point-set-topology

[mighty hull]

To start, I'd go back through your textbook and look up what theorems you were given about connected components and clopen sets.

Another way to think about connectedness is that a topological space is connected iff the only clopen sets are the empty set and the entire space.

If S is a topological space and K ⊆ S is clopen then S\K is open (because K is closed), but:

1. K∩(S\K) = ∅
2. K∪(S\K) = S

So if K is anything other than S or ∅, K and S\K separate S.

[shrewd night]

Is there a nonempty perfect set in R with no rational number?

My instinct is saying yes but I can't prove it

Like how do you even start to prove it in a relatively simple way

[rancid umbra]

create an uncountable closed set whose intersection with Q is empty. then apply cantor bendixon

[shrewd night]

I see

I'll just come back to this later

[shrewd night]

Meanwhile exercise 23 looks alright but what's the best way for a breakthrough?

[rancid umbra]

,rotate

[rancid umbra]

the hint

[shrewd night]

I know that exists but I have no clue how to use it

When I ask for help I actually have a go first

[opaque scroll]

I mean the hint just is the solution. So you just have to verify it, which amounts to writing down the definitions and knowing that a set in a metric space is open if it contains an open ball around each of it's points.

[shrewd night]

Oh I see thx

[steep sand]

if I have a topological group G acting continuously on a G-set X which has the discrete topology (X is an abelian group here, but the action is not compatible with the sum), are the sets of the form {g in G: g(a_1)+...+g(a_n)=g(b)} open? I was thinking you can vary g(b)=c and write it as a union of ({g in G: g(a_1)+...+g(a_n)=c} \cap {g in G: g(b)=c}, the second component is open because its a translation of a stabilizer, and for the first one you write it as union in c_1{g in G: g(a_1)=c_1} \cap {g in G: c_1+...+g(a_n)=c}, and again the first component is open, etc., and you iterate, obtaining an arbitrary union of finite intersections of open sets, and so it´s open?

like, in general, if i have a topological group G acting continuously on a G-set X (also with a group structure, but not compatible witht he G-action) with the discrete topology, is something like {g in G: equation 1 depending on specifying the action of G in some elements = equation 2 depending on specifying the action of G in some elements} always open? by saying its the union of all possible values making the equations true, and then using that every {g in G: g(a)=b} is open?

[low flame]

im having a hard time understanding what i should do with 9.7

9.6 is straight forward

[steep sand]

what is the interior of an open set? what happens to limits under continuous functions?

[low flame]

the interior is all the points that have nbhds completlyh within the set.

[prime elbow]

Hint: can you show that f^-1(Int A) \subset Int f^-1(A) ?

Provided f is continuous

[low flame]

so its the opposite of the closures inclusion

[prime elbow]

Yes

Same question

[low flame]

what books that from? munkres?

[prime elbow]

No basic real analysis by Sohrab

[prime elbow]

Is it viro? Right

[low flame]

yes it is 🙂

[prime elbow]

Chapter 9, good

[low flame]

yeah im falling behind my class, but im not doing this for a grade so it doesnt matter.

because im doing viro, and its a workbook haha

[prime elbow]

Good

[low flame]

so the reason for the cls f^-1(A) subset f^1(clos(A))

is because that the closure of A is the largest closed set that contains A.

therefore the closure of the inverse of A may be a random set thats smaller than the actual closure of A right?

[prime elbow]

Not necessarily true last statement

[low flame]

im confused then

[prime elbow]

Let x in cl f^-1(A) now you have to show that x in f^-1(Cl A)

So you have to show that f(x) in Cl A

Now how can you show that?

What are the equivalent definitions you know about cl A?

[low flame]

i just remember its the smallest closed set that contains A, IE the intersection of closed sets that contain A

[prime elbow]

Here is the one more x in cl A if and only if for any open set containing x intersect with A

So assume that now

You can prove that if you want

[low flame]

no it makes sense,

[prime elbow]

Okay

[low flame]

it has to 🙂

[prime elbow]

Now you have to take any open set containing f(x) it must be intersect with cl A

So when you take the open set f(x), U

So f^-1(U) is open right

[low flame]

yes

its continious

[prime elbow]

And it is an open set containing x

[low flame]

yes

[prime elbow]

Now x is in cl f^-1(A)

What does it mean ?

Take any open set containing x it must intersect with f^-1(A)

[low flame]

then take the closure of that

and its all open sets x intersecting with CL f^-1(A)

[prime elbow]

So you have a point in f^-1(U) which intersect ^-1(A)

[low flame]

x is in the closure of A, it means that for every point x in the set there exists a nhbd N(x,r) r >0 such that N(x,r) cap A is not empty.

thenthe other way around is, take the intersection of nbhds, and you will get the closure of A

would this be right?

[low flame]

an i see and thats the def of closure right

[prime elbow]

Yes

So you get your result, right?

[low flame]

haha yes, i see it 🙂 thank you sir

[prime elbow]

Wait I am not sure about it

Can you explain more?

[low flame]

well its the converse direction you dont like right?

[prime elbow]

I don't get it

[low flame]

can i do something like if x is not in A, then there exists a nhbd that doesnt intersect A, which would contradict our assumption of N(x,r) cap A

[prime elbow]

You mean x is not in cl(A) ?

[low flame]

yes, sorry

[prime elbow]

Let me do it again, you know cl A is the intersection of all closed sets which contain A

Now we have to show that x in cl(A) if and only if every open set containing x intersect with A

So let first prove that x in cl(A) then every open set containing x intersect with A

[low flame]

yes, that means that every point x has a N(x,r) where r > 0 such that N(x,r) cap A != \empty

[prime elbow]

Not particular neighborhood, every open set

And this is true in topology

So we cannot assume metric stuff

[low flame]

im sorry i dont understand what you mean by the last bit. but i ahve to go i have a meeting and then a job fair 🙂

[prime elbow]

No problem

[heady skiff]

can I have a hint for c please, intuitively I know I want to express it as a preimage of T in some way

[iron bolt]

can you show that the subset of Y where x_1 is a c if x_0 is a b is closed? if you can, how does that help you?

[heady skiff]

do you mean only for the coordinates 0 and 1?

i guess if that's the case, then W is equal to the union of all those subsets

which are closed

or wait no maybe W is the intersection of all thos epoints

[iron bolt]

yep

if it was the union it wouldn't work, since unions of closed sets are not always closed

but it's the intersection

[heady skiff]

cool! I'm having a bit of trouble proving this - for each $n \in \mathbb{Z}$ I'm letting $U_n = \prod_{k = -\infty}^{n - 1} X_k \times {b} \times {c} \times \prod_{k = n + 2}^{\infty} X_k$ and my claim is that $W = \bigcap_{n \in \mathbb{Z}} U_n$

[gentle osprey]

okeyokay

[iron bolt]

oh, that's because it's not actually true

[heady skiff]

okay i was able to do it by showing that the complement is open

[iron bolt]

you don't want every consecutive pair of characters to be "bc" - you just want it to be "bc" if the first character is a b

[heady skiff]

would this work?

