woven sinew]

If you insist on being strict about it, geometry is about angles, lengths, areas, volumes, and curvatures of geometric objects. Geometry cares about when things are preserved under isometries, see e.g. Theorema Egregium. Topology cares about when things are preserved under continuous transformations. In this sense, topology is a more general framework than geometry.

But in reality the two subjects are interlinked, and I doubt any advanced researchers will care about making a strict distinction.

[grave solstice]

more or less, yes

[gentle girder]

Neither but people who care about one probably won’t turn their nose up about a conversation about the other, i don’t like the lines that people draw between the branches of math

[hoary kayak]

Hi that would be great thanks could you DM me an invite?

[shy phoenix]

can anyone point me to a proof of or a resources about the universal property of quotient spaces?

[ebon galleon]

https://topology.mitpress.mit.edu/ is a good reference. Chapter 1 discusses subspaces, products, and quotients in a direct construction and also through their characterizing universal property

(also it's free!)

You need virtually no category theory for the first few chapters if that's a concern

[tidal cedar]

not if you bought a physical copy

[ebon galleon]

[warm quiver]

I think you need to restrict to the subcategory with objects vector bundles and morphisms vector bundle isomorphisms. You don't get a map between frame bundles unless the vector bundle homomorphism sends frames to frames which is equivalent to the homomorphism being an isomorphism.

[umbral panther]

The question was about inner product spaces. Those morphisms will necessarily be injective

[warm quiver]

oopsies

[radiant walrus]

I was wondering if my proof of this was okay

Since X is an incomplete metric space, then, by Theorem 1.6-2, there exists a complete metric space X^ and there is a bijective isometry A from X to a subspace W of X^, with W being dense in X^. This X^ is unique except for isometries. X^ being unique means that it is the same set as in the proof for 1.6-2, that being the set of equivalent classes of Cauchy sequences in X. Given x,y E X^ and a scalar a, define x + y = [(xn + yn)] and ax = [(axn)], where (xn) E x and (yn) E y. We'll show that, given Cauchy sequences (xn),(yn) E X and a scalar a, then the sequences (an),(bn) defined by an = xn + yn and bn = axn are also Cauchy. Given e > 0, there is an N such that n,m => ||xn-xm|| < e/2 and ||yn-ym|| < e/2, therefore ||an-am|| = ||xn+yn-xm-ym|| = ||(xn-xm)+(yn-ym)|| <= ||xn-xm|| + ||yn-ym|| < e/2 + e/2 = e, thus (an) is a Cauchy sequence. If a = 0, then (bn) is constant and thus a Cauchy sequence. If a != 0, then there is an N such that n,m > N => ||xn-xm|| < e/|a|. Then, ||bn-bm|| = ||axn-axm|| = ||a(xn-xm)|| = |a| ||xn-xm|| < |a| * e/|a| = e. Therefore, (bn) is a Cauchy sequence, which means that x + y, ax E X^. Since addition and scalar multiplication of X^ is defined as addition and scalar multiplication of X, and X is a vector space, then X^ satisfies all properties of a vector space, thus making it a vector space. Define the norm in X^ to be ||x|| = lim ||xn||, where (xn) E x. This limit exists due to lim d(xn,yn) = lim ||xn-yn|| existing if (xn),(yn) are Cauchy sequences, so you can always take yn = 0 for all n to make it lim d(xn,0) = lim ||xn-0|| = lim ||xn||. Since the norm in X^ is defined in terms of the norm in X, then it follows all the properties of a norm, showing that it is a norm. Furthermore, d^(x,y) = lim d(xn,yn) = lim ||xn-yn|| = ||x-y||, so d^ is induced by the norm, therefore X^ is a normed space, and, since it is complete, it is a Banach space.

For reference

@paper wedge Since you've been helping me with this for a while now, I was wondering if you'd be okay with it this time since you know the context x)

[shy phoenix]

so can i use the universal prop of quotient topologies to tell when a map f:X->Y induces a homeomorphism on some given quotients of X and Y? rn i can show that this is true if f is an open map

[radiant walrus]

What am I supposed to do here? This book does not define what continuity means for functions with two variables

Will it work if I fix one of the variables to an arbitrary value and show that the whole function is then continuous?

Like, "Let x be fixed. We'll show that the function T(y) = x + y is continuous for all y"

[unreal stratus]

Unfortunately no

[unreal stratus]

But this definition always works

[radiant walrus]

By taking x = (x,y)?

[unreal stratus]

Sure, though that is slightly ambiguous notation

[radiant walrus]

But um

[unreal stratus]

But yes, addition is a map X x X -> X

X x X is a metric space

As is X

[radiant walrus]

But um

I don't think I have seen anywhere in this book about mentioning or in the exercise sheet that X x X is a metric space if X is

Yeah it doesn't

[elder loom]

think about what what the open sets of the domain and codomain are

[radiant walrus]

I don't know what the metric on X x X is. Surely it has to do with the metric on X, no?

Well, the first one is X x X and the second one is K x X, where K is the field X uses

Just to be clear, a normed space always has the distance function being d(x,y) = ||x-y||, right? Like, d is always induced by the norm, yeah?

Because I've seen that there are vector spaces which are also metric spaces but without d being induced by the norm

[hexed steppe]

it depends on the context

[radiant walrus]

Well, I'm using the fact that a distance induced by the norm has the property d(x, y) = d(x+z, y+z) for any z

So, does this work to show that addition is continuous?
`Given e > 0, let x',y' E X be such that d(x',x) < e/4 and d(y',y) < e/4.
d(x',x) < e/4 => d(x'+y',x+y') < e/4
d(y',y) < e/4 => d(x'+y',x'+y) < e/4
d(x+y',x'+y') + d(x'+y',x'+y) < 2e/4
d(x+y',x'+y) < e/2

d(x'+y',x+y) <= d(x'+y', x'+y) + d(x'+y, x+y') + d(x+y',x+y) < d(y',y) + e/2 + d(y',y) < e/4 + e/2 + e/4 = e`

Like, I picked an arbitrary pair (x',y') such that d(x, x') < e/4 and d(y, y') < e/4

But if there's a metric on X x X, let's say, d1, then I don't really know what d1((x',y'), (x,y)) would be like

Since the question said it's with respect to the norm, I could probably rewrite it as
`Given e > 0, let x',y' E X be such that ||x'-x|| < e/4 and ||y'-y|| < e/4.
||x'-x|| < e/4 => ||x'+y' - (x+y')|| < e/4
||y'-y|| < e/4 => ||x'+y' - (x'+y)|| < e/4
||x+y' - (x'+y')|| + ||x'+y' - (x'+y)|| < 2e/4
||x+y' - (x'+y)|| < e/2

||x'+y' - (x+y)|| <= ||x'+y' - (x'+y)|| + ||x'+y - (x+y')|| + ||x+y' - (x+y)|| < ||y'-y|| + e/2 + ||y'-y|| < e/4 + e/2 + e/4 = e`

Does this work?

