point-set-topology

[gritty widget]

Yeah unless I'm remembering wrong

[prime elbow]

so every point of V is path connected to v

to x

[gritty widget]

yeah

[prime elbow]

and any information about the whole space?

because i can add one point to totally disconnected set

oh so we need X which is connected and X\ {a} is totally disconnected for some a in X

[gritty widget]

Yeah

Not all such spaces fail to be locally connected, by the way. But the space I'm thinking of is quite a popular example of this, and is not locally connected.

[prime elbow]

Right now, brain is not working

But it seems interesting

[kind marlin]

this is really degenerate but ||would it work to have a space with 2 points x and y, such that only the empty set, {x}, and {x, y} are open? or more generally, a space where every point except 1 has a neighborhood of just itself, and the remaining point's only neighborhood is the entire space?||

i have no context idk if theyre supposed to be hausdorff 😵‍💫

[prime elbow]

How do I show X has an upper bound property?

[gritty widget]

That is a space with a dispersion point, yep!

[lucid ocean]

show that without it, X isn't connected

that you can find a way of "splitting X by inching along" iirc

[kind marlin]

Oh my god I remember doing that question

I probably did it badly but it was so annoying 😭

Wait omg I just realized there’s a much better way

||find a subset bounded above without a lub, and consider the union of the open sets (-inf, x) for each x in the subset, and then a separate union of the open sets (y, inf) for each upper bound y of the subset, and show that this is a nonempty disjoint cover of the space||

[prime elbow]

I am thinking one thing to show that if A is bounded above but has no least upper bound then A is clopen

[kind marlin]

The solution I originally did on my homework was to construct a new space, add lubs to the space, and show the old space is equivalent to a subspace topology on the new space LMAO it was so messy

[fallow salmon]

You seem like a really advanced topologist no?

[prime elbow]

Sorry?

[fallow salmon]

I mean you sound like you know a lot of topology

[prime elbow]

no

[kind marlin]

I feel like it’s easier to find a pair of nonempty open sets directly here tbh, but i would be interested in an alternative method if u see a direct clopen approach :0

[prime elbow]

someone give me hint that show that set of all upper bounds of A is clopen

[kind marlin]

It’s easy to see that it’s open for an A without a lub, try covering it with basis sets

you could similarly create an infinite intersection of closed sets, again using the fact that there’s no lub

[prime elbow]

yes

[kind marlin]

That’s just equivalent to covering the set of elements less than all upper bounds of A with open sets

And then you’re done

[prime elbow]

so that will be open. because say S is set of all upper bounds of A, now let x in S, since A has no least upper bound so there exists y in S such that y < x, so take (y, \infty) \subset s

right?

[kind marlin]

yup

[prime elbow]

now i have to show that set is closed

let x \in X\S. Since x is not upper bound of A so there exists a in A such that x < a, now take open set (\infty, a)

thanks @kind marlin

[kind marlin]

Is the one you were thinking of different? This one seems kind of boring 😭

[prime elbow]

Part a,

First I am showing injectivity, so if x ≠ y, that means x < y (wlog) therefore f(x) < f(y), because order preserving.

Now to show continuity, take the basis element in Y is (a,b), since f is injective so it will be (f(x), f(y)), now the inverse image of this will be (x,y)

Similarly for the basis element [a,b)

And f is injective, surjective order preserving so f^-1 should order preserving

correct?

How lines help here? infinitely uncountable lines passing through a given point of R^2, right?

[gritty widget]

||It’s the one point-compactification of the rationals||

[kind marlin]

Ohh

[kind marlin]

take two points in R^2, if we can draw a line between them directly then we’re done bc that’s an obvious path. But if the line goes through a point in A then we can’t do that. But we can try to draw a line in another direction, and eventually make a turn towards the other point such that neither line segment goes through a point in A

[prime elbow]

This space is locally connected, right?

[kind marlin]

i think so, was one of the requirements not being locally connected oops

[rapid olive]

hello, i would like some help in understanding why the last statement here is true

for context

and C(X,Y) is the set of cts functions from X to a metric space Y

it says it's true by definition of d_∞, but I don't really get it

[ruby delta]

the completeness of C(X,Y) is inherited from the completeness of Y

because for every epsilon, you can construct a new function f, where f(x) defined to be some point chosen from the epsilon-neighbourhood of f_n(x)

(completeness isn't needed for the existence of an f for every epsilon, you need the completeness to guarantee that f converges)

[rapid olive]

ok but it doesn't seem justified that the f_n uniformly converge to f

[ruby delta]

...by construction

[rapid olive]

hm

[ruby delta]

and by definition of d_oo(f,g)

[rapid olive]

ah nvm I'm seeing it now

the fact it's a cauchy sequence in C(X,Y) gives like
an eps bound on how far f can be from the f_n, uniformly

apologies

[ruby delta]

lol its good

[gritty widget]

Just learned about regular open sets(open subsets which are the interior of their closure), are they ever really useful?

[prime elbow]

product of path connected space is path connected, right? because if (a,b) and (c,d) in our product space then since we have path between a and c, b and d, then just map to its product function

[quartz horizon]

yep, though you should justify that the product function is continuous

[prime elbow]

yes i verified it

sorry i am not consistent to category

[quartz horizon]

nice

(wdym by consistent to category)

[prime elbow]

i mean i am not studying category

[quartz horizon]

oh cat theory you mean

that's alright

[prime elbow]

if A is connected space of X, how do i think int A is connected or not?

[rancid umbra]

are you asking for intuition?

[tender halo]

in pointset if there is no apparent reason that something is true there probably will be a counterexample

[prime elbow]

Yes

[rancid umbra]

intuitively, taking interiors is going to remove limit points that aren't in the interior, so it could potentially disconnect the space

whereas taking closures will add points that are close to the set

[prime elbow]

Yes

So how do I make a counterexample?

[rancid umbra]

i think an example is R^2 - (Q x Z). this space is path connected, but it's interior is R^2 - (R x Z) which is a union of infinite horizontal strips.

not really sure how to construct a large class of examples

[prime elbow]

Yes they are a union of infinite horizontal strips, but how do they make them disconnected?

[rancid umbra]

draw R^2 - (R x Z)

just think about what it looks like

[plush folio]

Isn't the union of two closed balls in R^2 a counter-example too? Both with radius 1, centered at (-1, 0) and (1, 0) respectively, so they're touching at the origin

[prime elbow]

I think so

[lament steppe]

Im a little confused by this question. Do they mean:

1.) given an x and an r, there is an r1 and r2 that satisfies the conditions (for all given x and r... so there is a new r1 and r2 for a given r and x)
2.) for all the x's and all the r's simultaneously, there is 1 r1 and 1 r2 that satisfies the conditions?

Im guessing its #2, but for a topology using metrics, I assume r can get infinitely small, so for any r1 and r2 that works for a subset of r's there is probably a smaller r such that r1 and r2 dont hold.

Hmmmmm

[austere flare]

It's #1; r1 and r2 are dependent on x and r

The quantifers are "for all... there exist..." rather than "there exist... such that for all..."

[lament steppe]

Ah interesting... that does make sense and I never noticed the difference in ordering of statements. Im guessing this is very conventional in most texts?

[austere flare]

The order of the quantifiers is actually really important (as you'll see later on when you learn about uniformly continuous functions; I'm assuming this is for a course in analysis/metric spaces). For example, the statement "for all integers m, there exists an integer n such that n > m" is true, but the statement "there exists an integer n such that for all integers m, we have n > m" is definitely not true

[lament steppe]

It's actually self study via John Lee's Introduction to Topological Manifolds... but yes. Thanks for pointing this out.

[austere flare]

Ah sorry, I must've misunderstood your question then

[prime elbow]

Union of R × (n,n+1), n over Z, right?

[prime elbow]

And if A is connected, not necessarily that bd A is connected.

Let A = R × [0,1] in R^2, so it is a path connected therefore it is connected.

Now bd A = R × {0,1}, since it is R×{0} union R×{1}. Both are open in subspace topology and empty intersection therefore bd A is not connected.

I first found that bd A is not path connected, so now I am thinking that which condition implies that if Y is not path connected then Y is not connected.

[rancid umbra]

yes

[rancid umbra]

what does this last part mean? there are connected spaces which are not path connected.

[quartz horizon]

Topologists sine curve for example

[prime elbow]

i know

i am saying which condition on connected space imply it is path connected

[rancid umbra]

locally path connected i think

maybe even something slightly weaker than that holds

[prime elbow]

I need an example where path component is not the same as connected component

[ruby delta]

that sounds about right

no just checked it's wrong, wtf

[opaque scroll]

Any space that is connected but not path connected will do

Topologists sine curve as already mentioned for example

[quartz horizon]

wait is it wrong

i'm pretty sure connected + locally path connected implies path connected

[ruby delta]

wikipedia says its wrong but I coulda sworn its a result from my point set class

yea

oh wait I think i misread the wiki page

[quartz horizon]

since connectedness is essentially a kind of topological induction principle

one way to translate local things to global things

[unreal stratus]

Ye

[unreal stratus]

Wdym by this?

Or do you mean if stuff is locally true then being connected means it holds globally

[quartz horizon]

connectedness gives you a strategy to show that a subset S of a connected space is the whole space

[unreal stratus]

Sure okay now I see what you mean by the 2nd thing

[quartz horizon]

you show it's nonempty, open and closed

nonempty is analogous to the base case

and showing it's clopen is analogous to the inductive step

if you've ever heard of "induction on the real numbers" this is essentially it

https://math.stackexchange.com/questions/2244723/generalization-of-real-induction-for-topological-spaces

[unreal stratus]

Ye noice

[quartz horizon]

compactness, too, is a kind of induction/recursion principle

[prime elbow]

Yes it is true

[prime elbow]

Yes

[ruby delta]

sounds like an instance of noetherian induction?