For each $n \in \mathbb{Z}$, let $U_n \coloneqq \prod_{k = -\infty}^{n - 1} X_k \times {b} \times {a, b} \times \prod_{k = n + 2}^{\infty} X_k$, which is open in the product topology on $Y$. We claim that
[
Y \setminus W = \bigcup_{n \in \mathbb{Z}} U_n
]
which is open as the arbitrary union of open sets in $Y$. Let $(y_n){n \in \mathbb{Z}} \in Y \setminus W$. Then there exists $n \in \mathbb{Z}$ such that $y_n = b$ and $y{n + 1} \neq c$; thus $y_{n + 1} \in {a, b}$, so $(y_n) \in U_n \subseteq \bigcup_{n \in \mathbb{Z}} U_n$. Conversely, if $(y_n){n \in \mathbb{Z}} \in \bigcup{n \in \mathbb{Z}} U_n$, then $(y_n){n \in \mathbb{Z}} \in U_m$ for some $m \in \mathbb{Z}$. Thus $y_m = b$ and $y{m + 1} \neq c$; it follows that $(y_n)_{n \in \mathbb{Z}} \in Y \setminus W$. Thus the claim is proved, and since $Y \setminus W$ is open, $W$ is closed.

[gentle osprey]

okeyokay

[iron bolt]

that would work, yes

the argument as an intersection of closed sets works basically the same, except the set in the middle is not {b} ⨯ {a,b} but its complement in {a,b,c}^2

then it's closed and W is an intersection of those

as always, many paths to the same answer

[prime elbow]

For part c, X is isometrically embedded in X* *, I don't understand what it means?

I found a function f:X -> X* *such that x -> [x], where [x] denotes the equivalence class of x with respect to a given relation, p(x,y) = 0.

Now this function is well defined and p(x,y) = p* *(f(x), f(y) ).

[rancid umbra]

that’s exactly what it means: there is a function f : X —> Y such that d_X(x,y) = d_Y(f(x), f(y))

[prime elbow]

Does my function work here ?

[rancid umbra]

yes

[prime elbow]

Can you help me to understand d part ?

[rancid umbra]

17 part a?

[prime elbow]

I know that real numbers is a special part of d

[prime elbow]

Done

[rancid umbra]

oops, misread

what are u stuck on?

[prime elbow]

First I don't understand the hint here what it means < < x_n,m> >

[rancid umbra]

the elements of X* are equivalence classes of cauchy sequences

[prime elbow]

Yes Cauchy sequence of X

[prime elbow]

Yes

But how can I show that X* is complete metric space

[rancid umbra]

you need to show that any cauchy sequence of equivalence classes of cauchy sequences is convergent. try to unwind the definitions

[prime elbow]

But for showing it is convergent should I know where they converge ?

Is there any use of part a ?

[prime elbow]

A metric space X is compact if and only if every collection F of closed sets with the finite intersection property has a non-empty intersection.

I proved this one. Is it true in arbitrary topological space?

I think yes because in my proof I just used set theory not metric space property

[red yoke]

Yes

[prime elbow]

Thank you

[rugged nimbus]

yo can someone plis expalin to me the sequential definition of a limitpoint

[paper wedge]

wdym

a point is a limt point if there exists a sequence in the set converging to it?

[rugged nimbus]

of a set

[paper wedge]

a limit point of a set is a point ( not neccesairyly in the set ) ssuch that every nbd of this point intersects the set in more than one point (iirc)

[prime elbow]

Any hint to construct a metric space which is totally bounded but not compact ?

[opaque scroll]

A metric space is compact iff it is totally bounded and complete

[prime elbow]

Yes so I thought about it and took a metric which is not complete, Q or open interval but they don't work, now I need different metric space which is not complete

[opaque scroll]

Why do you say they don't work

[prime elbow]

Q is not totally bounded, right?

[opaque scroll]

Well, not all of Q. But like (0, 1)

[prime elbow]

Oh wait this one totally bounded?

Oh wait

Intuitively yes if you give me the distance then I divide the interval in sub interval

Right?

[opaque scroll]

Yeah that works

[prime elbow]

Thank you ❤️

[paper wedge]

i know that in general the number of sheets is the same

for lpath-connected coverings

in the sense that you have a bijection sending one fiber to the other with a map

now with showing that it is finite

wouldn't any evenly covered nbd of p^-1(p(x)) cover X?

and hence has a finite subcover by compactness?

[iron bolt]

nope

for a very simple example, the identity map on an a compact space X is always a covering - the fibers only consist of single points then, and you can easily pick an evenly covered neighbourhood that is not the whole space

the fibers do have a special topological property though that together with compactness can get you the result

actually, the proof I was thinking of does require X to be Hausdorff as well. not sure about the more general case

[paper wedge]

yeah

i thikn it has t odow ith closed sets being compact

singletons being clossed

something like that

ig thats what ur thinking of

here is something but i thought the U V_b,x cover the space already

[iron bolt]

what I was thinking of (I guess I can spoil it now since it doesn't actually solve the exercise) is: the fibers are always discrete, and if {x} is closed they're closed too, so as closed subspaces of a compact space they're compact

[paper wedge]

https://math.stackexchange.com/questions/728279/why-is-the-fibre-of-each-point-compact?noredirect=1&lq=1

[iron bolt]

and compact discrete spaces are finite

so I guess the precise fact I was thinking of is that coverings where X is compact and Y is T1 have finite degree

[paper wedge]

check out the answer by stefan hamcke

its the kinda the same key point " the sheets cover the space "

[paper wedge]

but theres this

[iron bolt]

yeah, I thought you only meant the preimage of a single evenly covered neighbourhood

[iron bolt]

this for example works with all such preimages instead though, which is something I didn't think of and indeed an open cover

[paper wedge]

i meant like

take p(x)

take an admissible nbd

then the sheets of that admissible nbd cover Y

[iron bolt]

X?

[paper wedge]

i think i messed up

the naming

cover the covering space

yeah X

in my problem notatino

[iron bolt]

yeah, I don't think so

[paper wedge]

yeah take X-->X with f(x) = x ig

as u sad

said

but ig

[iron bolt]

finally understood this one - it actually works without any assumptions other than compactness, I think

so if you're looking for a working solution, that's one

[paper wedge]

yeah

thank you!