Both x' and y' are arbitrarily close to x and y, respectively, independent of each other

[night shard]

does anyone have an idea of how to show that R^2 / the image of a simple path that is not a loop, is path connected/connected (in this case I understand they're the same thing)

[hexed steppe]

is that true

[night shard]

sorry a path which is not a loop

that's probably an important distinction

[hexed steppe]

what about a line

[night shard]

I think so, I guess I shouldn't assume but it feels true

[hexed steppe]

i mean its false if you allow loops

[night shard]

yeah but I meant without loops

[hexed steppe]

i guess probably the parameter needs to be in a compact interval

otherwise you have lines

[night shard]

I thought by default a path was a continuous function from [0,1]

[hexed steppe]

idk

i guess

[night shard]

well that's what I mean in any case

[hidden crag]

What if you have a self intersection somewhere

[night shard]

ok well let's say it's also simple

[hidden crag]

Remind me what that is again

[night shard]

not self intersecting I think

oh wait no

it's a bit more complicated than that

well let's say it doesn't intersect itself

(although I think simple is probably enough in this case)

[hidden crag]

Is this part of a bigger problem

[night shard]

https://mathworld.wolfram.com/SimpleCurve.html

[night shard]

other sources have different definitions though

[night shard]

yeah

I'm trying to show that if you have 3 simple paths in R^2 from point A to B, then one of them (let's say p_2) will be in the interior of the closed loop formed by the other two, and p_3 will be exterior to the closed loop formed by p_1 and p_2 + p_1 will be exterior to the closed loop formed by p_2 and p_3

if I prove the question I will be done

[tiny obsidian]

this isn't true

[night shard]

what's the counter example

[hexed steppe]

why would any be in the interior

[hexed steppe]

of the loop enclosed by the other two

yeah lol

[night shard]

sorry they also can't intersect one another

I keep forgetting these details because I've been thinking about this for a while now

[hexed steppe]

..

i mean then like

probably just cite jordan curve thm

[night shard]

I am

[hexed steppe]

no i mean

[night shard]

but it's not proven trivial after that point

[hexed steppe]

isnt this

a corollary

of jct

[night shard]

how

[hexed steppe]

idk

[night shard]

bruh

[hexed steppe]

it just seems like it

or you can like

[night shard]

I mean you definitely cite jct

[hexed steppe]

fix a coordinate system

and then talk about the induced ordering on paths

like one will be “leftmost”

[night shard]

what is the induced ordering on paths

[hexed steppe]

orient the endpoints so theyre on the y-axis

then the x coords

are ordered

strictly ordered

by the nonintersection

that should be enough to give a proof

[night shard]

If you're defining leftmost as "the path whose leftmost point has the lowest x coordinate" the problem is that a path can be both leftmost and rightmost

[hexed steppe]

how?

and no

i am not defining it that way

pointwise leftmost

[hexed steppe]

you need to parametrize them via a common set of times

like [0,1]

or something

[night shard]

ok

[hexed steppe]

hmm that picture is concerning

[night shard]

lol

that's what I thought too about 5 hours back

[hexed steppe]

maybe you can fix situations like that by somehow modifying a path

[night shard]

as I said before though, I've put enough of the proof together that you only need to show the complement of a simple non-loop path in R^2 is path connected

[night shard]

how

[hexed steppe]

not sure immediately

[hexed steppe]

yeah idk how to prove this

though it does sound plausible

[night shard]

I've been doing a little reading around this, and I think this question would most likely come up under planarity theory

because I feel like knowing that the graph with two vertices and one edge does not have a face is pretty important if you want to prove that stuff rigorously

[paper wedge]

wdym?

[radiant walrus]

Like, I was wondering if the proof for that was correct, because I'm not 100% sure

And also for the thing below that, because it got me confused

[woven sinew]

There are different equivalent norms on the product space, and you can just pick one of those.

A couple of nice ones are $|(x,y)| = |x| + |y|$ and $|(x,y)| = \text{max}(|x|,|y|)$.

[gentle osprey]

OneTrackPony

[woven sinew]

They will induce the same topology on X x X. The 'Euclidean' norm also works.

[radiant walrus]

But why am I allowed to pick one?

[naive trench]

I think you mean norm(x-y)

[woven sinew]

It's because the norms I gave are equivalent, and if a function is continuous in one norm, it is in another as well.

Also, they induce what is usually assumed to the product topology on a product.

[radiant walrus]

So, if it's continuous under one norm, then no matter what other norm I pick, it will also be continuous there?

[woven sinew]

Not no matter what, but these specific norms are equivalent.

[radiant walrus]

But how can you be sure that X x X doesn't have a weird norm

[woven sinew]

Because then they better tell you about it, otherwise you have no chance 😬

[radiant walrus]

The book says nothing about this

[woven sinew]

It's very strange. What book is it?

[radiant walrus]

The Kreyszig one

[radiant walrus]

Like, isn't this good?

You can just pick delta to be e/4

[woven sinew]

So strange. Usually authors mention how metrics or norms work on a product space.

[woven sinew]

So you just use the sum of the norms here (also called the l1 norm)? Should be fine then.

I would probably be more explicit with when I'm on the product space and when I'm not.

Like begin with $|(x,y)-(x',y')|$ and then something similar to what you already did.

[gentle osprey]

OneTrackPony

[radiant walrus]

Like, there's this

🤷‍♀️

[woven sinew]

Is it his engineering book?

[radiant walrus]

Introductory Functional Analysis

[woven sinew]

Even stranger then 😆

Maybe he just forgot about it.

[quartz horizon]

I remember this being a lemma in proving the Jordan curve theorem

At least the proof I’ve seen

[radiant walrus]

Well, I did prove this, which is similar

[woven sinew]

Yeah

But idk, he stills owes to tell you what distance you can use on X x X, it's hard to guess.

[radiant walrus]

I guess I can skip the problem

But yeah, with the norms you gave me, I could show that they are continuous

I already know in my head how the proof is gonna look like, so

[woven sinew]

If you want. Or use either the l1 or the max norm, this is what was intended with the problem anyway.

[radiant walrus]

Does the fact that addition and scalar multiplication is continuous get any use?

Or is it just for the fun of it

[woven sinew]

I wonder, is the book good otherwise?

[radiant walrus]

I've been told the book is really good

For learning Functional Analysis

My teacher wanted me to study it

It's a book from the 70s

[woven sinew]

Yeah, it gets a lot of use. There's a concept called topological vector spaces, which is both a vector space and a topological space, and where addition and scalar multiplication are required to be continuous. There's a rich theory of these spaces.

[radiant walrus]

That's wild

Here we only work with metric spaces

But not being able to calculate a distance while it still being there sounds weird to me

[woven sinew]

You got a point. 😆

[radiant walrus]

Like, if the distance still exists, what's stopping you from assigning it a value?

[woven sinew]

Topological spaces generalize metric spaces. You can say that while metric spaces have the concept of distance, topological spaces have the concept of "nearness". But in a topological space, you cannot necessarily distinguish the nearness of points to the same extend that you can in a metric space - because there is no exact distance.

[prime elbow]

How is the interior of B the union of all open sets contained in B?

[radiant walrus]

What does it mean for two points to be near?

[radiant walrus]

First thing that comes to my mind

1. Show that any point in the interior is in an open set of B
2. Show that any open set of B is contained in the interior of B

[woven sinew]

Loosely speaking if they are in the same open set. In a topological space, all you know are the open sets, and how near points are are determined by what open sets they are in.

[tiny obsidian]

the way to approach this depends on what definition of interior you're going from

[radiant walrus]

So how are closed sets defined when you only know about open sets?

[tiny obsidian]

As the complements of open sets

[rain ether]

By your definition, every point is near to another one

[woven sinew]

By definition a closed set is the complement of an open set.

[radiant walrus]

Right

I forgot that

[tiny obsidian]

there's a 'hierarchy' of sorts where there are smaller (subsets) open sets

in general though nearness is a very vague notion on topology that isn't formal at all

[radiant walrus]

That's wild ngl

[tiny obsidian]

just the intuition from metric spaces sometimes getting carried over

[radiant walrus]

But I think I get it!

[prime elbow]

They define interior in a point of view of neighborhood

[radiant walrus]

So the limit is like, getting smaller and smaller open sets?

[woven sinew]

I don't know, depends on what you mean. 😄

[prime elbow]

I have a result that if x belongs to the interior set of B then there is an open set U which contained in B

[tiny obsidian]

can you give this definition?

[prime elbow]

Let B be subset of X then point x belongs to X is called an interior point of B according to B is a neighborhood of x

[tiny obsidian]

Ah, I've not encountered this one before

[prime elbow]

U be a subset of X is called a neighborhood of x belonging to X if there is an open set V with x belongs to V and V is contained in U

[radiant walrus]

How are open sets defined?

[prime elbow]

They just tell it an element of topology defined on X

[radiant walrus]

Ah, I see

Right

I remember that

Oopsies

Max is already looking at me weird

[gleaming warren]

All good, i just thought it was funny: How is interor defined? How is nbhd defined? How is open defined?

[prime elbow]

I don't understand in metric space and in the topology space this open set definition is equivalent?