[quartz horizon]

What’s that?

[ruby delta]

a version of transfinite induction where the thing you are inducting has some sort of noetherian property

I don't fully understand the technical definition for what it is myself

I just know that I've done it a few times, apparently

[quartz horizon]

The most general form of induction I know is “well-founded induction”

[rancid umbra]

the other way i like to extend local results to global ones using connectedness is the chain connected characterization

[quartz horizon]

What’s that?

[quartz horizon]

Huh I see

How is this useful?

[warped shore]

Does this help with showing path-connectedness? i forgot

[slender glen]

maybe this isn't all that well-defined of a question, but are there any "tame" properties we can impose on a totally disconnected space that forces it to be discrete

other than just "being discrete" ofc

[rancid umbra]

yes. this lets you stitch together paths

and i think it allows for a good proof slightly weaker statement of local path connectedness

[warped shore]

icic yeah i vaguely remember seeing chain connectedness in some context

local path connectedness makes sense

[gritty widget]

I'm gonna attempt a proof for this, but spoiler it for others cause it seems fun to try for yourself.

Definition:A space X is chain-connected if, given any open cover U, and any two points x, y, there is a sequence of elemetns of the cover U_1, ..., U_n such that:

x in U_1

y in U_n

U_m intersect U_(m+1) is non-empty

Theorem:This is equivalent to connectedness.

~Connected -> ~Chain Connected

||Proof:Let X be a disconnected space with components U, V. Then, {U, V} is an open cover without such a chain. Hence, disconnectedness implies not chain-connected.||

~Chain Connected-> ~Connected

||Now for the other direction. Let X be some space, and suppose we have points x~=y, and an open cover with no chain. Then, let U_1 be the set of all open sets in the cover which contain x. Then, let U_(n+1) be the union of all open sets in the cover which intersect U_n. Clearly, each U_n is disjoint from any element of the open cover containing y. Let U_inf be the union of U_n for all n. Then, suppose g is some point of closure of U_inf. By definition, every neighborhood of g intersects U_inf, so does the open cover element containing it, say at some point u. u then belongs to some U_n for finite n. Assume n is minimal. Then, since the open cover's element containing g intersects U_n, the open cover element is a subset of U_(n+1) and hence g is an element of U_inf. U_inf contains all of its points of closure, and is a union of open sets, so it is clopen, and not the entire set(since it misses y), therefore X is disconnected.||

[prime elbow]

I want to prove that if Y is compact then the projection X × Y -> X is a closed map

Any hint?

First I don't understand how I take a random closed set of X × Y, how it looks like

[gritty widget]

Well, it's the finest topology such that projections are continuous, so in that sense the topology on it is the "smallest" topology such that, for a closed subset A of X, AxY is closed, and so is XxG for closed G of Y, then close it under arbitrary intersection(So AxB for closed A, B, is closed, and any such intersection of those)

[alpine nest]

Maybe first think of how you'd prove it in the specific case of something like X = R, Y = [0,1]

And then try to rework your argument so it works for a general case.

[alpine nest]

Also it shouldn't really matter how it looks like; you have some general characterizations of closed sets (complement is open, or for every convergent net of elements, the limit is also in your set)

[rancid umbra]

nit pick for not connected implies not chain connected, U and V may not be components. otherwise, it’s good. the way that i would prove this direction would be directly: if V U W is an open cover of X by non-empty sets, then there is a chain from v in V to w in W by assumption, and so V and W have non-empty intersection. since the open cover of X by V and W was arbitrary, then X is connected.

i haven’t read your proof for not chain connected implies not connected. but i think there are some flaws. it shouldn’t be this complicated. my hint would be to instead consider the set of all points reachable from x via chains in the open cover. this set is clopen.

you can modify this into a direct proof as well, by defining an equivalence relation on X after fixing an open cover of X, by saying that x ~ y iff there is a finite chain in the open covering from x to y.

[rancid umbra]

you are going to need the tube lemma iirc

[gritty widget]

Shouldn't the set I constructed be the set of all points reachable from x by a chain?

[rancid umbra]

idk. i don’t wanna read through all of that when you can just say, let C be the set of all points reachable from x via a finite chain in the open cover

i hope that’s fair at least

[gritty widget]

Yeah that's reasonable

I just didn't think of that so I tried more explicity to construct it lol

[rancid umbra]

yea, i really like this characterization of connectedness tho. big fan of it

[gritty widget]

I could've avoided the whole "contains all points of closure" by going the more simple, equivalent route of, "every point outside of it is obviously not a point of closure"

[gritty widget]

Me too! First time I've heard of it lol.

[rancid umbra]

yea, any open neighborhood of such a point from the open cover is contained in the complement

[rancid umbra]

it’s gives a lot of efficient proofs of local to global properties, for example,

• connected & locally path connected implies path connected,
• locally constant functions are constant on connected components,
• homeo(diffeo)morphisms act transitively on connected topological(differentiable) manifolds, and
• locally euc hausdorff space is second countable iff paracompact and has countably many connected components

[gritty widget]

Neat!

Is there any similar(not in the snnse of also involving chains, just like, a novel approach) definitions for compactness? I only know of like, the usual open cover one, the "every collection of closed sets with the finite intersection property has non-empty intersection", and "every ultrafilter has at least one limit"

[quartz horizon]

Interestingly this is an if and only if

Though proving that is fairly hard iirc

[gritty widget]

In general(e.g. Y is compact iff it's true for every X) or in particular(e.g. XxY->X is closed iff Y is compact, no matter what X I pick?)

[quartz horizon]

Yeah

Oh sorry

Y is compact iff it’s true for every X

[gritty widget]

Ic

That's neat

Sort of like how X is Hausdorff iff the diagonal X into XxX is closed?

[quartz horizon]

https://ncatlab.org/nlab/show/closed-projection+characterization+of+compactness

[quartz horizon]

Yeah!

[rancid umbra]

i’m not aware of one. i always took the perspective that compactness was kind of the nicer definition and that the chain connectedness characterization brought it closer to the niceness of the open cover characterization

[gritty widget]

Yeah that makes sense

[gritty widget]

I still haven't really found a case where the closed set definition is useful

The third one interestingly makes a nice definition for Compact+Hausdorff:Every ultrafilter having exactly one limit

[quartz horizon]

There are two ways I like to think about compactness

One is net-compactness

Every net has a convergent subnet

Which is a straightforward generalisation of sequential compactness for metric spaces

The second is as an induction/recursion principle

I like the latter because it shows how compactness allows you to extend “locally defined things” to “globally defined things”

[gritty widget]

Can you elaborate more on this? Sounds interesting!

[quartz horizon]

https://www.drmaciver.com/2015/03/topological-compactness-is-an-induction-principle/

I think it’s useful to draw an analogy with natural numbers

Induction for natural numbers says that to show $\forall n . P(n)$, it suffices to show $P(0)$ and $\forall n . P(n) \implies P(n + 1)$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

The compactness analog of this is the following

You’ve got some predicate on the topology of X, which maps open sets to truth values

To show $p(X)$, it suffices to show that $\forall x \in X, \exists U \in \tau, p(U)$, which is the “base case” - your predicate holds locally

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

And that $\forall U, V \in \tau, p(U) \wedge p(V) \implies p(U \cup V)$, which is the “inductive step” - your predicate is stable under finite unions

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

One example of such a predicate is “this continuous function is bounded on U”

[gritty widget]

That's very neat!

[quartz horizon]

You also have a recursion formulation

For the natural numbers, you can recursively define functions

The most general way to do this is as follows

Suppose you have a family of N-indexed sets, X_n

And you have functions f_n : X_n -> X_(n + 1), as well as a chosen element of X_0

Then there’s a uniquely specified function f : N -> U_n X_n such that f(n) is in X_n, and f(n + 1) = f_n(f(n))

Note that if you set X_n to be “set of proofs of P(n)” for some predicate P, you actually recover the induction principle!

There’s a similar formulation for compactness

Suppose you have a family of sets indexed by topology tau, S_U for U an open subset

Suppose you have a function S_U x S_V -> S_(U u V) for each pair of open sets U, V

And for every point x, you have an open set U_x containing x, as well as a chosen element of S_(U_x)

Then compactness tells you that you can obtain an element of S_X

This is done by finding a finite subcover of X, and combining the elements of S_(U_i) for each U_i in the finite subcover using the “pairing” functions

The moral of the story is - if you want to define something globally, compactness says you only need to define it “locally”, and ensure you can define it on finite unions

Which is a “base case” and a “recursive step”

E.g. continuous functions on a compact metric space being uniformly continuous makes sense to me this way

Because of continuity you can “locally” define a delta, and then make this work over finite unions

[prime elbow]

Yes, I got it.
Thanks

Let F be closed in X × Y.
Claim: π(F) is closed.

Now let x in X\π(F) implies for any y in Y, (x,y) not in F.

Since F is closed so we get the basis element U_y × V_y such that x in U_y and y in V_y and U_y × V_y \subset F^c.

So for all y, we get U_y × V_y, take the union of such open sets, it will cover {x} × Y.

Now use the compactness of Y, we will get a finite subcover, say Y \subset union V_i, i = 1, 2, ...,n.

Now take U = \cap U_i, i = 1,2,..,n.

Now x in U, and take any z in U, so z in U_i for all i = 1,2,..,n.

So take any y in Y, it will be in some V_i, say V_j, so z also in U_j implies (z,y) not in F. It holds for any y, i.e., z \in X \π(F).