[low flame]

im confused by this

i looked at the hint and i was even more confused

isnt there a jump discontunity in f(x)

but I guess the map itself is continious?

[river granite]

the idea of "jump discontinuities" works for real functions, with the usual topology on R

the induced topo on Z_+ is a whole different thing

[low flame]

ive kinda looked at the f mapping and said it maps everything cleanly

{0} -> {0}, {2} -> {1/2}

so its one to one and surjective, i think so its continous

also its the discrete topology

[river granite]

this doesn't immediately imply it's continuous -- but yes the induced topo in~~ both Z_+ and f(Z_+) ~~ (e: correction, Z_+ only) should coincide with the discrete topo

[low flame]

and then i see with the g function how you get abunch of values approaching zero, which means there is multiple g^-1 which can map to zero?

[river granite]

no, 0 has only one preimage

under this fn

[low flame]

the hint says that the singleton 0 is not open in the topology of f(Z+)

and i dont understand it.

[river granite]

yes I was also thinking of this just now

think of the definition of subspace topology

how do you induce a topo into a subspace

[opaque scroll]

What are the open sets in R that contains 0?

Are there some that only contains 0 and no other elements from f(Z+)?

[low flame]

to be honest im not enterily sure, does this imply the subspace topology?

[river granite]

yes. I'm asking if you recall this definition

since that's the key behind that hint as jagr says

[low flame]

its some set intersected with another

i can go look, but thats what i remember off the top of my head, you take a subset, and it has to intersect i think the parent set?

[river granite]

yeah, it's the set of intersections of the subspace with open sets in the "parent" topology

so in this case if $\tau_\bR$ is the topology in the real line, we might define [\tau_{\bZ_+}\coloneqq \bZ_+ \cap \tau_\bR = {\bZ_+\cap U:~ \text{$U\subseteq \bR$ is an open set}}]

[gentle osprey]

derivada.schwarziana

[river granite]

same thing if we wanted a topology on f(Z_+)

[low flame]

okay im ingesting this

[wise mist]

hint

[opaque scroll]

X is always a closed subset of X. So you might try to construct counterexamples to A and B with X=Y

And it should be enough to consider X as subspaces of R

[wise mist]

what about the clopen one

[opaque scroll]

Is X open in X?

[low flame]

so is it that that in f(Z+) theres gonna be a lot of values approaching zero, so it wont have a nbhd of only zero?

[wise mist]

🥶

alr

its so simple

[river granite]

that's kinda the point yes, try to actually write out what the set {open sets in f(Z_+)} is

[low flame]

okay, im gonna go for a walk now and think on it, i appreicate you help 🙂

[river granite]

no prob

[opaque scroll]

If you want it to be harder you can add the assumption that X is complete.

[wise mist]

do i put d = min(1, |x-y|)

[opaque scroll]

Yeah, so still not that hard I guess

[wise mist]

i mean if X=Y then i have to put some bound on metric or the X right ?

for the distance in Y is atmost 1

is this the only sort of counter ?
via tweaking the metric ?

[opaque scroll]

Well, I guess that depends what metric you start with. For the original question something like (0, 1) with standard metric works.

If you also want it to be complete, something like bounded functions with supremum-norm would work.

I guess the main take away is that A, B, C all depend on the topology, and not really in the metric.

And restricting the metric to be less than 1 doesn't change the topology, as it only depends on "small" open balls

[wise mist]

at most 1 doesn't mean in some case it has to be 1 right?

[opaque scroll]

Nope

[wise mist]

i mean imagine sup|f(x)| = 1

i can say it's at most 2 ?

would that be logical to say

[opaque scroll]

It would be correct.

Though perhaps misleading

[wise mist]

ok

[wise mist]

for this interval the distance never reaching 1 but arbitrary close to 1, so we can say at most 1

that was my concern

i mean what else would it be, i am such a blind

[low flame]

something like (1/n - e, 1/n + e) \cap f(Z+)?

[river granite]

yes all of these for each n are open sets, but that's not the whole induced topology on f(Z_+)

actually you aren't missing much -- just the neighborhoods of 0

so the induced topology looks like, {1/n} for each n and then the sets {0, ..., 1/(n+1), 1/n} for each n

and unions of those ofc

[low flame]

AH okay it makes good sense now

[heady skiff]

in proving b => a, could we let any $y \in X \setminus V$, and since $y \in X \setminus V$ there is no sequence $(x_n) \subseteq V$ with $(x_n) \to y$, otherwise this would contradict b, so we have $\epsilon > 0$ such that $B(y, \epsilon) \cap V = \varnothing$, so $B(y, \epsilon) \subseteq V$

[gentle osprey]

okeyokay

[heady skiff]

I feel like this doesn't work, since just because (x_n) doesn't converge to y doesn't necessarily imply that we can find a neighborhood that doesn't contain points of x_n

because sequential criterion is just saying that eventually all the points have to be in it

[opaque scroll]

Think about the sequence of sets B(y, 1/n)\cap V

What would it mean if they are all non-empty

[heady skiff]

ye i was able to get it

[heady skiff]

For the $\leftarrow$ direction, could we say the following: Suppose for contradiction that $x \notin \overline{A}$, so that there exists some closed set $C$ with $A \subseteq C$ but $x \notin C$. Then $x \in X \setminus C$, which is open, and $X \setminus C \cap A = \varnothing$, which contradicts our assumption

[gentle osprey]

okeyokay

[mighty hull]

Yes, if x is in not in the closure of A then there's an open neighborhood of x that doesn't intersect A. The contrapositive of that statement: if every open neighborhood of x intersects A then x is in the closure of A.

Do the contrapositive in the other direction, too.

[sonic crane]

Whats ur guys’s definition of closure of a set?