[gleaming warren]

In metric spaces you define openness with these open balls,and you can prove that open sets fulfill the definition of a topology. For a topology you make these properties axioms

[radiant walrus]

A subset M of a metric space X is said to be open if it contains a ball about each of its points.

[prime elbow]

Means in Topology an open set is just an element of that Topology?

[woven sinew]

Mathematician: "Your Honor, before we continue, can you define 'good' and define 'bad'?"

[gleaming warren]

Yes

[prime elbow]

Oh

[gleaming warren]

You just say i regard these sets as open

[radiant walrus]

Tbh, I thought open sets were something you could define, like how you can define how lines work, but that they must follow a certain property(ies)

[gleaming warren]

(Ofc they should fulfill reasonable properties, thats why you have the topology axioms)

[radiant walrus]

Tbf I haven't studied topology. I don't know these abstract axioms

[prime elbow]

So how can I prove that?

[gleaming warren]

Its not that hard to understand, if you know metric spaces for instance

[gleaming warren]

Prove the two inclusions

[tiny obsidian]

If we explicitly write this dependence of $U$ on $x$ by writing $U_x$ for the open this gives such that $x\in U_x\subset B$, you can show that the interior is a subset of the union (because in particular it includes the union of the $U_x$, and hence all $x\in B$)

[gentle osprey]

Edward II

[tiny obsidian]

the other inclusion is also reasonably straightforward

[radiant walrus]

I know it's not hard. It's just that I haven't studied it so I don't know them

[tiny obsidian]

I meant to reply to this above

[prime elbow]

And which properties hold for open set in topology is it hold for metric space ?

[tiny obsidian]

(and I messed up the grammar due to an edit in the middle)

[prime elbow]

Yes

[tiny obsidian]

a metric space is a specific case of a topological space, so anything that holds in topology holds in a metric space as well

the other way is not true, because metric spaces are really really nice

[merry geode]

I find it interesting how topological axioms came down being quite simple

[tiny obsidian]

(e.g. they satisfy a bunch of extra properties called separation axioms, which e.g. mean for any two disjoint closed sets there are disjoint open sets containing them)

[gleaming warren]

Even stronger, every metric space is normal

[merry geode]

Even better, every metric space is metrizable

[woven sinew]

There's even the concept of perfectly normal spaces - they always make me suspicious.

[tiny obsidian]

more niceness is that you get a bunch of implications in topology become equivalences in metric spaces, and also a bunch of properties can be characterised using sequences

[prime elbow]

Normal what?

[merry geode]

It is a normal space

But maybe not completely normal

[tiny obsidian]

topology terminology is old, terrible, and inconsistent

[prime elbow]

I don't know how to show this, I think it helps ,the set of interior points is open

[merry geode]

I think one can use the universal property of union regarding to inclusion
EDIT: Sorry, maybe this does not work..

[prime elbow]

How?

[merry geode]

Sorry that I wrote it opaquely, but

You want to know if union of U_x's is included in some set V

Can you identify necessary and sufficient conditions directly following from this?

[prime elbow]

I know that the interior of B is a subset of the union of U_x but I do not figure out how to show that int B is the union of all open sets contained in B

What if I assume let V be an open set in B then V be a subset of the interior set of B , I think from there I will get the result

I think the union of U_x will be an open set which is also contained in B

Then by let V= union of U_x I will get the result, right?

[merry geode]

Ah yeah, that works

[prime elbow]

So that's same way or different? Because I didn't understand this

[merry geode]

Ah, so I am taking V to be the interior of B.

[prime elbow]

Oh

[merry geode]

Then you do have U_x \subset V, right?

[prime elbow]

Yes

[merry geode]

Then union of U_x should be contained in V as well

(By logic)

[prime elbow]

Yes

But how do I show that it is a union of all open sets contained in B because I worked on U_x how all open sets?

The Union of U_x is a subset of union of all open sets contained in B , so maybe that's work

[merry geode]

If you can show that any open set is contained in the interior V,
then the union of all open set is also contained in V.

[quartz horizon]

Mhm

[prime elbow]

And what is V?

[prime elbow]

Is it not work?

[merry geode]

As I said earlier, I gave a name V to the interior of B.

That is, my V is the interior of B.

[quartz horizon]

It’s a general fact that if $(A_i){i \in I}$ and $(B_i){i \in I}$ are two indexed families with $A_i \subseteq B_i \space \forall i$, then $\bigcup_{i \in I} A_i \subseteq \bigcup_{i \in I} B_i$

[gentle osprey]

Pseudonium

[merry geode]

That sounds like the reverse direction of what we want

[prime elbow]

But it will work? Because first we showed that int B is a subset of union of U_x then we know that union of U_x is a subset of union of all open sets contained in B.
So , we have int B be a subset of union of all sets contained in B but union of all sets contained in B is a subset of int B.
Therefore, int B is the same as the union of all open sets contained in B.

[rain ether]

There's no need to go through the union of U_x to show that the interior of B is equal to the union of open subsets contained in B.

[prime elbow]

Is it not correct?

[rain ether]

It's fine, but a bit convoluted imo

[prime elbow]

So what's your point?

[rain ether]

You can show the equality by showing both inclusions directly. An interior point of B has an open neighborhood contained in B, so it is in the union of open subsets contained in B. Conversely, an element of the union is contained in some open subset contained in B, which is an open neighborhood of that element, so it is an interior point.

[prime elbow]

Thank you, I will look into it

[hexed steppe]

agreed

idk the language but ill read a bit today if i have time

[radiant inlet]

Guys, can you help me pls

Proof that a simplicial complex of dimension n can be embedded in $\mathbb{R}^{2n+1}$

I don't have any ideas how to prove it, maybe we need to use the fact that the simplicial complex is compact. This exercise is right after the definition

[gentle osprey]

Pumpon

[eager vigil]

Hey, I'm teaching a brief introduction to algebraic topology - a lecture of only 90 min-ish -, and I wanted to ask y'all what ur favorite relatively basic problems are?

[tidal cedar]

one of the fixed point theorems or Jordan's curve theorem

[umbral panther]

Use covering spaces to prove Picard’s theorem on essential singularities

Use covering spaces to prove that a subgroup of a free group is free

[last topaz]

Borsuk-Ulam (at every moment on the surface of the Earth, there is a pair of antipodal points where the temperature and the humidity are the same.)

[quartz crater]

Hello I hope this is the right place to ask this about; in a proof im doing i'm unsure about some parts— ill just show what i have so far and highlight the parts later. \

So given that every convergent seq. in a set $M \subseteq \mathbb{R}^n$ has its limit in $M$, i want to show that $M$ is closed. (i use the def. $\partial M \subseteq M$)

So take an $a \in \partial M$. For each integer $k \geq 1$ we can pick a point $x_k \in M$ from the neighborhood $B(a,1/k)$ (this is fine right?). Since $| x_k - a | < 1/k \quad \forall k$ then $x_k \to a$ as $k\to \infty$, so $a \in M$ as promised and we've therefore shown that $\partial M \subseteq M$ (closed).

[gentle osprey]

Aslan

[quartz crater]

I havent really seen a good motivation or argument why the seq. can be choosen this way, as i feel uneasy doing so. Could anyone explain any possible fallacies when this would be done (in this type of proof) if thats possible? Thanks.

[quartz horizon]

I mean, what’s your definition of boundary?

Personally I don’t really see anything wrong with this proof..

[quartz crater]

$a\in \partial M \iff$ each neighorhood of $a$ contains at least one point from $M$ and atleast one from $M^C$

[gentle osprey]

Aslan

[quartz horizon]

Right, that’s what I thought

In that case

I think your proof is perfectly fine!

[quartz crater]

I see, im still in the early stages of these types of proofs so i tend to overthink most parts without ability to reason it down to something concrete and i suppose i just need more experience. But take for example the construction of (x_k), could this in some way been avoided? or this the only reasonable way to easily show its convergent?

[quartz horizon]

Well, I don’t think so?

The only thing you know about M is that it’s sequentially closed

So you have to use that at some point

[quartz crater]

Hm yeah, i think i might not just be used to such constructions. Is it a reasonable reaction to have?

[quartz horizon]

Sure, if you’re not used to it!