Hence, X\π(F) is open.

[prime elbow]

One direction complete

Now I have to show if it holds for every X, then Y is compact

[rancid umbra]

this is surprisingly more difficult than you would expect

i would suggest reading through the nlab page after maybe thinking about it for a bit

[prime elbow]

I see

[gritty widget]

Can two regular closed sets share a boundary?

[kind marlin]

[0, 1] and [1, 2] in the [0, 2] subspace topology?

[gritty widget]

darn that was simple

dangit

[kind marlin]

simple is good i think :D

[dawn frigate]

To prove that the interior of closure of A is not contained in closure of interior of A is A equals Q a good counterexample

[rancid umbra]

where are you confused?

[dawn frigate]

Whats the interior of Q

Is it empty set

I was thinking it was empty set but I couldn't justify why

So for closure of Q it is R since Q is dense in R right?

[ruby delta]

it is sufficient to prove that no open interval is contained in Q

[dawn frigate]

Thanks

[umbral hamlet]

Had a quiz today about quotient topologies and I wanted to make sure I understand

Let X be a topological space and R be an equivalence relation on X. Let Y be the set of equivalence classes of R and f: X -> Y be the standard projection.

1. If the quotient topology on Y is the discrete topology, is the topology on X discrete?

2. Same but indiscrete

I want to start with 2 I think since understand that better

Even if Y has the discrete topology, the preimage of Y isnt necessarily just X right?

[quartz horizon]

Both are false I think

[umbral hamlet]

Yeah thats what I said

[quartz horizon]

What if Y is a singleton lol

[umbral hamlet]

Wait would it matter if it isnt?

[quartz horizon]

Oh idk I was just looking for the stupidest example

[umbral hamlet]

Fair enough

[quartz horizon]

The only topology on a singleton is both discrete and indiscrete

Doesn’t matter what X is

[umbral hamlet]

I mean the equivalence could be really stupid right

[umbral hamlet]

Ah good point

[quartz horizon]

Yeah in this case equivalence is “everything relates to everything else”

[umbral hamlet]

But like let's say the relation just sorts X into one thing like u said

Then X can have any topology at all

[quartz horizon]

yeah

[umbral hamlet]

Alright cool

I got an 87% on the small quiz but im pretty sure one of the questions was wrong

Smth about comparability of box vs product topology in finite vs infinite factors

[quartz horizon]

It’s always fun spotting typos

Ooh ok

[umbral hamlet]

Box is finer than product for infinite but equal for finite

[quartz horizon]

Yep

[umbral hamlet]

I got those two right but the third question was about when box = prod in finite

And I said that was false

I only got 2/3 on that section tho so gonna have to talk to the prof

[quartz horizon]

Wait what

[umbral hamlet]

Oops misspoke

I meant infinite

[quartz horizon]

Oh

[umbral hamlet]

Box is never equal to prod for infinite factors

[quartz horizon]

Well idk if that’s true

What about a product of indiscrete spaces

[umbral hamlet]

Ah wait I should clarify

One of then was strictly finer the other was equal

So only one or the other can be true

[quartz horizon]

Hm…

[umbral hamlet]

But I still got 2/3 somehow despite answering logically

[quartz horizon]

But box and product are equal for indiscrete spaces I think

[umbral hamlet]

I'll keep that in mind then, I thought wrong

[quartz horizon]

Afaict a product of indiscrete spaces is indiscrete

In either topology

[gritty widget]

Consider X=(0, 1)U(1, 2) and the relation R such that the equivalence classes are the connected components

[quick delta]

Yeah it should be
Because your only non-empty open is the whole space
So every open in the basis of the {box, product} topology is the whole space

[gritty widget]

Fun motivation for the seemingly odd definition of product space:When I have a product space, say AxBxCx…, I want the preimages of open sets under projection to be open. But taking finite intersections of these gives us the all-but-finitely-many-are-the-whole-space open sets as a basis, but not necessarily infinite products of open sets.

[quartz horizon]

I like the universal property def

[gritty widget]

If the product of k many spaces satisfies the property that an intersection of k many open sets is still open, then it should agree with the box topology(I think?)

For example if the product is Alexandrov(https://topology.pi-base.org/properties/P000090), then it agrees with the box topology

Not sure in what cases a product would be Alexandrov though

[stuck geyser]

I am being kinda dumb right now

I can’t figure out how to show sigma(S/~) is closed via a preimage argument

Do I need to use Hausdorffness (and probably the diagonal being closed in the product topology) somehow for this one

[gritty widget]

Wait isn’t this kinda easy lol

I’ll spoiler it

[stuck geyser]

Just evading me right now usually I’m pretty good at point set

[gritty widget]

I was too naive nvm

Lmao

[stuck geyser]

I wonder if I can do something on sigma(S/~) x S/~

[gritty widget]

Let X be a topological space

We define a filter F on X to converge to a point x iff every neighborhood of x is an element of F

We define a filterbase G of F to be a subset of F such that the upwards closure of G is F

Let x be a point in X, and A a subset of X.

Then, x is a point of closure of A iff there is a filter F which converges to x, and a filterbase G of F such that every element of G is a subset of A.

Let x be a point of closure of A. Then, for all neighborhoods U of x, define the filterbase by letting V_U be the set UnA. The upwards closure of this then converges to x, because V_U is a subset of U for all neighborhoods U of x.

Conversely, let x be a point of A such that there is a filter F converging to x which has a filterbase of elements contained in A

Then, since there is no open set of x disjoint from A(trivially), x must be a point of closure of A.

Are there any like, minor flaws in this proof of equivalence of definitions of points of closure?

[prime elbow]

i don't get it, how it is true? say A and B is sub-basis element, how do i show f(A intersection B) is open?

[quick delta]

I believe that's false
Consider the map from the discrete space on {1, 2, 3, 4} to the indiscrete space on {0, 1} sending i to i mod 2
The discrete space on {1, 2, 3, 4} has subbasis {{1, 2}, {2, 3}, {3, 4}, {4, 1}}
Each of these map to {0, 1}, hence are open. The map is continuous as the codomain is indiscrete
But the open set {1} maps to the non-open set {1}

[prime elbow]

i see

thanks

[quick delta]

(0, 0, 1) + t(P - (0, 0, 1))

[prime elbow]

got it, thanks

[dawn frigate]

unions of closures is contained in closures of unions but otherway around doesnt work for infinite unions. And to prove unions of closures is contained in closure of unions since Ai is contained in unions of Ai union of closure of Ai is contained in closure of union of Ai right?

Right?

[zinc siren]

Yeah this is how I remember the definitions of the topologies on the quotient space, product space, disjoint union/coproduct,

[rancid umbra]

is this true?

take S = R^2 - 0 and x ~ y iff |x| = |y|.

S/~ = (0,oo). put sigma(t) = (t,0).
then pi(sigma(t)) = pi(t,0) = t, but the image of sigma is (0,oo) x {0} which is not closed in S

[stuck geyser]

Dude idfk

No errata, I spent like 3 hours on it

[rancid umbra]

mind checking this over?

[rancid umbra]

same lol

[stuck geyser]

https://cdn.discordapp.com/attachments/1284539400453361766/1399970271657721887/IMG_6111.gif

[stuck geyser]

Seems fine

[rancid umbra]

i’m at a barnes and nobles rn so i can’t do much more. maybe some other ppl can look at it too

[plush folio]

Maybe I'm missing something, but isn't (0, oo) x {0} closed in S since [0, oo) x {0} is closed in R^2?

[rancid umbra]

yea

alr

i was trying to think of examples

missed that detail

[dawn frigate]

Does this proof look right

[rancid umbra]

was not fine

[stuck geyser]

I’m making food lol

[stuck geyser]

Damn :c

Meh point set is a figure of our imagination anyway

Just take your meds and we won’t have to care

[dawn frigate]

4. Let R2 have the order topology induced by the dictionary order on R2. Let S1 ⊂ R2 be the standard unit circle. What is the better known name for the subspace topology induced on S1? use strips aka a1<x<a2 but make intervals small enough Prove your answer.

Idk what to do here

[ruby delta]

take a standard open set in R2, intersect it with S1

see what the result looks like

[dawn frigate]

Order relation where (x1,y1) less than (x2,y2) if x1<x2 and x1=x2 and y1<y2

So (axb) less than (c cross D) if a less than b and a equals c and b less than d

If -1<x<1 i get two intersections

[ruby delta]

hmmm okay think about it like this

the open sets induced by the dictionary topology looks like rectangles in R^2

where some extra stuff is happening on the boundary

think about that image

[dawn frigate]

Intersection is an arc or part of it

Am I getting anywhere

[ruby delta]

yes, getting close

so what does the arc look like

[dawn frigate]

Um like {a1<x<a2: x and y is rad(1-x^2)

Plus or minus the rad

[iron bolt]

this doesn't seem right to me

[ruby delta]

not all open sets, I was talking about the basis

should have clarified

[rapid escarp]

I have a general topology question:
So consider $(X,\tau_x)$, $(Y,\tau_y)$ as top spaces, and consider $(\tau_x,\subset)$ and $(\tau_y,\subset)$ as ordered sets. Question: does an order preserving map from $(\tau_y,\subset)$ to $(\tau_x,\subset)$ correspond to a continuous map from $(X,\tau_x)$ to $(Y,\tau_y)$? If not, how about Oder isomorphisms?

[gentle osprey]

mandaly

[rapid escarp]

The converse correspondence is true which is where this question came from.