My prof had as definition, all points not exterior to A

So like for me, by definition x being in closure means every open set containing x intersects A

[paper wedge]

A union the limit points of A

[hoary breach]

smallest closed set containing A

[heady skiff]

this seems to be the first definition i've encountered in most of the courses i've taken, but you end up proving every open set containing x intersects A is equivalent to it anyways later on so it all doesn't rlly matter

[heady skiff]

ye i did the other directoin

luckily this it not hw just proving all theorems in prep for midterm next week 😅

[iron bolt]

fun fact: much like how you can define topologies as some collection of sets satisfying certain properties and then just declare that sets in that collection are called "open", you can also define topological spaces as a set X with a map P(X) → P(X) satisfying certain properties and calling that the closure operator

because the closure operator allows you to recover open sets as the complements of its fixed points, so a topology is completely determined by its closure operator

and whenever a map P(X) → P(X) satisfies some simple axioms the complements of its fixed points form a topology whose closure operator agrees with that map

what I'm getting at is: depending on the definitions you use, the definition of closure might be as trivial as that of an open set :p

[round oyster]

This is the definition we were given :
x is in the closure set of $A$ if for all neighborhods $U$ of $x$, $U \cap A \neq \phi$

[gentle osprey]

MadAbdo

[round oyster]

Or, its the smallest closed containing A

[ocean canyon]

If M^n is a connected manifold for n > 1, and B is a regular coordinate ball, we know that M^n \ B is still an n-manifold with boundary homeomorphic to S^{n-1} (see e.g. problem 4.17, Lee:s "Topological Manifolds"). Is it the case that M^n \ B is still connected?

[snow gull]

Hey guys. I have to show that there exists only one topology such that the sets Bx are neighbourhood basis. The sets Bx have this behaviour.

How do I prove uniqueness of topologies?

[rigid perch]

this is my proof of "show that any closed subset Y of a compact set C in X is also compact". let V be an open cover for Y. Then, V union ({C\Y}) would be an open cover for C, since all sets in this collection of sets are open (by Y being closed), and clearly, for all x in C, x is either in C/Y or in Y. if x is in Y, then it gets covered by V. if it's in C/Y, then it gets covered by ({C/Y}). By compactness of C, V union ({C\Y}) has a finite subcover. Thus, taking elements in V of this finite subcover, we have a finite subcover of Y. is this proof correct??

[iron bolt]

yes. I don't know a nice elementary proof though

it's a direct consequence of the last part of the Mayer-Vietoris sequence

trying to prove it without heavy machinery like that might be a fun exercise

[ocean canyon]

oh, I see, thanks

I was trying to prove b) here, but then I suppose my idea is not the way to go

I'd be happy for a hint

though

^^

[iron bolt]

if your idea is to show that M_1 and M_2 are connected I think you're on the right track

[ocean canyon]

well, if you mean M'_1 and M'_2 then yes

that was my idea

[iron bolt]

I'd assume there is some elementary proof, I just didn't think about it for long

I'd say, assume that you have a clopen subset of M'_1 and see what you can say about it

[ocean canyon]

yeah I was trying to prove the thing I asked about, and since manifolds are locally path-connected we have connected iff path-connected, so me and a friend discussed whether we could show that M'_i = M_i \ B_i for B_i regular coordinate ball is still path-connected (since M_i is connected iff M_i is path-connected)

[iron bolt]

right, that approach works too

thinking about non-smooth paths directly is always a bit ugly since they can be so all over the place and thus hard to modify - that's one reason why all the algebraic topology stuff is so useful, to get arguments like that written down in a very clean way

but modifiying paths does actually get you what you want

[ocean canyon]

Oh, I see, thanks for the input; my energy levels are quite depleted, I might sleep on it (CET time zone, here). 🙂

[iron bolt]

I think it's clear visually if you think about what modification you need to do and draw a picture - it's just converting that intuition into a rigorous proof that might be a bit hard

[ocean canyon]

Like, roughly, the idea would go like this (and this might be totally wrong). Take any pair of points p,q in M'_i. If there exists a path not going through B (in the original manifold M_i), then this gives us a path in M'_i. If not (i.e. every path between p and q goes through B), then you'd want to "nudge" the path "around" the ball B?

[iron bolt]

yep

[ocean canyon]

or wait, hmm, what about the boundary

of B

[iron bolt]

you're not removing the boundary, right?

[ocean canyon]

youd want the path to go along the boundary perhaps, of B; i.e. youd perhaps want to connect p to q by going from p to the boundary of B, and then along the boundary to another boundary point, and then to q (very very roughly, or something like that)

[ocean canyon]

right

[iron bolt]

yeah, that works actually

I was thinking of only modifying the parts of the path that are in B, it could potentially still enter and exit the boundary of B an infinite number of times

but ofc you're right, in the middle of the path you can do what you want

[ocean canyon]

well, ok, I guess the issue is that a path from the boundary of B to say q might still go through B, so I don't think that solves it (atleast how I thought about it; perhaps you thought of something different)

[ocean canyon]

right

I am not sure what it would "technically" mean to modify the parts inside B such that one still gets a path, ig

[iron bolt]

yeah, the path from p to q can cross ∂B an infinite number of times - if you pick any t with γ(t) ∈ ∂B, you could still have γ enter B both on [0,t] and [t,1]

so you can't just pick any point in the preimage of ∂B - ||you need to pick the leftmost and rightmost one||

can you see why that is possible?

[iron bolt]

writing down a different path and showing that it's also continuous

[ocean canyon]

hmm, right, this is what my friend said (he is a lot more topology-knowledgeable than mean). iirc his idea was that B is open, and a path f is cont. so f^{-1}(B) is open, so its complement is closed, and since in [0,1] it is bounded, so compact (heine-borel)

so one would want to pick the first point that a path f "enters" the boundary of B (and I suppose the last point, as you pointed out)

[iron bolt]

yep, that's exactly the right idea

you don't need Heine-Borel for compactness btw - closed subsets of compact spaces are compact

[ocean canyon]

ops, yes

you are right (this is prop. 4.36.(a) in Lee, iirc)

[iron bolt]

[ocean canyon]

hmmm, wait,

maybe different ed?

[iron bolt]

maybe

[ocean canyon]

I think I have 2^nd (E-book)

[iron bolt]

I have a pdf of the second edition, but all the topology is in the appendix

anyway

[ocean canyon]

wait, all the topology in appendix (!?), are you talking about the book "smooth manifolds"?