But really, you proved it exactly how I would

[quartz crater]

We were shown the construction part of the proof and I just used it without really thinking until now, so thanks this makes me more confident!

[fading fern]

how would you go about doing it for the first one? (if i see right, it's a looping tape with a hole on it)

[gentle girder]

you can find an immersion (just do it visually in your head) from the surface of genus 2 to your guy in the picture (you actually can do this with an isotopy of immersions if it makes it easier to visualize, think of one handle “passing through” the other)

then since it’s an injective immersion and the surface of genus 2 is compact then you know you have a homeomorphism

[fading fern]

thanks smay!!!!! imma try to figure the others out with that

also not appropiate place for this but did you take a break?

alright i figured out the second i think

(the hole in hole)

[gentle girder]

yeah for compact things you can think about isotopes of immersions (think “pop science rubber sheet geometry, you can pass things through itself) then whenever you get something injective you know that you have a homeomorphism

[gentle girder]

my comment was about the hole in a hole

[fading fern]

oh

i thought you meant the 1st one had 2 holes

anyways imma say answer and i'll tell how i did it

i got that it's a circle glued on 2 points with a torus

wait that's wrong

[gentle girder]

wdym by glued on 2 points with?

all your guys are manifolds!

[fading fern]

i mean

glued on 2 points of a torus

[gentle girder]

can you draw it

[fading fern]

sure

wait a bit

[gentle girder]

whatever you draw better be a manifold

[fading fern]

i know the answer's wrong though

[gentle girder]

also, you can “cut things up” as long as you remember how you did it

for example you REALLY want to get a decomposition of the first guy

[fading fern]

i know it's wrong cause you know

circle is 1d

[gentle girder]

yeah that’s not a manifold!!

[fading fern]

me when no manifold: B(

yeah i made a mistake while doing it

so my idea on how to do this

was like drawing the "skeleton" of the shape

and then bending it to see if i get something nice

i'm struggling to do this for 3rd figure

[formal tide]

Is this right? It doesn't seem to me that "interiors cover the whole space" iff we have a pushout. E.g. [0,1] <- {1} -> [1,2] is a pushout I believe

so it should bi that the fundamental groupoid presevers some pushouts

[unreal stratus]

Well if the interiors cover X, then in particular the whole sets cover X

They didn't claim the other direction was true

[umbral panther]

Von kampen is not true for all push outs

[formal tide]

They say "2.6.1. says that the functor Pi preserves pushouts"

hmmm maybe I'm misinterpreting it

[umbral panther]

In practice people use van kampen and MV in cases where the intersection is closed, not open. But for reasonable spaces, there is an open neighborhood of the closed subset that deforms to it

[unreal stratus]

Oh sorry, yes that line is a bit unclear.

You should think of it as like

it preserves "nice pushouts"

As bw says, like often you don't need the full hypotheses here in 2.6.1 - for example if a CW complex is written as a union of two subcomplexes

(maybe modulo some finiteness issues? but i think that is true)

[formal tide]

I see, thanks

[formal tide]

btw do you know any counterexamples?

[umbral panther]

Glue together two Hawaiian rings at the baseline base point. An element of the free product of the fundamental groups is a finite word that goes back and forth between the two a finite number of times. Whereas the fundamental group includes paths that go back and forth infinitely many times, but making smaller loops each time

[formal tide]

Beautiful!

do you mind sharing where you found that example? (if you found it somewhere)

[umbral panther]

[formal tide]

lol aight, thanks still!

[umbral panther]

The Hawaiian earrings are almost a universal counter example. To build a counter example, we need an inclusion of a closed set that is not a neighborhood deformation retract. So infinitely many things are happening near that closed set. But only things about the fundamental group matter, so we need arbitrarily small loops. Hence the Hawaiian earrings

Actually, that suggests another example, because we don’t need arbitrarily small loops. The other aspect of the fundamental groupoid is objects. So if we have components arbitrarily close to the base point, that works, too. Cut the Hawaiian earrings in half by a line that cuts all off the loops. Consider it as a push out of the half on one side and the mirror image on the other, glued along {1/n} cup {0}

[formal tide]

absolutely no idea how to calculate the pushout for those groupoids (which I think are trivial?)

but if that works, it should also work with S1 right?

[umbral panther]

The groupoids don’t know the topology. It push out of groupoids would be the same for the infinite wedge of circles, where the circles all have the same radius. Which gives a free group with generators labeled by the circles

[red yoke]

Does Hawaiian earring have same group as infinite wedge

[unreal stratus]

Do you mean fundamental group?

Then no

[formal tide]

nope, not clear to me which group it is but you can have a path that covers infinitely many circles

e.g. divide the interval in n partitions of length 1/2^n, and map each interval [a,b) to cover a smaller circle than the last one

map 1 to the common point and done

[red yoke]

Yea

Though in any case the groupoid of earring ∨ earring still isn't the coproduct

[radiant walrus]

If T is a bijective isometry between the normed spaces X and Y, so the distance is preserved, d~(Tx, Ty) = d(x,y). Does this imply that T(x+y) = Tx + Ty or is there something else missing?

[hexed steppe]

how would it imply that

[radiant walrus]

I don't know. I thought it did

I tried to show it but got nowhere

[hexed steppe]

Tx = x+y

for some y

does not satisfy T(x+x') = Tx + Tx'

[radiant walrus]

[radiant walrus]

This took so many hours but it was really fun to do!

Here's my proof if anyone would like to check it out :3
Since X is a normed space, then, by Theorem 2.3-2, there is a Banach space X^ and an isometry A from X onto a subspace W of X^ which is dense in X^. Since this X^ is unique, we know that X^ is the set of all equivalent classes of Cauchy sequences of X. Let w E W- => w E X^, then, given e > 0, let (wn) E W be such that ||wn-w||X^ < e/n^2, for all n. We thus have that n,m > N => ||wn-wm|| <= ||wn-w|| + ||w-wm|| < e/n^2 + e/m^2 <= 2max{e/n^2,e/m^2}. Let (xn) E X be such that Axn = wn for all n. Thus, for n,m > N, ||xn-xm||X = ||Axn-Axm||X^ = ||wn-wm||X^ < 2max{e/n^2,e/m^2}. Let (x'n) E X be such that x'n = x_(n+1) - xn for all n. So, we have that x_(n+1) = x1 + sum j=1 n x'j. Furthermore, sum n=1 infinity ||x'n|| = sum n=1 N ||x'n|| + sum n=N+1 infinity ||x'n|| < sum n=1 N ||x'n|| + sum n=N+1 infinity 2max{e/(n+1)^2, e/n^2} = sum n=1 N ||x'n|| + sum n=N+1 infinity 2e/n^2 = sum n=1 N ||x'n|| + 2e sum n=N+1 infinity 1/n^2 < sum n=1 N ||x'n|| + 2e * pi^2/6. Thus, (x'n) is absolutely convergent. By hypothesis, x'1 + x'2 + ... converges to x E X. By increasing N if necessary, also making sure it's greater than 1, we have n >= N => ||(sum j=1 n x'j) - x|| < e/2 => ||x_(n+1) - x1 - x|| < e/2, which means that n > N => ||xn - (x + x1)||X < e/2 => ||Axn - A(x+x1)||X^ < e/2 => ||wn - A(x+x1)||X^ < e/2 < e, therefore (wn) converges to A(x+x1) E W. Since, given n > N, we have ||w - A(x+x1)|| <= ||w - wn|| + ||wn - A(x+x1)|| < e/n^2 + e/2 <= e/2^2 + e/2 < e. Since this is true for any e > 0 and that ||w - A(x+x1)|| >= 0, then ||w - A(x+x1)|| = 0 => w = T(x+x1) => w E W => W = W-. Therefore, W is a closed set. By Theorem 1.4-7, W is complete. By Prob. 4 of Chapter 1.6, X is complete.

Here are the Theorems I referenced

[formal tide]

My book says that a path to an infinite wedge sum VX_i can only be surjective onto a finite number of X_i because of compactness of the interval

I'm not really seeing why this is so (also this is false if X_i=*)

The X_i may not be open, so their preimage may not be open

am I missing something?