[ruby delta]

the issue I'm seeing that that the two things are fundamentally defined on different base objects

if you give me an arbitrary order preserving map, unless the topologies are very fine there are going to be multiple ways to extend it into a map on the base sets

[rapid escarp]

Yeah not looking for uniqueness

Just existence

Almost feels more like a set theory question tbh

[ruby delta]

so what's the map, then? how are we going to recover the map on the base set?

because if you want to make the map such that it induces the order preserving map through taking preimages then it is necessarily continuous by construction

[rapid escarp]

Yeah so the problem is, does such a set map exist

I don’t know how I would go about this problem

It does more and more feel like a set theory problem

If it isn’t true the follow up question would be, when does there exist a corresponding set map to the order preserving maps.

[iron bolt]

yeah, but the basis consists of the open intervals under the dictionary order, and an open interval under that order is not a rectangle

it's more like "product of an interval with R, with things going on at the boundary"

but really what ends up happening is that a set is open iff every vertical slice of it is

[dawn frigate]

Is this def of closure x in A closure iff for all open sets U contained in X such that x is in U implies U intersect A is nonempty

[unreal stratus]

There are many different equivalent definitions of closure. This seems to have a typo it (should be a comma instead of "implies" i guess) but otherwise equivalent

[dawn frigate]

Ok

[dawn frigate]

Thanks

I'm having trouble proving that pi1:X cross Y to X and pi2:X cross Y to Y are open maps

[iron bolt]

hint: you know a basis for the topology on X ⨯ Y, right? each element of that basis gets mapped to an open set - can you see why that is enough for the map to be open?

[dawn frigate]

The basis for XxY is {U cross V| U contained in X and V contained in Y}

So pi1(u cross v) is U

[iron bolt]

yep, almost - U and V are not just any subsets of X and Y but open ones

so π₁(U ⨯ V) = U is by definition open

the remaining question is just, how does it follow from that that π₁ is open? and well, you can actually show more generally that any map f : X → Y that sends all elements of a basis on X to open sets in Y is open

[dawn frigate]

Ok…was I close though

By using summations

So let W be union of basis elements?

Am I getting anywhere

[iron bolt]

yep. you want to show that an arbitrary open subset of X ⨯ Y gets send to an open subset of X, and an arbitrary open subset can indeed be written as a union of basis elements

[dawn frigate]

Union of basis elements in product topology

Thanks

[dawn frigate]

For proof that closure of A minus closure of B contained in closure of A-B. I started by let x is in closure of A minus closure of B. By def of closure for all nonempty open set U U intersect A is nonempty and exists nonempty open set such that U intersect B is empty set…I need to show that for all nonempty set U U intersect A minus B is nonempty

[kind marlin]

suppose a point x is in the closure of A, and not in the closure of B. then every neighborhood of x contains at least 1 point in A, and at least 1 neighborhood N of x does not contain any points in B.

consider an arbitrary neighborhood U of x, and consider the intersection I = U \cap N. This is still a neighborhood of x since we took a finite intersection of neighborhoods of x, and I is now a subset of both U and N.

this should help you finish the proof

[dawn frigate]

Thanks

[rancid umbra]

what book is this? also, does Hausdorff here mean compact + Hausdorff?

if so, then it follows quickly because compact subsets of Hausdorff spaces are closed, and S/~ would be compact if S is compact Hausdorff

[stuck geyser]

Compactness is introduced in a later section

[rancid umbra]

dude

[stuck geyser]

The book is Abraham Marsden Ratiu

[rancid umbra]

this has been in the back of my mind for the past couple of days

[stuck geyser]

I skipped it and now I’m working on this one

[rancid umbra]

pretty frustrating ngl

[queen prism]

🤔 when does hausdorff mean hausdorff + compact

isn't it usually compact that means hausdorff + compact

[rancid umbra]

ah, right

i am hallucinating

[stuck geyser]

I am shit out of luck on this one too

[queen prism]

ignore all previous instructions?

[stuck geyser]

I can’t for the life of me figure out how to use compactness

To show (i) => (ii)

It would be trivial if S is Hausdorff but it isn’t

[rancid umbra]

does closed eq rel mean that every eq class is closed?

[stuck geyser]

Projection map is closed

Trying to think if I can open cover S or S/~ somehow in a useful way to find an open neighborhood of any point outside of the projection of some closed set

@rancid umbra this is intended to be a review of point set but like I am actually shit out of luck

I am supposed to take a diff manifolds course next sem I am boned

[rancid umbra]

nah, this is more difficult than you will see in diff manifolds

usually lol

[stuck geyser]

The book involves Banach manifolds

[rancid umbra]

o

[stuck geyser]

But yeah I am going to ponder this problem

wait

Closed map lemma

[stuck geyser]

For some stupid reason I thought you can only show closed subsets of compact spaces are compact for Hausdorff spaces

It is in fact the other way around

because the complement of that closed subset can be added to the open covering of the closed subset to find a finite subcover :p

[rancid umbra]

compact subset of hausdorff is closed, closed subset of compact is compact

i think i figured the other one out

[stuck geyser]

Took me a second

[rancid umbra]

so sigma \circ pi is a retract retraction, to use some terminology from alg top

and i think i remember from hatcher

[stuck geyser]

it’s a section

[rancid umbra]

sigma is a section

but

if we put r = sigma \circ pi, then r^2 = r

[rancid umbra]

in here, i think he shows that any retract of a hausdorff space is closed

something silly we are missing

i can’t remember the proof

lemme look rq

okay, not sure if hatcher proves this actually

but uh

its not hard to show

consider the map x |-> (x,rx) and look at the preimage of the diagonal

in words, sigma(S/~) is the set of fixed points of sigma \circ pi

[stuck geyser]

So uhm

How is sigma \circ pi a retract

What’s it a retract of

[rancid umbra]

do you know what a retract is?

-# just asking, not trying to be snarky

[stuck geyser]

Like the retract of a given map

Might be brain dead rn, it’s 2 am

I should go back to it tomorrow :p, thank you for the help

Well technically later

[gritty widget]

Pretty sure they mean a retract of the inclusion

[stuck geyser]

OHHH

[gritty widget]

yeah, took me a sec too lmao

[rancid umbra]

sorry

but yea

like, sigma \circ pi is a retraction, and sigma(S/~) is a retract of S

really, sigma \circ pi corestricted to sigma(S/~) is the retraction

but meh

anyways, that should do it

[gritty widget]

This, completely unrelated btw, gives us an embedding of fundamental groups $\pi_1(S, s)\to\pi_1(\sigma(S/~,), s)$ too right? Since the inverse is just the induced homomorphism from inclusion

I always forget the direction so it might be the other way around idk

[gentle osprey]

Eclipso

[gritty widget]

Oh no it’s the other way around isn’t it

[rancid umbra]

this follows from the more general fact that equalizers of maps into hausdorff spaces are closed subspaces

[rancid umbra]

so yea, the fundamental group of image of sigma embeds into the fundamental group of S (with a chosen base point)

[gritty widget]

Alright thanks

[trim saffron]

What is topology about

[rancid umbra]

the study of topological spaces and continuous maps between them

[trim saffron]

What entails those things I want to take topology next year but wanna learn what’s about more

[hollow geyser]

Have you studied calculus or real analysis yet? Are you familiar with continuity or limits?

[trim saffron]

Calculus

[stuck geyser]

Topology is basically the de facto way to take structures and “tie them together” into larger ones

[balmy nexus]

There is a useful function that takes the segment [0,1) to the circle (which we will call S). It is even bijective and continuous.

You might know it by the name e^ix, or the simpler function of "starting from some point in the circle and walking along till you reach the same point".

This is a very useful function and has caused much rejoicing in many different fields.

Annoyingly, the inverse of this function is not continuous. You walk along the circle, starting from some point x, approaching x again - but on the segment you start from 0, approaching 1.
As far as our current knowledge suggest, 0 != 1, so that is a problem for continuity.

Further study will show you that you cannot make a bijective continuous function from the circle to the segment [0,1).

Why is that so? Can we relate how the circle and line are different, to the continuous functions between them? Topology will answer this question, and more. It's a fundamental subject and touches most areas of math.

[trim saffron]

Wow

[lucid ocean]

e^(2pi i x)

[random mirage]

ya and u study holes

[gritty widget]

Since a space is compact iff all ultrafilters converge, is there a method of compactifying a space by "making" all ultrafilters converge?

[quartz horizon]

Sounds like stone-cech

[gritty widget]

Oh apparently when X is discrete that is the stone-cech compactification lmao

[quartz horizon]

Yeah, the associated monad is the ultrafilter monad

[gritty widget]

what's a monad?

[random sinew]

lmaoo

[gritty widget]

This also makes me wonder about a related question. Say you have Q, there are lots of filters that ""should be"" neighborhood filters, e.g. the neighborhood filter of e, intersected element-wise with Q. Is there any like, way of making this formal? E.g. a definition for "should be a neighborhood filter", and maybe some way of adding filters(up to some notion of equivalence) to perhaps add some irrational number to Q with the same topology as if you just took the subspace of R, in some natural way?

There are ways if you have say a metric but is there a similar way with only a topology?

[queen prism]

perfection

[iron kite]

This sounds a little bit like the construction of R by Dedekind cuts on Q? Although I suppose that this one doesn’t involve filters.

[tender halo]

you can talk about ultrafilters on a dense subspace which have a nice property of having at most one limit in the whole space

it is not really clear what a neighborhood filter is without a metric or related structure on Q, Q is very very disconnected

i think you the thing that is the closest to what you thinking of is uniformities and Cauchy filters

[gritty widget]

Ic

[pulsar lagoon]

Stupid question, but a continuous group action is just one so that the map (g,x) -> g.x is continuous right?