[iron bolt]

yes

[ocean canyon]

oh, no, haha, this is "topological manifolds"

😄

[iron bolt]

oh lol. I thought "Lee's manifolds" was unique

[ocean canyon]

haha, I made the same mistake when I took differential geometry, but the other way around

I ordered top. manifolds

[ocean canyon]

easy to mix up

[iron bolt]

by the way, I don't know if Lee covers it, but in my experience things like why concatenations of paths are continuous are very often glossed over

there's a useful lemma you can show that allows you to easily prove those things rigorously: if your space is covered by finitely many closed subsets and a map restricts to a continuous map on each of those, then it is also continuous on the whole space

works more generally for locally finite closed covers too iirc

[ocean canyon]

that sounds useful, I was about to ask about that earlier

oh wait, is this not the gluing lemma? Anyways, I think I need to relax. But thanks a lot for the input! 🙂

[iron bolt]

oh, neat. I didn't know it under that name

[ocean canyon]

I think also there is some elementary analysis fact I am forgetting for why we know that a minimal/maximal such element exists, and not just infimum/supremum; is it that a continuous map on a compact set attains its maximal and minimal element, applied to the special case of the points along the boundary of B? Well, yes, I suppose so, since the boundary is closed, so f^{-1}(partial B) is closed, so compact (prop. 4.36.(a)); so we can look at f restricted to f^{-1}(partial B), which is still continuous (the rest. of a continuous map is continuous)

[iron bolt]

the lemma you're referencing gives you a minimum/maximum with respect to an order on the codomain

[ocean canyon]

ops, yes

[iron bolt]

here we just have an order on the domain and a compact subset of it

[ocean canyon]

you are right

[iron bolt]

you're right that it is related though: that analysis fact is basically just the combination of the fact that continuous images of compact sets are compact and compact subsets of R have a minimum and maximum

[ocean canyon]

As I understand it now, it is enough that f^{-1}(∂B) is cpct, together with (something I seem to have forgotten; cpct subsets of R contains their infimum and supremum). I don't see where continuous images of a cpct set is cpct factors into it (perhaps I misunderstood you)

[iron bolt]

what I meant is that "continuous functions from a compact set to R take on a minimum and maximum" follows directly from their range being compact and compact subsets of R having minima and maxima

it's not relevant to the exercise though

[dapper kestrel]

I'm having problems solving this 3rd question, because if i'm not mistaken this topology would be the (lower) limit topology and every subset would be clopen? If that's the case then, (I think) every set would be it's own closure because is the smallest set that fully contains it? But then I don't know how to prove that every set it's clopen because closed sets "should" look like this $(-\infty, a) \cup [b, \infty)$, which the latter subset could be built from just the basis but the first one I don't know how, infinite union of a set whose lower limit goes to minus infinity I guess.

[gentle osprey]

pablo

[dapper kestrel]

Asked this in the wrong chat lol

[mighty hull]

No, every subset of R isn't clopen in the lower limit topology.

In general, the closure in a finer topology is a subset of the closure in the coarser topology.

[dapper kestrel]

How would a closed set look like in that topology then

If I take the complement of any basis element I get $(-\infty, a) \cup [b, \infty)$ the union of two sets I can build like this: $\bigcup_{n \in \mathbb{N}} [-n, a)$ and $\bigcup_{n \in \mathbb{N}} [b, n)$

[gentle osprey]

pablo

[dapper kestrel]

And those are the arbitrary unions of open sets then open too

[mighty hull]

You're over complicating it. The closure of A is the intersection of all closed sets that contain A.

What's the smallest closed set that contains (a,b) in the lower limit topology?

[dapper kestrel]

I'm guessing [a, b] but I how can I prove that that is a closed set

[mighty hull]

You did (2), right?

Forget about the guess

Is the lower limit topology finer or coarser? What does that mean?

[dapper kestrel]

Yeah lower limit topology is finer

[mighty hull]

What does that mean?

[dapper kestrel]

It has more elements than the usual topology

[mighty hull]

It means it has more open sets, yes. Everything open in the usual topology is open in this topology, and some other things are open, too.

[rancid umbra]

precisely, T is a subset of T_l

[mighty hull]

So do you know that [a,b] is closed in the standard topology? How?

If its complement is open in the standard topology then its complement is open in any finer topology, by definition.

What is the complement of [a,b)? Is it open in the lower limit topology?

[dapper kestrel]

Thats what I was saying, if this isn't wrong it is open

[mighty hull]

Yep. So [a,b) is closed.

And it contains (a,b)

Is there a smaller closed set that contains (a,b)?

[dapper kestrel]

I dont think so

[mighty hull]

There's only one possible smaller set

What set is that?

[dapper kestrel]

That is closed?

[mighty hull]

There's only one possible smaller set containing (a,b)

Maybe it's closed. TBD.

What's the one set contained in [a,b) that contains (a,b)?

And is it closed?

[dapper kestrel]

I can't think of one?

[mighty hull]

(a,b) contains (a,b)

It differs from [a,b) by a single point, so it's the only one

Is it closed?

[dapper kestrel]

Oh

[mighty hull]

It could be closed, this is a different topology.

But you know right away it has to be [a,b] or [a,b) or (a,b] or (a,b) because the topology is finer, so it's a subset of [a,b] containing (a,b) and there are only four of those

[dapper kestrel]

The complement would be (-oo, a] U [b, oo), which I think it would be closed because it isn't a basis element?

The first subset

[mighty hull]

That's the complement, but for (a,b) to be closed you need the complement to be open

[dapper kestrel]

So (a, b) would be open

[mighty hull]

It can be both open and closed

[dapper kestrel]

It isnt closed then

At least

[mighty hull]

Anyhow, I'll leave it to you, but (a,b) is open because the lower limit topology is finer and it's open in the standard topology. It's not closed, which you can prove if you need to.

Which means [a,b) is the closure of (a,b) in the lower limit topology.

Not [a,b]

[dapper kestrel]

Ok but does this prove that (at least) every basis element is clopen?

[mighty hull]

Yes.

[dapper kestrel]

Awesome, thanks for the help :)

[low flame]

so i was thinking of doing something like this

$\phi_1 = \min {k \in \mathcal(N) | U_k \subset W_1}$ then I was thinking for everything after this I can do the following

$\phi(n)$ when n >= 2 can be defined as the following $\min {k \in N | k > \phi(n-1) \text{ and } U_k \subset (U{\phi(n-1)} \cap W_n)}$

is this a good way to define this function phi?

[gentle osprey]

RealTek

[woven trail]

is this x cross x (x,x)?

I mean, it would make sense, but I generally see the cross representing being between sets and not elements

[plush folio]

Yeah, it's (x, x). I prefer this notation, x cross x looks kind of weird

[woven trail]

same

[plush folio]

Like (1, 2) is a point in R^2, writing 1 x 2 \in R^2 is cursed

[sonic crane]

Lol yeah definitely cursed

[inner cosmos]

Can anyone recommend any good books I can cite for n-dimensional geometry? Dont want to cite like 5 books. Anything that deals with euclidean like properties of n spaces and their topology

[quaint relic]

Is it true that in a compact T0 space there's a closed point?

[mighty hull]

Yes. If X is your space, consider the set {Cl({x}) : x ∈ X} of all closures of singletons in X.

[heady skiff]

can I get a hint for b, in particular showing that T is contained in T_F? I know this is Urysohn's lemma, but having trouble applying it

[quaint relic]

I'm sorry but how does that help?

[mighty hull]

Call that set F. Order the elements of F by set inclusion. Does it have a minimal element? What would a minimal element look like?