[red yoke]

Pick an open neighbourhood of x in each Xi

Then you can construct a neighbourhood Ui of x in ∨Xi that contains only one Xi

Unless pathological topologies on Xi

This argument also implies that if each Xi is locally contractible at x, then the fundamental group of ∨Xi is precisely the free product

[formal tide]

Thanks!

[half plover]

Hey I wanted to get started with topology
Any introduction
Or book recommendations?

[formal tide]

Munkres' General Topology

[red yoke]

Bump

[plain quail]

Hey everyone, just wondering what the general strategy is for computing the closure of an open set. Say for example (0,1) in R with the usual metric. Intuitively its closure is [0,1], but how do I show that the closure is [0,1] precisely, no more and no less. I guess what im not sure on is how do I show im not missing any limit points.

[red yoke]

Show that [0, 1] is closed

[feral copper]

You show that [0,1] is closed, and then in this case you can list all subsets of [0,1] that also contain (0,1). Only [0,1] itself is closed.

In more general sattings, it may be more difficult, and listing all such subsets needs not be possible (e.g. closures of sets of matrices)

But in general, it goes like this: find a closed set containing your thing, and try to show it is the smallest such

[hexed briar]

Anyone know where I can find a proof of this? This is knot theory.

I know you can use this lemma to prove 2.2.7 but I’m kinda unsure how to go about it

[cerulean oriole]

If a continuous map has finite fibres is it proper? What if it has compact fibres?

[umbral panther]

A closed map with compact fibers is proper (if it is between reasonable spaces)

The inclusion of the open interval in the closed interval is an improper map with finite fibers

[cerulean oriole]

For compact fibres, according to https://stacks.math.columbia.edu/tag/005M it's true if the map is closed.

[cerulean oriole]

Sad.

Hmm, my question has to become more complicated then.

Consider a directed inverse limit of discrete topological spaces
X = lim_{<- i} Xl_i.
Suppose for some i_0, the map from X_j to X_i_0 has finite fibres for every j before i_0.
Then is the projection from X to X_i_0 proper?

(If there's a generalisation like e.g. removing the discrete assumption but replacing finite fibres with properness, that would be even cooler.)

[umbral panther]

You want a proper map to a discrete space? Sounds like a closed map to me

[cerulean oriole]

Ah wait

No

They're not discrete anymore

[umbral panther]

X_i_0 sounds discrete

[cerulean oriole]

Yes

[cerulean oriole]

I need to correct this question, which may take some time to figure out.

[umbral panther]

Being proper is local on the target. If for every point of the target there is a neighborhood such that the map from the preimage of the neighborhood to the neighborhood is proper, then the whole map is proper. If the target is discrete, then the neighborhoods are single points and properness is just compactness of the fibers

[cerulean oriole]

I see.

BTW are you requiring proper maps to be closed?

I know nothing about proper maps but Wikipedia tells me that's a convention that varies.

[umbral panther]

If you assume Hausdorff, I don’t think there should be any ambiguity in the definition of proper

[limber sandal]

I am having a slight crisis, what is the negation of X is Hausforff?

[
\exists x,y ( \forall U,V \not\subsetneq X \text{ we have } U \cap V \neq \emptyset )
]

[gentle osprey]

spikey1421

[limber sandal]

But what does $ U \notsubsetneq X $ mean ?

[gritty widget]

something definitely went wrong in your negation. in fact, your original definition was probably wrong too, since there's no mention here about U and V being neighbourhoods of x and y

the negation of a statement like "for all x in X, P(x)" is not "there exists an x not in X such that not P(x)", as you seem to have done, but rather "there is an x in X such that not P(x)"

maybe i should have said there exists instead of for all in my example

"X is not Hausdorff" means "there exist x and y in X such that, for every pair of neighbourhoods U and V of x and y (respectively), U and V are not disjoint"

[limber sandal]

Thank you, this was silly, I think I need some sleep.

[sullen nimbus]

how does this answer work here
https://math.stackexchange.com/questions/1064/perfect-set-without-rationals

what is meant here by zero dimensional

[gritty widget]

lebesgue covering dimension / topological dimension

[sullen nimbus]

how does being zero dimensional compact metric space without isolated points enables it to have disjoint homemorphic copies of C

[gritty widget]

C x C has uncountably many disjoint homeomorphic copies of C because you can take slices, and then it follows that C does as well because C and C x C are homeomorphic (by a theorem about zero dimensional blah blah blah spaces)

[sullen nimbus]

where can i find the theorem

[gritty widget]

i googled it to make sure and i found it

[sullen nimbus]

i found it too

thanks

[hollow geyser]

Does the converse proof look good?

[limpid fern]

makes sense

you mean W is open in

oh nvm yeah thst works

lol

[tribal palm]

i love the fact that this generalises to: for any equivalence relation ~ on X, X/~ is Hausdorff iff ~ is closed in X x X

[quartz horizon]

Oh I didn’t realise that

That’s nice

[tribal palm]

i must admit i have absolutely no idea how to intuitively think of "~ is closed in X x X"

[unreal stratus]

Yes, though imo the cleanest way to prove that the graph of a continuous function (into a Hausdorff space) is closed is to appeal to this proposition :)

Probably a good exercise to see how you can get that from this

[limpid fern]

~ \subseteq X \times X so it's less cursed

[formal tide]

Let A be nonempty, if A x B is homeo to A x C can we conclude that B is homeo to C? Doesn't seem to follow from the universal property of product

[gritty widget]

counterexample: ||A = R x R x R x ...., B = R, C = R^2||

[fading vale]

U can't even conclude this for manifolds up to diffeo (A x B diff to A x C doesn't imply B homeo to C)

[formal tide]

nice, ty

[gritty widget]

another counterexample: ||B a point and A and C both the cantor set||

(just because it came up in here earlier)

[umbral panther]

Great example to know:
A = R
B = S^2 minus 3 points
C = torus minus 1 point

[formal tide]

that's kinda the same since the cantor space is 2^N

[unreal stratus]

Nice

[ebon galleon]

Another silly example: take A = N, B and C any finite/countable sets. Give them all the indiscrete topology.

[haughty cedar]

idk where to ask that, but the question is ?kind of? about topology.

So, I have a distribution space $D'(\mathbb{R}^n, \mathbb{C})$ and a derivative operator:
$\partial^{\alpha}: D'(\mathbb{R}^n, \mathbb{C}) \rightarrow D'(\mathbb{R}^n, \mathbb{C})$

Bright side of mathematics in his video claims that this derivative is linear and continuous. Linearity is trivial, however im not sure how is the continuity even to be shown here.

How is the continuity defined in the space of distributions?

[gentle osprey]

MINECRAFT(rust)_LOVERBOY

[haughty cedar]

My guess is that convergence is defined as:

$T_n \rightarrow T \iff \forall \alpha \in \mathbb{N}^k lim_{n\rightarrow \infty} sup_{f \in \text{test functions}}{|\partial^{\alpha}(T_n - T)(f)|} = 0$

[gentle osprey]

MINECRAFT(rust)_LOVERBOY

[haughty cedar]

nvm got it

[wispy veldt]

convergence in D* is defined by its weak* topology , so Tn--> T if for all phi in D(Omega) we have Tn(phi)-->T(Phi) in R and the continuity of the derivative follows from how a sequence of distributions limit exists and is a distribution (by banach steinhause)

also this would have been better in #advanced-analysis for future reference

[haughty cedar]

so generally by "derivative is continuous" one would mean that
$T_n \rightarrow T \implies \forall \phi \in D: lim_{n \rightarrow \infty} |\partial^{\alpha}(T_n(\phi) - T(\phi))| = 0$

[haughty cedar]

T_n -> T is by this definition

[gentle osprey]

MINECRAFT(rust)_LOVERBOY

[gritty widget]

ssssshhhhhhhhhhhhhhh it looks cooler

[wispy veldt]

yeah pretty much, sequential continuity is equivalent to continuity in D*

[stuck geyser]

Let D be a compact convex set in R^n, and let f be a continuous function D -> D. Does the image of D’s boundary necessarily have to be a subset of the boundary of D’s image

Actually I kinda found a contradiction in my head

[steel glen]

you can just take f to be a constant map to an interior point

oh i misread, my bad

thought you were asking if f(bd D) had to be a subset of bd D

[woven sinew]

Nah, you can do $D=[-1,1]$ and $f\colon D\to D$ with $f(x) = x^2$. Then $\partial D = {-1,1}$, and $f(\partial D) = {1}$, but $f(D) = [0,1]$, and so $\partial f(D) = {0,1}$.