[quick delta]

Yeah, cts as a map G x X -> X

[pulsar lagoon]

bet

thank you

[dawn frigate]

If tau is topology generated by subbasis for product topology on X cross Y then why are all elements of script S where script S={pi1^-1(u)| u in tau x} union { pi2^-1(v)| v in tau y} open in product topology.

[rancid umbra]

do you know what a subbasis is?

[dawn frigate]

Subbasis for topology is collection of Script S in power set of X such that union equals X

[rancid umbra]

the product topology tau is constructed to be the smallest topology containing Script S, and so every element of Script S is open in X x Y. the open sets of X x Y (that is, the elements of tau) are unions of finite intersections of elements in Script S.

[dawn frigate]

So how is it the smallest topology containing script S is it due to subbasis condition where it is arbitrary unions of arbitary intersections…is that why every elemtm of Script S is open in X cross Y

So every S in script S is in topology so every script S is open right

So my question is why is product topology tau constructed to be smallest topology containing script S

[ruby delta]

the short answer is that this definition fits the idea of what the product of two "things" should look like

[quartz horizon]

To an extent the universal property made this clear for me

[gritty widget]

Yeah

Specifically, the topology on the product topology is the one with the least open sets, such that the projection maps XxY to X and XxY to Y are continuous, so the only open sets in XxY are the ones required, e.g. preimages of open sets in X and Y, and their finite intersections, and their arbitrary unions

[quartz horizon]

Mhm - and it’s constructed so that continuous functions A -> X x Y correspond to pairs of continuous functions A -> X, A -> Y

So it lets you package together a pair of continuous functions, and also unpackage it

In a sense your observation is: open subsets of X x Y correspond to natural transformations Hom(-, X) x Hom(-, Y) -> Op(-)

How do people typically like to motivate the topological definition of continuity?

[novel acorn]

the derived category of opers!!!

[crisp lintel]

Start from real or metric continuity and observe that the notion depends only on the open sets and not the actual distance

alternatively, you can remark that no matter how small a neighborhood we choose around a point in the image of a point f(x), there is always some open neighborhood of x mapping into it

which is basically just the definition restated pointwise

[radiant stone]

i usually start with functions f;R\to R or f;R^2\to R with some observations

[gritty widget]

Imo “given x, A, if x is a point of closure of A, f(x) is a point of closure of f[A] too” is motivation enough, you just have to prove it’s equivalent

Basically seems like the most intuitive notion of continuity

[quartz horizon]

Mhm mhm, right

[gritty widget]

If x is “infinitely close” to a subset A, then f(x) is to f[A] too, e.g. it preserved “infinitesimal” distances.

[quartz horizon]

I guess you can phrase this in terms of filters kinda

[quartz horizon]

I think my issue is that I don’t really have good intuition for infinitesimals

[crisp lintel]

You don't need to work with infinitesimals really

You can actually reformulate the axioms of topology in terms of a "touches" relation between points and sets I believe

we say x is close to A if x is in the closure of A

then f is cts if and only if whenever x is close A, f(x) is close to f(A)

[gritty widget]

Yeah a set with a Kuratowski closure operator inherits a topology

[quartz horizon]

So I’ve heard

[tender halo]

i like the nbhd base definition the most personally

that for nbhd of the image A of a point x you can find a nbhd of the x whose image fits inside A

[quick crane]

metric spaces

[bright acorn]

If $X$ is a compact topological space and $S \subseteq X$ is a sequentially closed subset of $X$, does that imply $S$ is compact as well?

[gentle osprey]

MisterSystem

[bright acorn]

I think not, but need a counterexample

[rancid umbra]

is [0,\omega_1] compact?

i think [0,\omega_1) works if it is

[kind marlin]

does sequentially closed mean the same thing as sequentially compact

oh i found the wiki page

[tender halo]

every successor ordinal is compact yea

[rancid umbra]

\omega_1 isn’t a successor ordinal tho

oh

my bad

lol

\omega_1 + 1

[unreal stratus]

Lol

[rancid umbra]

-# i always forget the order of addition for ordinals

[umbral hamlet]

Start with the transfinite

[umbral hamlet]

How can I show that the taxicab metric on Rn induces the same topology as the standard one?

Do I just show that theres an taxicab open ball inside and outside of a standard open ball?

[iron kite]

More precisely, you need to show that for every p in some standard open ball U, U contains a taxicab open ball centered at p; and that for every p in some taxicab open ball V, V contains a standard open ball centered at q.

This is automatic if you can show that dist_taxicab(x,y) \leq C dist_std(x,y) \leq D dist_taxicab(x,y) for all x,y, where C, D are positive constants.

(Proof: exercise!)

[umbral hamlet]

Im not sure munkres proves this

Maybe I have to myself

Especially considering that the inequality you gave is smth proved in a prerequisite for my topology course

[iron kite]

I didn’t learn topology from Munkres, so I can’t know about what he does. How does he solve questions of this kind?

[umbral hamlet]

Ah nvm

It actually just follows from a lemma about bases

We know that the set of all metric epsilon balls is a basis

So what you said would essentially just prove that both topologies are finer than each other

Meaning theyre equal

[queen prism]

what on earth is this and which part of it has to do with topology

[rain gyro]

You’re right😢 but there doesn’t seem a set channel and topo is very set theoretic 😭

[queen prism]

you can try asking in #proofs-and-logic

idk

[rain gyro]

Thanks for guide! 🥰 Set theory, set arithmetics, my course has an emphasis on set arithmetics and this is too hard. so basically it’s arithmetics but too hard

[umbral hamlet]

For a moment I thought u were talking about my question and I was very confused lol

[queen prism]

topological taxi

[prime elbow]

I want to check Q and { x in Q | x > 0 } are homeomorphic or not, with subspace Topology with respect to standard Topology on R.

How do I start this? Intuitively i think they are homeomorphic.

Also I am looking for an invariant property which holds in one but not in second one.

[tender halo]

easiest solution is

they are both countable metric spaces without isolated points

so both are isomorphic to Q

easier solution is both are order-isomorphic to Q (being countable dense unbounded linear orders) and their topologies are generated by their orders so they are homeomorphic

the latter is the actual solution probably

its hard (read: annoying) to make a constructive homeomorphism between the two without gesturing towards the back-and-forth argument

[prime elbow]

So is it the result?

[rancid umbra]

think this works

it restricts to a homeo Q -> (0,oo) \cap Q

clearly injective since it’s increasing.

for p/q > 0,
if p/q < 1, put x = 1 - q/p < 0
if p/q > 1, put x = p/q - 1 > 0

so this map is surjective

[tender halo]

huh i guess not that hard

[rancid umbra]

yea

my first try was e^x

but that doesn’t work for obvious reasons

and rational functions won’t work because they have roots and poles

so the next best thing is to stitch together rational functions before their vertical asymptotes

[umbral hamlet]

To show that the proposition that a quotient map from a T2 space is T2 is false, it would suffice to find a space in which two points in the same open set which get mapped to different equivalence classes, correct?

[cerulean oriole]

It would suffice to find a top space X and an equivalence relation ~ such that X is T2 but X/~ (with the quotient topology) is not.

So you would want x, y mapped to different equivalence classes but such that you cannot find any disjoint open sets containing them which are closed under ~.

[umbral hamlet]

The classes are closed under ~?

[cerulean oriole]

No, the open sets.

[umbral hamlet]

Im not sure what that means then

Is it just that all elements of the set get mapped to the same class?

[cerulean oriole]

You want to find X and ~ and x, y such that (i) X is T2 (ii) x ≁ y (iii) you cannot find open sets U, V in X such that (a) x in U, y in V (b) U, V are disjoint (c) U, V are closed under ~ in the sense that u in U, u ~ v ⇒ v in U and similarly for V.

(The reason is that open sets W of X/~ are precisely those subsets whose preimage in X is open, and the subsets U of X which are equal to the preimage of some subset W of X/~ are precisely those U which are "closed under ~".)

[cerulean zealot]

in this theorem, \beta(X) is the stone-cech compatification of X
i have a few questions, 1. regarding a proof and 2. regarding the uniqueness in the statement

1. could one proof be to notice that \beta(X) is a normal space, and therefore apply tietze's extension theorem with the closed set being X embedded into \beta(X) and the map being f? this was not given in the book so i'm wondering if this works
2. i don't understand why there exists a unique extension. for example, since \beta(X) is normal, it is completely regular, so take some point x not in X and define a map g : \beta(X) -> [0, 1] that sends X to 0 and x to 1. Then, the map g + f agrees with f on X but it is clearly a different extension. Where did I go wrong?

[tender halo]

X is not closed in \beta X so you cannot define g

in fact, quite the opposite, X is dense in \beta X, so its very not-closed

denseness is what ensures in a unique extension, that is, if you have a map f: A -> Y and A is a dense subset of X then there is at most one extension of f to a map X -> Y

(if Y is hausdorff)

that is due to the fact that if f and g are two maps from X to Y where Y is hausdorff then the set {x | f(x) = g(x)} is closed in X

[cerulean zealot]

ah i see

but why is X not closed in \beta(X)? I thought there exists an embedding \Phi: X -> \beta(X) so wouldnt \Phi(X) be closed?

oh oops.

well i get it now

[umbral hamlet]

im wondering if i phrased what i was thinking correctly

This is a proof that if an element is in the closure of A then there exists a net converging to that element

[cerulean oriole]

It's correct except for the last half a sentence. If you want to show that (x_a) converges to x, you need to show that for all open sets U containing x, there exists an a in S such that for all b in S such that b ≥ a, x_b lies in U. Informally, that U contains eventually all terms of (x_a), rather than just some term of it.