Also, consider compactness in terms of the finite-intersection property: https://en.wikipedia.org/wiki/Finite_intersection_property

[heady skiff]

as a sanity check, is the half open interval (a, 1] for a > 0 open under the subspace topology on [0, 1]? Since (a, 1] complement = [0, a] which is closed under subspace topology on [0, 1]

[quaint relic]

Ahhh you use Zorn to get that the set F has a minimal element using the finite intersection property?

[mighty hull]

Yes, the FIP tells us the intersection of everything in each chain is non-empty. That means each chain is bounded (below), so you can apply Zorn's lemma.

And then you have to prove that what you get from Zorn's lemma looks like {x}, for some x.

[mighty hull]

Correct. The open sets in the subspace topology look like the subspace intersected with open sets from the whole space.

In particular, (a,2) is open in ℝ, so (a, 2) ∩ [0,1] is open in the subspace topology on [0,1].

[quaint relic]

Thanks I didn't think of Zorn sooner

[heady skiff]

are the basis elements in R^2 under the product topology of the form (a_1, b_1) x (a_2, b_2)

[sonic crane]

yea

[heady skiff]

hm i'm a bit unsure though because doesn't the product topology (at least for finite products) have basis elements of the form U_1 x U_2 for U_1 open in T_1 U_2 open in T_2 blah blah blah

so U_1 open in R wouldn't necessarily look like an interval right, but an arbitrary union of open intervals

[sonic crane]

(a_1, b_1) x (a_2, b_2) stuff can generate all of that tho

also a basis for product topology can be basis of factors

U x V with U and V open are a basis but also Bu x Bv with Bu and Bv basis elements is also a basis

u can make all the U x V's with the basis products

[heady skiff]

i see, i think that makes sense

how would you achieve that tho? bc say we have U x V with U = arbitrary union of open intervals (and same with V)

i don't think we can write U x V = (a_1, b_1) x (c_1, d_1) U (a_2, ,b_2) x (c_2, d_2) x ..... where U(a_i, b_i) = U and U(c_i, d_i) = V right

or am i getting this wrong

[sonic crane]

Hm i dont think i understood what u wrote

[heady skiff]

i guess i'm confused about ths

[sonic crane]

((0,1) U (2,3)) x (0,1) = ((0,1) x (0,1)) U ((2,3) x (0,1)) for example

Products confuse me too. They suck

[heady skiff]

anyone?

[iron bolt]

I'm assuming the problem is that you don't know which two sets to apply the lemma to, right?

if you have an open set U, there is one closed set that immediately comes to mind

then to apply the lemma you just need to think of another closed sets that is disjoint from it. that's less easy - there's no canonical choice there

so what you can try instead is to apply the lemma to all possible sets, and see what you can construct from that. you will need to take a union at some point

[devout sorrel]

when we're taking the product topology, can we have empty sets at finitely many coordinates in the product and the full spaces in the rest of the coordinates, and this set will be open?

i.e. on $\prod_\alpha X_\alpha$, $\prod_\alpha U_\alpha$ such that $U_\alpha=\varnothing$ for $\alpha=n_1, \dots, n_N$ and $U_\alpha=X_\alpha$ otherwise

[gentle osprey]

CoolShot

[devout sorrel]

how do you even write down an element belonging to this set?

i think this set should not be well defined?

or wait

that set is just the empty set if even one U_a is empty

yeah idk what i was thinking i forgot how regular cartesian product works..

[vital river]

is the set of limit points the same as the boundary?

because I have seen formulations of the closure as $\bar A = A\cup \partial A$

[gentle osprey]

Kakaka

[vital river]

where $\partial A$ is the boundary of the set A

[gentle osprey]

Kakaka

[alpine nest]

That's an equivalent definition/characterization of the closure, but the fact that $A\cup B$ and $A\cup C$ are equal doesn't imply that $B = C$.

[gentle osprey]

Outsider

[alpine nest]

(also please don't ask the same thing in multiple channels)

[prime elbow]

Let K be a compact subset of metric space X and {U_i | i in I } are open cover for K. Prove that there exists eps >0 such that any ball B(x, eps) is contained in some U_i.

Yes I think eps is Lebesgue's number but I want to prove the existence of Lebesgue's number, maybe it is better if we prove for a subset E which has a diameter diam (E) < 2eps, then E contained in some U_j

I don't see how contradiction works here

[nocturne matrix]

https://math.stackexchange.com/questions/1830929/in-t-1-space-all-singleton-sets-are-closed

Why does the author of the answer claim that U is the complement of {x}? Isn't U just a subset of said complement?

Oh no nvm I got it; it can't be a subset because no point other than x are left out of that union

[iron bolt]

if you let r(x) be supremal radius such the B(x,r) is contained in a set from that cover, is there anything helpful you can say about that as a function of x?

[heady skiff]

ya, that being the complement, but kinda tricky finding out the other set

i guess i'll ponder some more for a bit and then giv eup

idk maybe singletons in U

disjoint from the complement of U

[iron bolt]

that would work I think

you just need to figure out what to do with all of them

[heady skiff]

yeah so I know if i can express it as an arbitrary union of finite intersections of preimages of open sets in [0, 1] then I'm done

i think

[iron bolt]

yep

[heady skiff]

so maybe define a function for each singleton in U or something

which maps each singleton to 0

and then consider the intersection of their preimages of [0, 1]??

well my idea was to send all of them to 0

via a continuous function f_u

for each u in U

and then look at f_u^{-1}({0}) but that's not an open set in [0, 1]

wait nvm it is

(0, 1] is its complement

right?

[iron bolt]

nope, that's closed

[heady skiff]

wait

oops i'm tripping lmao

bruh this is an exam problem and i've spent 3 hours on it

[iron bolt]

any open set in [0,1] where you know that the preimage is contained in U?

[heady skiff]

well the open sets are of the form (a, b), [0, b), (a, 1], [0, 1], and to map elements to those sets I want to use urysohn's lemma...

but urysohn's lemma only guarantees elements sent to 0 and 1, so I'm having trouble here

i was thinking about this earlier on

oh wait

[0, 1/2)

and we can send element sof X/U

to 1

so they can't live in [0, 1/2)

or the preimage of it

[iron bolt]

yep

[heady skiff]

LET'S GOOO

ok now the problem is trying to define it for each u

because that would give a possibly infinite intersection

[iron bolt]

even [0,1) works, it just can't contain 1

[heady skiff]

oh wait i don't even need to define it on all of U

or for each singleton

because the preimage must contain U

wait i don't know if it must equal U th

o

[iron bolt]

the preimage is contained in U, but could be a lot smaller

since parts of U can get mapped to 1 too

what can you do with open subsets of U like that to get closer to it?