[gentle osprey]

OneTrackPony

[gritty widget]

can you clarify which inclusion between ∂f(D) and f(∂D) you're asking about? "image of D's boundary" and "boundary of D's image" are ambiguous

or at least confusing

i would assume you're asking whether f(∂D) has to be a subset of ∂f(D), since this reading is a bit less awkward to me

[stuck geyser]

I realized it was wrong anyway so it doesn't matter lol

[formal tide]

Why are non-(locally path connected) spaces unreasonable?

[plain raven]

A locally path connected space is the disjoint union or coproduct of its connected components. For this reason it suffices to consider the pieces "one at a time", so to speak - most results about the homotopy theory of locally path-connected spaces are obvious and easy consequences of theorems about path connected spaces, just applied to the disjoint union of a family of such spaces

A space which is not even locally path connected is hard to study. The path components of the space will not be 'components' in the usual sense, i.e., they will not be clopen.

[hollow geyser]

I spent too long on this problem. Please tell me it's good.

[hidden crag]

Using that limits are unique in Hausdorff spaces would be significantly shorter

[hollow geyser]

how do you mean?

[hidden crag]

Nvm I misread the problem

Thought it was the good ol dense set one as soon as I saw hausdorff and f=g

Actually this should work the same

[hollow geyser]

If what I wrote looks satisfactory, could you share what you mean a bit more? I don't follow

[hidden crag]

I haven’t read what you wrote I was thinking about a different solution

What you wrote looks good you’re basically doing the same just using a different characterization

What I was thinking is taking an element of the closure, using a sequence that converges to it and then using that limits in Hausdorff spaces are unique and sequential continuity

[hollow geyser]

I have not learned about sequentual continuity yet

[hollow geyser]

Wow I don't get this at all 🥲

[hollow geyser]

I don't follow on what you are saying.

[merry geode]

Closed as being in graphs

[tribal palm]

yeah ~ is a subset of X x X, which satisfies the properties of an equivalence relation

[limber umbra]

Here's a fun exercise.
Given an equivalence relation $R \subseteq X \times X$ that is a closed subset, this implies that the quotient $Y:= X/R$ is T1. Does one need extra conditions to ensure that it's also Hausdorff?

[gentle osprey]

koaaa@甜瓜乱墜

[limber umbra]

BTW does anyone know why textbook authors stick with stating the bounds like $(m-1)$- and $(n-1)$-connectness implies relative $(m+n-2)$-connectedness with the homotopy excision theorem?
I mean one could have just said $m$, $n$ and $(m+n)$.

[gentle osprey]

koaaa@甜瓜乱墜

[unreal stratus]

Oh I'm saying that like
You used the fact that graphs are closed in order to prove the diagonal is closed
But you can also go the other way and it provides a clean proof that a graph is closed

[glossy talon]

Yeah this is okay

[glossy talon]

You have probably

f X->Y is sequentially cts if a_n in Y converging to a implies f(a_n) converges to f(a)

[glossy talon]

Does this still work?

Wouldnt we want the converse?

Oh no we wouldnt

Thats pretty concise I think

I forget to think in sequences a lot

[heady magnet]

Suppose I have topological spaces $X,Y$ and an equivalence relation $\sim$ on $X$. I can define the relation $\sim'$ on $X \times Y$ by $(x,y) \sim' (x',y') \iff x \sim x' \text{ and } y=y'$. Is it then true that $$([x],y) \mapsto [x,y]$$ is a well-defined homeomorphism between $(X/\sim)\times Y$ and $(X \times Y)/\sim'$ ? It feels almost trivial but I have a lot of issue with showing that the map is continuous. My issue is, with $\pi:X \times Y \to (X/\sim) \times Y, (x,y) \mapsto ([x],y)$, showing that $\pi^{-1}(U)$ is open implies $U$ is open. This problem came up when I was trying to define a homotopy on a quotient space.

[gentle osprey]

instagram

[chilly lynx]

hey guys

[steel glen]

maybe it should suffice to show that (X x Y)/~’ is equivalent to (X/~ x Y)/id

[chilly lynx]

What do you think of this proof?

[heady magnet]

this feels very intuitive but I dont see how that would show that its a homeomorphism

[chilly lynx]

i need proof that tau is space topology

[heady magnet]

I think there has to be a step where we deal with the product topology

[steel glen]

since (X/~ x Y)/id is homeomorphic to X/~ x Y, you’d get a homeomorphism by transitivity

the idea is that (X/~ x Y)/id and (X x Y)/~’ generate the same equivalence classes (i believe)

[heady magnet]

yes they do

sort of

[steel glen]

there has to be some flattening but they are essentially equal

[heady magnet]

there is a nasty thing where they are not the quotient of the same space

[steel glen]

yeah

[heady magnet]

would that show the topologies are the same though ?

[steel glen]

let me give it some thought but it should

[heady magnet]

thank you

[steel glen]

so

(X/- x Y)/id is homeomorphic to X/~ x Y trivially

[heady magnet]

yea

[steel glen]

ok i see what you’re saying about needing to use the product at some step

[heady magnet]

yea right

its driving me crazy whahaha

im just trying to make a homotopy lol T_T

its easier to when $U$ is a product

[gentle osprey]

instagram

[heady magnet]

one idea i had is to first prove the case in hwich $\pi^{-1}(U)$ is a product

[gentle osprey]

instagram

[heady magnet]

and then use that its the topology generated by those kinds of opens or something

[steel glen]

i was thinking of showing (X/~ x Y)/~' being homeomorphic to (X x Y)/~'

you then can just make the chain
X/ x Y = (X/ x Y)/id = (X/ x Y)/~' = (X x Y)/~'

[heady magnet]

hmmm

i dont see why considering (X/ x Y)/id is helpful

[steel glen]

the idea is that
(X/ x Y)/id and (X/ x Y)/~'
are the exact same space

and
X/ x Y and (X/ x Y)/id
generate the same topology

[heady magnet]

hmmmm

the fact that its not easy to prove makes me afraid it might not be true

am i goign insnane whahhaha

[steel glen]

i don't think it is difficult to prove, i just think we are avoiding actually looking at the product and quotient topologies (unless you already tried that)

[unreal stratus]

Unfortunately this is not true in general

[steel glen]

never mind then

[unreal stratus]

The product of a quotient map with another map needn't be a quotient map

[heady magnet]

no way

alright

ok

[unreal stratus]

However, it is true if uh

one of the spaces is locally compact

[heady magnet]

kabooom

AWESOME

[unreal stratus]

So like, in practice this will be true if you're working with manifolds and things

[heady magnet]

Y is compact

ok so how do i prove

waitt

ill show you what i actually want to prove

[unreal stratus]

Now I want to find a counterexample lol

[heady magnet]

let me type out the lemma rq

Let $\tilde{A}$ be a space and let $A$ be a quotient of $\tilde{A}$ with quotient map $\pi_A$. Suppose $\tilde{F}:\tilde{A} \times [0,1] \to B$ is a map such that for all $a,a' \in \tilde{A}$ such that $\pi_A(a)=\pi_A(a')$, and for all $t \in [0,1]$, $\tilde{F}(a,t)=\tilde{F}(a',t)$. Then there exists $F:A\times [0,1] \to B$, such that $\tilde{F}(a,t)=F(\pi_A(a),t)$.