[umbral hamlet]

oh right

luckily that should be okay because of the fact that i chose a directed set

so N_a+1 is in N_a by how i defined the directed set

of course the way i worded it is still wrong but its an easy fix is what i mean

[cerulean oriole]

? I mean, it is OK, but not because your index set is directed - that has to be true to define the notion of a net converging in the first place.

One does need to actually sit down and show the strengthened claim at some point for the proof to be complete.

[umbral hamlet]

yes

sorry i didnt mean that the proof is okay as is

[cerulean oriole]

(Just mentioning - there is no a+1.)

[umbral hamlet]

i shouldve instead said that the fix should be easy-ish because it follows from the fact that the index set im using is directed

[umbral hamlet]

oh, i thought that such a thing was okay because of how ordinal arithmetic works

though i guess maybe A doesnt necessarily have to be an ordinal

likely im just not wrapping my mind around these definitions enough

[cerulean oriole]

It's very valid to wrap your mind around a definition by using it.

[umbral hamlet]

would the better way to say this to be that i can just pick two different open sets in my set of all open sets containing x and ill find an open set thats contained in both?

i guess that makes sense because the reverse inclusion ordering would be bounded above by the empty set

[cerulean oriole]

That is directedness, yes. (And their intersection is such an open set.)

[umbral hamlet]

im having a serious amount of trouble answering part b

[crisp lintel]

Come up with a candidate for the limit of each sequence, then analyze the open neighborhoods of that candidate wrt each topology

[rancid umbra]

what is the uniform topology

[umbral hamlet]

i forgot to mention that we dont need to consider the uniform topology

only product and box

it was obvious that they all converge to 0 in product

but idk about box

0 is the only candidate for all of them it seems

[crisp lintel]

Yeah so then you just want to identify the open neighborhoods of 0 in the box topology

and apply the definition of sequence convergence

[umbral hamlet]

I guess i can show x converges to 0 by considering products of open balls of radius 1/n

Well no but that doesn't cause my open set to be arbitrary

[rancid umbra]

maybe consider something more general:

if X is a top space and T and T' are two topologies on X with T' finer than T, then if a sequence converges in (X,T), it must also converge in (X,T')

[umbral hamlet]

Im just not sure how im supposed to show that my specific sequence elements are inside an open set i know very little about

[umbral hamlet]

Oh wait does this just automatically do it

Box is finer than prod right

[rancid umbra]

for some of them

yes

[crisp lintel]

This is the wrong way around no?

[umbral hamlet]

Thats what I was thinking

[crisp lintel]

stronger topology = more difficult convergence

[umbral hamlet]

Either way its still pretty trivial that all of these sequences converge in prod

[rancid umbra]

T' has all the same open sets as T, with potentially more

[crisp lintel]

More open sets = less convergent sequences

[rancid umbra]

oh yes

my fault

had it backwards

for some reason i was thinking only about one open set

[crisp lintel]

I always forget which way things go

[rancid umbra]

real

[crisp lintel]

although I just think about the indiscrete vs discrete usually and it clears things up

[umbral hamlet]

Im still pretty lost

I mean i guess if I have an open set containing 0

[rancid umbra]

i just wasn't thinking lol

[umbral hamlet]

Then a subset of that set is some epsilon ball centered at 0

[crisp lintel]

well 0 here is the 0 sequence

[umbral hamlet]

Yes im just saying it simply

[rancid umbra]

which sequence are you considering?

for the box topology

[umbral hamlet]

All of em

Since im pretty sure all of em except for w are easy to consider

[crisp lintel]

I would consider infinite products of small open balls

[rancid umbra]

are you just asking what the open sets in the box topoogy look like?

[crisp lintel]

Since those differentiate the box topology from the product topology

[umbral hamlet]

Nah its prods of opens

[crisp lintel]

for example the first sequence would not converge in box

[umbral hamlet]

But arent I supposed to show that it works for any open set

[crisp lintel]

yea

[rancid umbra]

okay, and you actually know a basis for the box topology too.

[crisp lintel]

But to show that it fails to converge

[umbral hamlet]

Ah good point

[crisp lintel]

its enough to find a single one

[umbral hamlet]

Im somewhat convinced that at least two of these sequences converge in box

I should be more specific and say that im struggling to confirm convergence

[rancid umbra]

yea, so if you take some open set U containing 0, you can find a collection of open intervals (an,bn) containing 0 such that the product of all of them is contained in U.

[crisp lintel]

Another thing to watch out for is that in the box topology you can have a big product of open sets that shrink as they go

which I think is gonna screw convergence on at least one of these

but yeah if you can show the definition of convergence for any product of opens thats enough

[umbral hamlet]

I see

By the definition of basis

[rancid umbra]

mhm

[umbral hamlet]

Agh I should've known that

Its painful how aware I am of that fact and still was oblivious about it in retrospect

I guess thats just me being new to topology

[rancid umbra]

ye

i miss patterns when the yoneda lemma is used all the time

so i feel you lol

[crisp lintel]

topology can be tricky there's a lot of subtleties and weird stuff that happens

[rancid umbra]

but that's why its cool

if you want to check later, i think that ||y and z converge in the box topology and w and x do not converge in the box topology, they should all converge in the product topology||

also i found this kind of interesting

[umbral hamlet]

its by far the most abstract math course ive done so far

people say abstract algebra is where that comes but i found group theory fairly straightforward

but topology is just insane

[crisp lintel]

Yeah I often see people try and jump into topology early and often people will say it has no or little prereqs

and while that is technically true I think most of the time it isn't a great idea

[rancid umbra]

thats because it doesn't lol

[umbral hamlet]

im currently in my second year of uni but i think i have the prereqs for it

taking this class alongside analysis 2 was a very good idea

[crisp lintel]

Yeah 2nd year uni is good

[rancid umbra]

i think it's best if you try to generate lots of examples and keep them in your back pocket for point-set

[crisp lintel]

If you've done a proof based analysis class ur pretty good for topology

[umbral hamlet]

theres an undergrad version of this class offered but i did the grad one because i wanted a challenge and a challenge i got

[rancid umbra]

and relate things to spaces that you already know, like R^n

[crisp lintel]

Yeah examples are everything (really in any field of math but especially once things start getting abstract)

[umbral hamlet]

it helps a ton for examples

if i didnt do analysis i imagine that R omega would be screwing with me a lot

but ive been able to sort out most of my confusions with infinity, at least the ones that come up in the classes im in

[rancid umbra]

what were your confusions with oo?

[umbral hamlet]

i suppose they werent as much confusions than just being unaware of how mathematicians deal with it in calculus

but after analysis 1 thats of course clear to me now

and extending that knowledge to R omega was pretty straightforward

[crisp lintel]

the one thing to be careful of is that in topology you kinda have to unlearn all of your intuition from R^n

or at least sideline it and build new intuition

[umbral hamlet]

true

fortunately this hasnt been an issue for me yet

learning the separation axioms wasnt too bad

and recognizing that a lot of spaces weve looked at so far in topology dont meet the properties we take for granted in R^n

[rancid umbra]

in what sense?

[umbral hamlet]

i found that sequences being able to have multiple limits was very cool

[umbral hamlet]

i interpreted it as losing Tn properties

[crisp lintel]

separation and also just a lot of other things

[rancid umbra]

hmm. okay, sure. i guess i thought of it as learning when to apply reasoning from R^n instead of unlearning it and starting from scratch

[crisp lintel]

yeah sure that's maybe a better way to think about it

[rancid umbra]

or like, when things generalize, or when they break from nice R^n intuition

[umbral hamlet]

this is moreso how it came off to me

[hallow oxide]

helo guys

how do you solve this

and if someone can help me with topology for today and tomorrow

i will be very grateful

[wide kayak]

what does derived set of R^n mean?

[hallow oxide]

@wide kayak i have read theory and all but when it comes to prove smthg i dont really know how to start

[hallow oxide]

it mean accumulation set

[wide kayak]

that's normal if you're starting out higher math

[hallow oxide]

yeh like i can prove calc and normal algebra stuff but topology is kind of new to me

[wide kayak]

ok, the set of all limit points or accumulation points of S is S'

[hallow oxide]

yes yes

[wide kayak]

ok, let's try to prove S' is closed

[hallow oxide]

lets go

[wide kayak]

any ideas for how to show a subset of R^n (or any set) is closed, in general?

[hallow oxide]

yes

we show that R^n-X is open

[wide kayak]

nice, yes

now R^n is kind of special in this situation

[hallow oxide]

oh

[wide kayak]

because the open sets of R^n (in the standard topology) come from the distance function in R^n

the metric

so we can use that

[hallow oxide]

yes yes ik that too

ohhh

[wide kayak]

although I suspect it's not even necessary

[hallow oxide]

damnn

why tho

wait someone's calling me

can u explain all

ill read it after coming back

[wide kayak]

actually, there's a standard result that instantly takes care of this part of the problem

I don't know if you've proved it already so we have "access" to using it

eh, let's just do this directly

so consider the complement, R^n - S'

we want to show it's open

for this it suffices to show that for each x in R^n - S', there is an open neighborhood U containing x, with U contained in R^n - S'

which you should think about after your call

I believe all parts of this problem hold if R^n is replaced with a general space X

I guess except for part (a), for that X needs to be T1 (?)

[hallow oxide]

yes

[hallow oxide]

ohhh

let me try

[hallow oxide]

damn thank you @wide kayak

[wide kayak]

sure, happy to help

[sweet arrow]

any hints

[rancid umbra]

how are you defining Ext A?

if Ext A is the complement of the closure of A, this should follow from expanding definitions; it should be tautological

[sweet arrow]

i tried that but lwk got stuck

[rancid umbra]

where are you stuck?