[heady skiff]

sorry wdym by this?

does this have to do with T being regular maybe

since normal + hausdorff => regular

[iron bolt]

you want to construct U from those subsets to show that it is open in T_F

for each u in U, you already have constructed an open subset of U that contains u

open in T_F I should say

[heady skiff]

damn I really want to repeat this construction for all $u \in U$ to get $\bigcap_{u \in U} f_u^{-1}\bigl([0, 1)\bigl)$ or something but I know this wouldn't work since it would be an infinite intersection

[gentle osprey]

okeyokay

[iron bolt]

you're really close, only caught up in one simple detail xD

[heady skiff]

i don't know, take some finite subset of U??

then consider intersection over that finite subset of the functions f_u?

[iron bolt]

intersecting sets makes them smaller, not larger

[heady skiff]

uhhhh

[iron bolt]

the intersection might even be empty

you want to do something else with all the sets instead (and yes, for all u)

[heady skiff]

oh unions lmao

bruhhhh

every time i consider unions it happens to be intersections and vice versa 😭

okay lemme see this

oh wait yeah this works

FINALLY

thank you so much

[mighty hull]

You could also use the universal property of the initial topology, I think.

It's the unique topology with the property that, for any topological space $S$, a function $g: S \to X$ is continuous if and only if $f \circ g: S \to [0,1]$ is continuous for every continuous function $f: X \to [0,1]$. Then prove that the topology on $X$ has that property.

[gentle osprey]

Cufflink

[mighty hull]

Something like, for $s \in S$, take a neighborhood $U$ of $g(s)$. Urysohn's lemma then gives you a continuous function such that $f(g(s)) = 0$ and $f(x) = 1$ for all $x \in X \setminus U$.
\
Then because $f \circ g$ is continuous, look at the preimage of $[0,1)$ under $f \circ g$.

[gentle osprey]

Cufflink

[heady skiff]

oh that's neat, I didn't think of that

i guess it makes since that it has a universal property then

[mighty hull]

Any time you hear the word "initial" or "final", there's a universal property at hand. Same with "smallest blah blah" and "largest blah blah"

https://ncatlab.org/nlab/show/separation+axioms+in+terms+of+lifting+properties

[shrewd night]

I used the hint they gave

But I can't seem to picture what this E set would look like

[mighty hull]

TBH the contrapositive is easier to prove, here.

But say you have a limit point x of E. What can you say about any neighborhood of x?

And remember that each F_i is closed.

[shrewd night]

I have a feeling it has something to do with, well obviously every neighborhood of a limit point must contain a point in E but something would be wrong if that were to be the case

[mighty hull]

Keep in mind that F_{i+1} is a subset of F_i.

[fierce lily]

Question: 1. Can I treat this long line as a real numbers line? 2. How can I prove this line is not metrizable?

[iron bolt]

the long line is not the real number line, no

it's longer

[fierce lily]

Then how will you describe it as a topological space

[iron bolt]

in the text you're using, how was the long line defined?

[opaque scroll]

omega_1 x [0, 1) with the order topology for example

[fierce lily]

it does not give me any def in my text.

[fierce lily]

omega_1x [0,1) refers to?

[iron bolt]

rip. weird that it gives that as an exercise than

but I guess it might just be assumed knowledge

[fierce lily]

our instructor style is pretty free

[opaque scroll]

[0, 1) is the half open interval, omega_1 is the first uncountable ordinal

x product with lexicographic ordering

[iron bolt]

I think it's a slightly tricky exercise if you haven't seen ordinals before, because I think you do need some amount of induction over ordinals to really get started

the most important thing to know is basically that for every point x > 0 in this long ray (let's call it L), the interval [0,x) is homeomorphic to [0,1)

but the space as a whole isn't, because it is too long

by 0 mean (0,0) ∈ L here btw. it's the first point in that ray

the long line is technically not this ray, but what you get by gluing two copies of it together at 0 - much like R can be viewed as two copies of [0,∞) glued together

but it doesn't really make a difference for most topological properties

[opaque scroll]

I guess you could also just remove 0

[iron bolt]

then you only get a half-long line

[opaque scroll]

Right, I see

[mighty hull]

Strongly recommend this book: https://en.wikipedia.org/wiki/Counterexamples_in_Topology

What theorems about metrizability do you know?

[mighty hull]

Any theorem of the form "Here are five different characterizations of Concept X and in a metric space they're all equivalent" is one possible avenue towards proving a space isn't metrizable.

For example, if the space satisfies one of those characterizations but not another.

[fierce lily]

for metrizable, I know that it must be a topology induced by the distance d on a set X, namely any basis for this topology is of the form B_d(x, epsilon) for each x in X. that's the thing I know only I guess

[fierce lily]

about this part, can you point out the page of this book?

[mighty hull]

What book?

You mean Counterexamples in Topology? I was just sharing that as a catch-all reference when you're looking to understand what spaces have properties X and Y but not Z and why.

[mighty hull]

What book are you using?

[fierce lily]

I am using james munkres.

[mighty hull]

Is this an exercise from the book or something your professor gave you?

[fierce lily]

it is the practice our instructor gives

[mighty hull]

Without pulling a rabbit out of the hat, I have two suggestions:

1. Find all instances of "long line" in the index of Munkres and look at what he says
2. If there's any special notation Munkres uses to talk about the long line, look that up in the index, too
3. Find a digital copy of Munkres and search through it for "metrizable space", "are equivalent", "long line", etc.

Countexamples in Topology is nice and would help here. For example, the index has this chart. The numbers along the left-hand-side refer to specific topological spaces.

Also, Munkres has several exercises about the long line that give more context. It might be helpful to do those, but you should at least look at them.

[long finch]

how do i know wtf im doing

[sonic crane]

Faaax bro

Preach it bro

[gritty widget]

b part, claim: G is not required to be open.

Let the open ball B(x, r) where r > 0. If B(x,r) has non-empty intersection with G then G cup Ext(G) has non-empty intersection with B(x,r).

If B(x,r) has an empty intersection with G imply x in Ext(G). Hence G cup Ext(G) has a non-empty intersection with B(x,r).

In both cases, G cup Ext(G) has a non-empty intersection with an open ball.

Since Ball is arbitrary thus it would intersect with every non-empty open set.

[gritty widget]

To prove: The finite union of nowhere dense subset of metric space M is nowhere dense.

Proof: We will prove the case when n = 2. Assume A and B are nowhere dense. By definition of nowhere dense, M\cl(A) is non-empty open dense subset of M.