[gentle osprey]

instagram

[heady magnet]

Is this true and how do i prove it

I just want to prove it is a homotopy so baddd

If this is not true i have no idea how to create homotopies from quotient spaces

[heady magnet]

there is probably many ways to state it but the idea is: if i have a homotopy that at every time respects a equivalnce relation then i get a homotopy from the quotient

[unreal stratus]

Yes this is true precisely cause of the result I mentioned

I'll uh write C for A as I don't have tildes lol

But uh

You have your map C x I -> B

Use the universal property of quotients to get (C x I)/~ -> B

[naive trench]

If I have a continuous function such that f(D)cD the domain. Then x in D and a open ball B(x,ε) then I will have f(B(x,ε))cB(x,ε) so I want to express the image of the ball as B(f(x),smth) but Idk how to take the radius from the ε. All euclidean

[unreal stratus]

And then the theorem to turn that into A x I -> B

because I is locally compact

[heady magnet]

yea

sweet

sooo

how do i prove that theorem -_0

[unreal stratus]

||if you're worrying about homotopies then don't worry about local compactness and pointset||

Jk uh

I honestly am unsure lol, I mean

[heady magnet]

you have a reference or anything ?

i feel like this is a problem that would come up often when defining homotopies

[unreal stratus]

Well not really, because once you know you can just pass through the quotient like this you just remember that

But uh I can find a reference

[heady magnet]

thank you very very much

do you think its in hatcher ?

i think i am allowed to reference it

[unreal stratus]

https://math.stackexchange.com/questions/1933967/cartensian-product-of-quotient-maps here is a nice counterexample

[heady magnet]

sweet

[unreal stratus]

I doubt it's in Hatcher, as it is pointset topology

But uh

[heady magnet]

maybe ill just paly it cool and say it like its obvious LOL

[unreal stratus]

lol this is the paper which proves it

Uh where is it

right at the end

lemma 4

[heady magnet]

wow

thats old lol

thank you

[unreal stratus]

np

I remember worrying about the same thing cause like

People do this without justification kinda lol

But if you're just producting by the interval / a manifold / a "nice enough" CW complex then it's fine

by nice enough it's that you need the cells to be uh locally finite

so e.g. a finite cw complex is fine

and most spaces you will encounter are like this

[heady magnet]

I think im just gonna use it and then ask the proffesor about it

thank you very much

[unreal stratus]

np

[earnest epoch]

So, Heine-Borel is an equivalency

I think the necessity is the easier part

So, we let's first go for the situation that $U\subset \mathbb{R}^n$ is unbounded

[gentle osprey]

mophra

[earnest epoch]

show that it's not compact

[quartz horizon]

I think it’s easier to do it directly here

[earnest epoch]

just a subset of R^n

[quartz horizon]

I.e. show that if $K \subset \mathbb{R}^n$ is compact, then its closed and bounded

[gentle osprey]

Pseudonium

[earnest epoch]

I don't know what theory knowledge she has and this way is possible when just working with the definition

also can you try to formulate the definition of an open set in R^n

[quartz horizon]

Mhm, right..

[earnest epoch]

you can use the same definition in metric spaces, yes

can you define interior point

there is one

importantly for every element of the set

not just one

concretely for every element x there exists an epsilon>0 such that the epsilon ball centered on x is still in the set

can you come up with sets that fulfill this?

[quartz horizon]

Mhm

[earnest epoch]

that is tautological

[quartz horizon]

Oh as in

Come up with examples of open sets

Is what we want

[earnest epoch]

obviously the thing that fulfills the definition fulfills the definition

yeah, concrete examples

what kind of sets have you encountered in this context

or to make it simpler, in just R

can you name a subset of R that is open and one that is not?

think of intervals

hmmm, let's ask some questions then

is Q open in R?

Q being the rationals

why do you think that

give a reasoning

or change your answer with a reasoning

can you make one

also that is not the negation

the negation of "there exists one with this property" isn't "there exists one without this property"

let's skip this for now and get back to it after easier examples, this was more advanced

is the interval (0,1) open?

[quartz horizon]

if it is, why?

how would you show it’s open?

Why is that?

For example, why is 0.1 an interior point?

[earnest epoch]

make one

be more concrete

[quartz horizon]

Mhm, which neighbourhood?

[earnest epoch]

do you know interval notation

[quartz horizon]

So, what’s the definition of a neighbourhood?

Sort of, but that works for now

So, can you find an open interval centred at 0.1 that is contained within (0, 1)?

And if you can, please tell us what it is

Can you describe it?

We can tell you what notation fits your description

Mhm!

By the way, wanting to draw a picture is a good instinct that you should try to keep

[earnest epoch]

what exactly are you gonna do?

[quartz horizon]

Oh, um, do send us the picture..

Mhm, right

So, could you tell us where your green region starts and ends?

Like, which specific numbers would work

Right!

The notation for this is

(0.05, 0.15)

Much like how the notation (0, 1) refers to all numbers x with 0 < x < 1

Does that make sense?

That’s fine

To recap, we wanted to show that 0.1 is an interior point of (0, 1)

This means that we wanted to find some open interval centred at 0.1 that was contained within (0, 1)

We managed to find one - the interval (0.05, 0.15)

This shows that 0.1 is an interior point of (0, 1)

Make sense so far?

So lemme ask you a different point

Is 0.99 an interior point of (0, 1)?

Why?

Yep! That works

So we’ve shown that 0.99 is an interior point of (0, 1)

It does

Well, that’s what we want to check!

We’ve only checked 0.1 and 0.99 so far

And, er, it’s gonna take a long time if we keep working through specific points

An infinite amount of time, in fact

So what we need to do is the following

Given any point of (0, 1), which we’ll call x

Is x an interior point?

And why?

This doesn’t show that x is an interior point though

Remember, when we showed 0.1 was an interior point, we found an open interval centred at it contained within (0, 1)

Namely, (0.05, 0.15)

A similar story for 0.99

If you want to convince me that x is an interior point, you’ll need to do something similar

What do you mean by the “next element ahead”?

Like, what I’m expecting to see is “x is an interior point because the open interval [BLANK] is centred at x and contained within (0, 1)”

Where you need to supply the BLANK

So sure, you may have convinced me that 0.999 is an interior point

But you haven’t convinced me that x is

I want you to be concrete

Which element?

I think a good way to think about this is the following

What you should be trying to do is construct a function

This function takes in an element x

And provides me with an open interval, centred at x, contained within (0, 1)

I should be able to substitute any value of x within (0, 1), and get back a specific interval

Sure!

A function is (at your level) anything that accepts an input and produces a unique output

So for example, the output for 0.1 could be (0.05, 0.15)

The output for 0.99 could be (0.989, 0.991)

But what I want is - what’s the output for x?

This is why I’m asking you to be a little more concrete - I’d like to actually use your proof to get open intervals

Hmm

You see, I’m not so sure this works as a function

For example, if I plug in 0.7, what am I supposed to get?

But what does that notation mean?

Ah

This doesn’t make sense as a real number, unfortunately

Mhm?

This is a good idea

However, it depends on your definition of being centred at x

For example, if I plug in 0.1

I’ll get (0.05, 0.55)

I’m… not so sure that’s centred at 0.1

[earnest epoch]

that is not a thing

[quartz horizon]

So I don’t think this function quite works

[earnest epoch]

we can also just drop the centred

[quartz horizon]

If we did that, then (0,1) would work!

I think we should keep centred

[earnest epoch]

neighbourhood is even weaker than that

but maybe centred is instructional

[quartz horizon]

I think it is!

So let’s stick to it for now before potentially causing confusion

Mhm

[earnest epoch]

functions are not needed here

[quartz horizon]

Yes

[quartz horizon]

I’m not sure what else you’d suggest at this level

[earnest epoch]

I don't see how a function even helps

no

[quartz horizon]

There are sets with no maximal element that aren’t open

For example, the whole numbers

[quartz horizon]

I think it’s a good way to think about solving the problem of showing that an arbitrary point of (0, 1) is an interior point

[earnest epoch]

that doesn't involve a function tho

[quartz horizon]

What I’m suggesting is constructing a function f(x) = (a(x), b(x)) that gives an open interval centred at x contained within (0, 1)

Or equivalently a function epsilon(x) such that x +- epsilon(x) is contained within (0, 1)

[earnest epoch]

(x-epsilon,x+epsilon)

in other words

[quartz horizon]

Yes, but I’m emphasising the x dependence

[earnest epoch]

0<x-epsilon<x+epsilon<1

[quartz horizon]

Yes!