[sweet arrow]

i got cl(A)^c = union of subsets B of X s.t B subset A or B not closed

[rancid umbra]

ah. cl(A) is the intersection of all closed subsets containing A

its complement is then the union of all open sets whose complement contains A, or phrased differently, the union of all open sets contained in X/A

[sweet arrow]

I feel like this is wrong

[rancid umbra]

yes, it is wrong

[sweet arrow]

did i do the compliment wrong of the set

the intersection gets flipped to union

[rancid umbra]

yes. and the complement of a closed set is an open set

[rancid umbra]

this is how you interpret the complement of cl(A)

[sweet arrow]

ah ok yes i got it, it's just the notation was confusing me

thank u c²

[pulsar lagoon]

What is a totally disconnected group action and why are they nicer than continuous group actions?

[crisp lintel]

I haven't heard of totally disconnected group actions, are you sure you don't mean properly discontinuous?

[pulsar lagoon]

oh yeah properly discontinuous

my bad

[crisp lintel]

I think discontinuous in the name is kinda bad terminology since it sounds like the action is "bad"

but really its just a way of introducing some discreteness to the action

properly discontinuous means every point x has an open neighborhood that contains only x from the orbit

I believe that the main appeal is that it makes the quotient space nicer

if you want to identify points that belong to the same orbit in the quotient, having orbits with accumulation points is going to make things kinda nasty

[rancid umbra]

if X is a G-set with G a properly discontinuous action, then the quotient map X -> X/G is a covering space projection, and cocompactness of X is equivalent to compactness of X/G if G acts properly discontinuously.
i may be missing some mild assumptions on X for these to work out

i think lee and hatcher discuss the usefulness of this when classifying covering spaces

[iron bolt]

there's a few common definitions of "properly discontinuous" I think, that agree for free group actions but conflict in general

two I've seen are "every point x has a neighbourhood U such that gx ∈ U only for g = id" and "every point x has a neighbourhood U such that gx ∈ U only for finitely many g"

in the first case properly discontinuous group actions are in particular free, in the second case they at least have finite stabilizers

your definition agrees with the second one in the case of effective group actions, but allows properly discontinuous group actions to have arbitrarily large stabilizers in the noneffective case

[rancid umbra]

yea, there isn't a good well-agreed upon notion for properly discontinuous action

its also kinda funny that you can say a continuous properly discontinuous group action

i don't like the terminology that much

[near egret]

Is there a name for a topological space in which any point can be taken to any other by an automorphism? Spaces with this property are naturally based spaces in a unique way, so this seems like a useful notion for algebraic topology.

[gritty widget]

I think “homogeneous” is used for this sometimes, but it also might be defined to be a topological group with that property? Not sure

[pulsar lagoon]

yeah the context I have seen it in was with quotient spaces of manifolds but I could never really understand why they're so nice

[pulsar lagoon]

I need to go through hatcher ong

[crisp lintel]

Think about the integers acting on the circle by integer-radian rotations. The orbits are dense, so the quotient is really nasty

If instead you act on the circle by a nice fraction of 2pi, then the quotient space I believe is just the circle again which is a lot nicer

a properly discontinuous actions means you have a neighborhood of a point that isn't identified with that point in the quotient so intuitively I like to think that this makes the quotient look kind of "smooth"

if you have an accumulation point in an orbit then in the quotient you have to identify all those points which kinda makes the quotient "jagged" in some sense

[pulsar lagoon]

yeah I think that makes sense

Thank you

[balmy nexus]

im looking to do some reading about something that classifies points in a topological space according to their local behavior.

For example if we look at the closed ball B1 in R^2 together with an isolated point then we have 3 "types" of points - (the) isolated point, a point on the border of the ball, and points on the interior. I guess you could formalize this as eqvuilance classes based on having local homeomorphic neighborhoods or something.

Any ideas?

[tender halo]

well there are several purely pointset characteristics of points, namely character and pseudocharacter

[sly geyser]

ok so from what I understand

would it be correct to say

that, for example

for the box topology on R^N

the only cauchy sequences are those with the same value after a certain point

basically trivial limits

[queen prism]

does your proposition hold when N = 1?

"cauchy sequence" doesn't really make sense without a metric though (or at least more structure than just a topology)

[sly geyser]

N is the natural numbers

[queen prism]

oh

[sly geyser]

ok fine I can say R^w it doesn't really matter

[sly geyser]

can't you just replace "|f(a) - f(b)| < k" with "a and b are in open set K"

[queen prism]

do you wanna write out your definition more explicitly?

[sly geyser]

hmm actually I'm having a bit of trouble

as there's no point to center K around

ok fine it may only really work for metric spaces

now that I think about it

can't you basically create a new topology T U {t} with a limit for like any sequence?

just make it so that it's the topology created starting from the open sets of T and the sets {f(m):n>=m, m in N} U {t}

honestly at this point what are the conditions for there to be unique limits in a topology

hmm, so if there's a case of a non-unique limit, it's not hausdorff?

as any two open sets of the two limit points P_1 and P_2 would at the very least have an intersection of {f(m) : m >= max(n_1, n_2)}

is there being a non-unique limit point stronger than hausdorff or the same?

oh wait

sorry I mixed things up a lot

lemme reword it

is there being a non-unique limit point stronger than not being hausdorff or the same?

[gritty widget]

A space is Hausdorff iff all convergent filters/nets have unique limits

For sequences in particular, maybe not? Not sure

[sly geyser]

just a question, why are nets specifically used?

[gritty widget]

Sequences don’t characterize spaces, or lots of their properties, very well. They do for some spaces(e.g. first countable spaces)

Nets, however, do

Nets/Filters completely characterize a topology

[kind marlin]

i think this is a non-hausdorff space with unique limits

take the set R U {p} for some external point p

use standard topology on R, coupled with the collection of p U (cocountable sets in R)

this should be a topology, since finite unions of countable sets are countable

this isnt hausdorff, since every neighborhood of p only misses countably many points, so it always intersects every neighborhood of all points in R

but sequences have unique limits: we can never have a sequence limiting to p because R - (points in sequence) U {p} is a neighborhood of p

and for any two reals, we can find disjoint neighorhoods, so a sequence can never limit to two distinct reals

[unreal stratus]

Apparently yeah not true for sequences cause you can use e.g. cocountable topology of an uncountable set X, and then any convergent sequence is eventually constant - indeed, suppose (x_n) converges to x. Then X ( {x_n} \ {x}) is an open neighbourhood of x and so the sequence has to eventually be x

This is quite cute actually

Oh lol jinx ig

LOL

[kind marlin]

oop

no that seems like a pretty different topology tbh another example :D

[unreal stratus]

It's very similar at least in that there's the cocountable topology aha

[kind marlin]

oh thats true the p is throwing me off i guess

wait is the cocountable topology hausdorff?

[unreal stratus]

Not unless you space is countable, as otherwise in fact any two nonempty opens intersect nontrivially!

[kind marlin]

oh wait right we want non hausdorff spaces

[woeful coral]

what is the difference between a basis and a base of neighbourhoods?

[kind marlin]

i mixed up the direction

[woeful coral]

which definition builds on second countability?

[unreal stratus]

Firstly, base and basis are synonyms.

But moreover a base of neighbourhoods is of a point, but a base is a thing for the whole space

[unreal stratus]

This is having a countable base

[woeful coral]

are the definitions of every open U is a union of basis elements

and every x \in X and U \ni x there is B such that x \in B \subset U

[unreal stratus]

Those are equivalent

[woeful coral]

equivalenmt

so a base of neighbourhoods is a basis

[unreal stratus]

Well usually one says a base of neighbourhoods of a point right

[woeful coral]

of a point for all points

[unreal stratus]

Yes, if you take the union of neighbourhood bases over all points then you get a base

[woeful coral]

oh I see how these are equivalent

and second countability is countable basis for the entire space, but first countability is only for each point

[kind marlin]

i think as a concluding thought

T2 -> unique limits -> T1

but none of these reverse implications hold

[gritty widget]

If we have a space Y, and a (set) function F:AxA to Y, we can give a topology to AxA by letting a set be open iff it’s image is. Is there a way to extend this to a topology on A, such that the product topology of AxA agrees with the given one?

[cerulean zealot]

Could someone explain to me why the word "larger" is used here? It seems like \beta(X) should intuitively be smaller because every embedding from X into Y has to pass through \beta(X).
I'm familiar with the example that nets are sometimes created by ordering open neighborhoods of a point by reverse inclusion; is it the same idea here?

[rancid umbra]

if U and V are such that F(U) and F(V) are open in Y (wlog F is surjective) how do you guarantee that F(U \cap V) is open in Y?

[crisp lintel]

it maps onto Y kinda

iirc any compact hausdorff space containing X is a quotient of beta X

maybe only works for nice spaces

[crisp lintel]

containing densely

[cerulean zealot]

o i see this is cool

[kind marlin]

i think you have to define the topology on AxA as preimages of open sets in Y

[gritty widget]

ah

ic

[rancid umbra]

images won’t preserve intersections

[rancid umbra]

so unless F is a bijection, this process won’t define a topology on A x A

and in that case, you’re just transporting the top structure of Y to one on A x A

[kind marlin]

is injectivity sufficient?

[rancid umbra]

yea, but if you aren’t surjective, you might as well be

since you are corestricting to the image anyways to get the open sets

so they are equivalent

[kind marlin]

under the preimage definition:

i think this is still not going to work if we want each A to have the same topology

e.g. suppose F(a, b) = G(a), so fully independent of b. Then F-1 (S) = G^-1 (S) x X for any nonempty S, and just the empty set for empty S. if every open set in A x A needs to be a product with the empty set or the entire set in the second entry, while the first is allowed to have nontrivial sets, then there's no basis for A that would asymmetrically generate the open sets like that

maybe a more interesting question is if given a set function F: A x B -> Y, and defining the topology on A x B as preimages on Y, can we find topologies for A and B respectively such that the resulting product topology coincides with the preimage topology on A x B

i think... the answer to that is also no

thats actually also pretty obvious

if we just have {1, 2} x {3, 4} -> {5, 6, 7, 8}, let the open sets be {5, 6} and {7, 8}, and map

(1, 3) -> 5
(1, 4) -> 7
(2, 3) -> 8
(2, 4) -> 6

then we can kind of brute force that there's no topology on {1, 2} and {3, 4} that generates a compatible product topology, bc none of our open sets are products of a set in {1, 2} and a set in {3, 4}

[gritty widget]

Alright, thanks!

[rancid umbra]

another reason why is that maps into A x A are continuous iff its projections are, while in the other topology inhereted from preimages of Y, maps into A x A are continuous iff the composite with F : A x A -> Y is continuous

so these two topologies have different universal properties

the latter topology is called the initial topology on A x A induced by F

[woeful coral]

Does every topological space have an open path connected subspace?

Nvm, discrete spaces

[queen prism]

isn't every singleton an open path-connected subspace in that case?

take the constant path

[rancid umbra]

no, consider Q

it is totally disconnected, so its only path components are the singletons, but those are all closed (since they are components), and they can't be open since there are infinitely many rationals in any open interval of R

[kind marlin]

I’m kind of curious, are there examples of totally path disconnected spaces that aren’t totally disconnected

[rancid umbra]

like, path components are singletons but connected components aren't?

[kind marlin]

yeah

I see a definition for totally path disconnected in Wikipedia under the totally disconnected article but all the examples are just totally disconnected

[gritty widget]

Yep! There’s even some that are connected.

https://topology.pi-base.org/spaces?q=Totally+path+disconnected%2B~Totally+disconnected has a list of some

[kind marlin]

oh neat!

I would assume once you’re not totally disconnected you can just take the connected subspace?

[gritty widget]

That makes sense yeah, but actually all of the above examples, seemingly by pure chance of which spaces people decided to put in the list, are already connected.

[kind marlin]

interesting :0

I have like no intuition for what the cocountable and cofinite topologies represent 😭

I guess it’s maybe characterized by each point being in the closure of all other points, just ever so slightly away to be distinguishable

[gritty widget]

My intuition for it, for example for the cofinite topology on N, is just an open set is one that is "eventually full", e.g. there is some m such that for all n>m, n belongs to that set

If the cocountable topology is on, say, omega_1, then the intuition I have is similar.

[balmy nexus]

What do i google here? Im getting results for some group theory things.

[rancid umbra]

character + pseudocharacter + topology

[balmy nexus]

the smallest cardinality of a local base at x

[tender halo]

you can look at the first section of the handbook of set theoretic topology for a discussion of the various cardinal characteristics

[balmy nexus]

it's interesting but not what I was thinking about.
I was thinking about whether you can define which points are at the "edge" of a space from within the space.

This is a bit of a nonsense question because if you take an arbitrary space, you can embed it in a larger space so that the boundary of this set is whatever you want. But still, looking at the example of a closed ball, one can identify that the "boundary" points of the ball are different than the interior points.

In manifolds we usually say "yes, these points look like an open ball, and these points look like a half plane", and I was wondering if you could talk about the difference without passing the problem of characterising them to "being like something else"

[crisp lintel]

I think something that helps is to imagine that a set and it's closure should "look" the same to someone who hypothetically could imagine the space. After all, points in the closure of a set are supposed to be infinitesimally close to the set. For example, in the indiscrete topology every subset "looks like" the entire space, so the space looks like a huge blurry point. In the cofinite topology, every infinite set looks like the whole space.

[kind marlin]

yeah i think that's the rationale i landed on

i think the real issue is that i don't have a good intuition of what it means to be path connected

it's a stronger form of connectedness, and I guess at face value that feels like a sort of "super connectedness" to me, but in that sense it feels a little counterintuitive to me that you can have non-T2 spaces (or like, "super-not T2", in the sense of every two points violating T2) that are not path connected. sort of because of how "close" everything is to one another, to the extent that connectedness is basically guaranteed, but path connectedness is basically ruled out.

I guess it's because path connectedness cares about how that closeness is "structured"? idk, hausdorff spaces are so much easier to wrap my head around 😵‍💫

[hexed steppe]

its sort of like modeling connectedness on the topology of R

where there is an intermediate value theorem

it turns out IVT-ish behavior isnt really captured by connectedness alone

other stuff can break too like topologist sine curve but yeah

[rancid umbra]

this is somewhat related, but there is something called an end of a topological space. These are described as an "ideal boundary" of a space.

[quartz horizon]

Yeah like memorylessfunctor said, path-connectedness is about “probing” a space by [0, 1] to determine connectedness

There’s a nice sense in which path-connectedness is about probing into a space, whereas connectedness is about continuous maps out of the space

Connectedness is more about how your space embeds into other spaces

To be clear [0, 1] isn’t completely arbitrary, there’s a sense in which it’s the “universal” path space

[dawn frigate]

Whats an example of countably finite topology

[prime elbow]

You mean, you need finite topology?

[dawn frigate]

Countably infinite sorry

[wide kayak]

like one with countably many open sets?

we could take the empty set, R, and all open intervals of the form (-n, n) for n in N - {0}

[prime elbow]

Good one

[prime elbow]

<@&268886789983436800>

[ruby delta]

lance how do you have your own emoji

[red yoke]

Because lance is lance

[prime elbow]

Can we define such an order on R^2 such that it will give the usual Topology on R^2?

[prime elbow]

Hi Arki

[rancid umbra]

are you talking about how 2 -> I is weakly initial?

[quartz horizon]

I’m talking about [0, 1] being the universal bipointed space

[rancid umbra]

isn’t this the same thing? also, i don’t see how it can be properly universal if the path isn’t unique

did you mean universal in this sense?

[quartz horizon]

Yes

[unreal stratus]

If by order you mean total order (which seems reasonable) then no, because then removing a point would make it disconnected

[prime elbow]

I got it Man, how do you get it

[prime elbow]

Here, I used the FIP.

If \cal C is the collection of closed sets which have finite intersection property then I have to \cap F, F\subset \cal C.

Since f is closed mapping then \cal D = { f(F) | F in \cal C } then \cal D is a collection which has FIP and since Y is compact so there is y \in \cap f(F) | F in \cal C.

Also, f is surjective, implies that f^-1{y} is non-empty and y in f(F) for all F\in \cal C, so f^-1{y} intersection F is non empty.

And F intersection f^-1{y} will be closed in f^-1{y}.

How do I show the collection f^{-1}y intersection F has FIP?

[near egret]

I came across the term "bicompact" in a paper citing Alexandroff and Hopf's "treatise" for definitions. Does anybody know what this term means?

If it turns out to just be compactness, then what does compact mean in such texts?

[gritty widget]

https://ncatlab.org/nlab/show/bicompact+space this, maybe? But it’s not a property of topological spaces

[near egret]

No I saw this

All spaces involved have only one topology

The paper is "Rings of Real-Valued Continuous Functions. I" by Hewitt if that helps

[rancid umbra]

any finite collection of sets of the form f^{-1}(y) \cap F_i for i = 1 to n is written as f^{-1}(y) \cap (F_1 \cap ... \cap F_n). both sets are non-empty since \cal C has the FIP

then you should be able to conclude that ||since f^{-1}(y) is compact, f^{-1}(y) \cap \bigcap \cal C is non-empty, which implies that \bigcap \cal C is non-empty||

[lucid ocean]

compact used to mean sequential compactness and bicompact meant compactness iirc

[queen prism]

https://arxiv.org/pdf/1006.4131

[wide kayak]

super cool paper!

[near egret]

Ah thank you

[prime elbow]

But how is their intersection non-empty?

[rancid umbra]

y is in f(F) for all F in \cal C, so for a particular F, there is some x in F such that f(x) = y, and so f^{-1}(y) \cap F is non-empty

[prime elbow]

Yes

But how f^-1{y} \cap { \cap F_i } is non-empty, i = 1,2,..,n?

[rancid umbra]

each of the F_i are just one of the F's in \cal C

im just listing them since we only need to consider finitely many to prove the FIP

[prime elbow]

Yes

[rancid umbra]

and \cal C has the FIP

[prime elbow]

But their intersection does not need to be in \cal C

[prime elbow]

Yes

[rancid umbra]

am i missing something silly

yes, i think i am... apologies

[prime elbow]

No problem

[rancid umbra]

you will want to adjust your choice of $\mathcal{D}$. Instead, put 
$$\mathcal{D} = \left\{f\left(\bigcap C\right) : C \subseteq \mathcal{C} \text{ is finite }\right\}.$$
$\mathcal{D}$ has the FIP because for any finite collection of sets 
$$f\left(\bigcap C_1\right),\dots,f\left(\bigcap C_n\right)$$ 
there is an element of $x \in \bigcap C_1 \cap \dots \cap \bigcap C_n$, and so 
$$f(x) \in f\left(\bigcap C_1\right) \cap \dots \cap f\left(\bigcap C_n\right).$$
Now, since $Y$ is compact, there is some $y \in \bigcap \mathcal{D}$, and my first argument should go through the same.

[gentle osprey]

c squared

[prime elbow]

What if there is no C in \cal C, which is finite

[rancid umbra]

C is a subset of \cal C