[earnest epoch]

yes

[quartz horizon]

Or at least, give us an example of an epsilon(x) that works

[earnest epoch]

it's not unique

[quartz horizon]

Mhm

[earnest epoch]

yes

construct one

that doesn't work

[quartz horizon]

What if I plug in 0.9999?

[earnest epoch]

you have to keep in mind both 0 and 1

[quartz horizon]

(I know that strictly speaking constructing a function epsilon(x) is logically stronger than you need, but I think it’s still a useful framing of the problem)

Mhm

Would you like to simplify the problem a little?

No, I don’t think so either…

E.g. if I plug in 0.9, both answers are bigger than 1

Ok

Are you familiar with $(-\infty, 1)$?

[gentle osprey]

Pseudonium

[quartz horizon]

This denotes the collection of numbers x with x < 1

Or equivalently, “-infinity < x < 1”

Make sense?

Now, what id like you to do is

Show that this is open

I.e. find an epsilon(x) for this

Uh, ideally I’d like epsilon to be positive..

So we want (x - epsilon(x), x + epsilon(x)) to be contained in (-infinity, 1)

Lmk if you want a hint

As a hint, you’ve actually already written a correct answer in a previous message

:)

I mean, i suppose you could just due process of elimination…

Though ideally I’d want you to say an answer cause it makes sense to you

So if I plug in 0.9, what interval are you suggesting?

And for a general x, what interval are you suggesting?

You’re really close - I think you’ll find your left endpoint evaluates to (x + 1)/2

So it’s just an algebra error you’ve made somewhere

But having the right endpoint at (x + 1)/2 is a good idea

As a check - your interval should take the form (x - epsilon(x), x + epsilon(x)) - so if you add the left and right endpoints, you should get 2x

I don’t think this works currently

Hmm maybe my algebra is wrong lol

Lemme check

No lol you’re right

Sorry

Your left endpoint shoild evaluate to (3x - 1)/2

And I think it does

Excellent!

So you’ve shown that (-infty, 1) is open

Congratulations, that’s your first ever open set!

A note - the answer I had in mind was actually 1 - x

This would give you the interval (x - (1 - x), x + (1 - x))

Which is (2x - 1, 1)

I think the confusion lies in the following

1 is not in (-infinity, 1), right?

But (0, 1) is a subset of (-infinity, 1)

Nope!

After all, {0} is also a subset

And that’s not open

My point was more

There’s an endpoint of (0, 1) not contained in (-infinity, 1)

The right endpoint

But yet, (0, 1) does work as a subset of (-infinity, 1)

The right endpoint of (0, 1) is 1

It’s not!

This is just common terminology I think

Anyway, the point is that, you didn’t need to take the midpoint between x and 1

You could just take 1 itself, and that would also work

(2x - 1, 1) is a subset of (-infinity, 1)

It’d simplify your algebra too

Mhm

And yet, (0, 1) still works as a subset of (-infinity, 1)

Do you see why?

Ish, yeah

The condition for being in (0, 1) is that 0 < x < 1

Whereas the condition for being in (-infinity, 1) is that x < 1

So whenever you satisfy the first, you also satisfy the second

So with these open intervals, you are allowed to push the endpoint right up to the edge, if you want

The inequalities still work out

Now you do! :)

So, we’ve shown that (-infinity, 1) is open

Your next task - show that (0, infinity) is open

This is the collection of numbers x where x > 0

I.e. positive numbers

Mhm

I want you to give me a similar interval function

So if I give you 0.1, what interval are you suggesting?

Mhm

So for a general x, what interval do you suggest?

That works!

Again, you could also choose (0, 2x)

You are :)

Mhmmmm

It’s ok, takes a little getting used to

So, you’ve shown that (0, infinity) is open

So, we haven’t

But we can use our previous results to help

Do you have a better idea of how to make an interval function for (0, 1) now?

Try it out!

The issue is that this isn’t centred at x

If it was, the endpoints would add to 2x

It is!

Lmk if you want some help

If I plug in 0.9…

Mhm!

[stuck geyser]

sup bros, what are we looking at

[quartz horizon]

Have you heard of defining a function casewise? Or piecewise?

[quartz horizon]

We are trying to show (0, 1) is open

Using the definition that open means every point inside is an interior point

[stuck geyser]

isn't that circular

[quartz horizon]

Meaning there’s an open interval centred at the point contained in (0, 1)

These are the definitions we’re using, for now

[stuck geyser]

OH

Just draw it out :)

[quartz horizon]

Yes, but we’re trying to prove it formally

Or more accurately, I’m helping hyzae to

[stuck geyser]

mmm

[quartz horizon]

Yes

[stuck geyser]

I think the notion of trying to assign a function to everything isn't a way to go

because many classical notions of a function encountered in early math are nice and "smooth", visualizable

then when you have weird, unusual functions it gets confusing at first

til you kind of abstract it down

[quartz horizon]

Mhm, right..

[stuck geyser]

This is a struggle I had along with a lot of people I've talked to

though you often can chose between two things to work with, that's a simple concept to start working with especially with epsilonics :)

[quartz horizon]

I guess, you’re welcome to help too

[stuck geyser]

I am thinking of a way to try to give a slight nudge without flat out giving away the answer

[quartz horizon]

Right, then this explains your issue

[stuck geyser]

So we are working with R, how would you describe a "ball" in R in terms of it's center and radius

Specifically B(x;r), the ball around x of radius r. What will that open interval look like?

I'm asking for a formal description of the interval:

so B(x ; r) = (a,b), what are a and b? [Hint: x is in the middle]

The ball of radius r around x

in respect to R's metric, i.e d(x,y) = |x - y|

radius

yes, but I am asking what is a and b

if the interval has center x, and radius r

i.e what is (a,b)

that's b

what's a

x-r

so, in general a metric space is a space of literally anything, where you can describe the "distance" between two things

It's not always easy to visualize, usually isn't

If we have a point, like x, we can consider the "ball" of radius r, which is really the set of points with a distance less than r from x

but it's not always shaped like a ball in 2D space, because we can have different notions of distance

So you have to start thinking about the formal definition of it, and work with that. Took me a bit to get used to

So for R with our nice metric, we can say the ball around x of radius r is the interval (x - r, x + r)

so we want an interval (x + r, x - r) that fits in open interval (0, 1)

Also note that (0,1) is a ball of radius 1/2 around the point x = 1/2

@oblique briar What is the largest radius so that that interval fits in (0, 1). You have to make a choice between two values.

for our x

what is the distance from x to 0 and x to 1

We don't know what x is, I'm asking for an expression

in this case, the distance between x and 0 is d(x,0) = |x - 0| = x, but now I ask what is the distance from x and 1

d(x,1) = ?

if d(x,y) = |x - y|

[quartz horizon]

Hyzae, do you know what d is?

Mhm

But do you know how to evaluate it?

[stuck geyser]

no

[quartz horizon]

See, I feel like maybe the metric space route is not the way…

[stuck geyser]

That's how Rudin approaches it

as far as I'm aware

[quartz horizon]

That’s not necessarily a good thing..

[stuck geyser]

Because otherwise it's trivial because (0,1) is literally the open interval containing x lmao

[quartz horizon]

No

We require centred

That’s what we’ve been doing

[stuck geyser]

so you have to describe a radius that fits in (0, 1)

so the minimum of what two values?

[quartz horizon]

Ok now you’re just giving it away..

[stuck geyser]

is what I'm trying to ask hyzae

trying to get her to describe the two values

yes

Lets say a point x is in (0,1). What is the largest radius r so that a ball of that radius around x can fit in (0,1) {Hint: it is dependent on x}

[hollow geyser]

I have actually not. It doesn't seem to be mentioned in my little book. But I'll keep it in mind.

[hollow geyser]

So you're saying that I can prove "if the diagonal is closed, then graphs of continuous functions are closed"?

[stuck geyser]

no, that would be double x,

[quartz horizon]

Could I try my approach?

[stuck geyser]

sure, what concerns me is this is basic metric space terminology.

[quartz horizon]

I don’t think we need it atm though