Result: intersection of two non-empty open dense subsets is dense subset.
Thus, X(clA cup cl B) = X(cl(A cup B) ) is a dense subset in M imply A cup B is nowhere dense.

[heady skiff]

closed subsets of regular spaces need not be compact, right?

[iron bolt]

if they were, that would in particular mean that all regular spaces are compact

I don't think they are

[heady skiff]

yeah makes sense

does anybody have an idea about what the resulting picture/object would be?

i'm used to identifying ends of the unit interval so this is a bit of a weird construction for me

[iron bolt]

I mean, what are the points of Y? what are the open sets? it's a finite topological space (hope that wasn't a spoiler), so it shouldn't be hard to describe

once you have that we can talk pictures :>

[heady skiff]

ye part of the question was describe the open sets lol

working on that now

oh wait we just identify the whole entire open interval (0, 1) so it should be a 3 point space no

[mighty hull]

There's one point per equivalence class.

[heady skiff]

ye and there are three equivalence classes right

{0}, {1}, and (0, 1)

[mighty hull]

Mmmhmm

[heady skiff]

let me make sure i have definitions right - is the quotient topology on $Y$ the collection of all sets such that $U \subseteq Y$ is open if and only if $\pi^{-1}(U)$ is open in $X$?

[gentle osprey]

okeyokay

[iron bolt]

it's the set of all sets whose preimages are open, yes

[heady skiff]

okay, I guess i sometimes get it confused with quotienting out points of X

but i guess this is a more general formulation

[iron bolt]

which is equivalent to saying that under that topology, a set is open iff its preimage is

[heady skiff]

ah i think i got it now

oh wait so determining the open sets in X/~ under the provided equivalence relation is basically determining the open sets in a set of order 3

hmmm so I think the open sets should just be $X / {\sim}$, $(0, 1) \cup {0}$ and $(0, 1) \cup {1}$ no?

[gentle osprey]

okeyokay

[heady skiff]

wait no

wait yeah

[iron bolt]

almost

[heady skiff]

and empty set lmao

because if i'm not missing something, $\pi^{-1}\bigl((0, 1) \cup {0}\bigl) = [0, 1)$ which is open in $[0, 1]$ right?

[gentle osprey]

okeyokay

[heady skiff]

because 1 is just mapped to itself under the quotient map

[iron bolt]

that and some technicalities - the sets would be {(0,1),{0}} and {(0,1),{1}} since (0,1), {0} and {1} are elements of X/~, not subsets of it

[heady skiff]

so it cna't be in the preimage

oh yeah

[iron bolt]

oh, and there's one more set that's open

[heady skiff]

uhhhh i already said the whole space right

lemme think

it wouldn't be the singletons

not the collection of singletons either

[iron bolt]

why not?

[heady skiff]

well because the preimage of {0} is just {0} right

which is not open in [0, 1]

unless i'm missing something

[iron bolt]

{0} is an element of your space, {{0}} is a singleton in it :p

but yes

the other singletons are {(0,1)} and {{1}}

[heady skiff]

right my bracket game is p bad

[iron bolt]

it's fair, working with sets of equivalence classes like that can arguably get a bit confusing

[heady skiff]

so it would just be: $\varnothing, X / {\sim}, {(0, 1), {0}}, {(0, 1), {1}}, {(0, 1)}$ i think

[gentle osprey]

okeyokay

[iron bolt]

[heady skiff]

huh that wasn't as interesting as i though ti twould turn out

[iron bolt]

now to answer your original question about the picture: what you've done here is collapse the interior of the interval to a single point, right?

[heady skiff]

yes

[iron bolt]

so now you have three points, two still corresponding to the boundary of the interval and one representing the entire middle part

since the middle part of the interval gets in a sense infinitely close to the boundary - every neighbourhood of 0 intersects (0,1) - every neighbourhood of {0} in the quotient space contains the point (0,1)

similarly for the other end

[heady skiff]

okay wait i feel like this has to do with the next question they ask, which is is the quotient topology hausdorff

[iron bolt]

yep

is it?

[heady skiff]

no, i don't think so

yeah that makes sense i guess, let me try to formalize

thanks for the help!

[iron bolt]

I think there is some nice non-rigorous visual intuition to be found here btw, so I will still elaborate a bit :p

I think the most visual explanation of what an open set is, aside from the formalism, is that it is a set where for every element in it, the complement of the set can't get arbitrarily close to it

so the neighbourhoods of a point are exactly those sets whose complements don't get arbitrarily close to it

if a point x is contained in every neighbourhood of a point y, this means that {x} gets infinitely close to y - so in this sense, it means that x is infinitely close to y

the slightly wild part is though, this is not a symmetric relation

in the case of the exercise you're doing, (0,1) is infinitely close to both {0} and {1}, but not the other way around

soo... that shows you that non-T1 topologies can be wild I guess. but I still like that there's at least some visual understanding to be found, even if it's not the most intuitive

[iron bolt]

(if you want to how well it behaves exactly: you can show that this relation if reflexive and transitive, just not symmetric, and so it has more the vibes of a partial order than of an equivalence relation)

[heady skiff]

thank you! i'll read this in a bit once i finish this exercise

[devout sorrel]

say we have the set [a, b] subset X with order topology.
if A subset [a, b] is open (in [a, b]), then there exists an interval (c, d] subset A

can someone explain why this is true? i dont understand because (c, d] is not an open set in [a, b], right?

[iron bolt]

the empty set is certainly open and doesn't contain any non-empty interval, so there's probably some context missing from your question

[devout sorrel]

suppose non emptiness

[devout sorrel]

yes its a specific line from a proof in munkres im struggling with

suppose $B_0$ is a nonempty open set in $[a, b]$. we have a $c \in B_0$, then it follows from $B_0$ being open that there is a $d$ such that $(d, c] \subseteq B_0$

[gentle osprey]

CoolShot

[devout sorrel]

(with order topology)

[iron bolt]

c is probably required to be > a, and d to be < c?

[devout sorrel]

yes, sorry, we do have that

it reads like it claims (d, c] is open in [a, b] which i dont get

[iron bolt]

yep, it indeed is not

or at least, doesn't have to be

but there is an open set containing it that is contained in B_0

[devout sorrel]

if c = b it checks out but what about c < b

oh, b in B_0 also

[iron bolt]

I don't think that is even required

[devout sorrel]

yes but if c < b then say we have (d, e) subset B_0, if c > e it doesnt work correct?

[iron bolt]

do you know what specific neighbourhoods of points under the order topology look like? like, in R^n, if a point is contained in an open set then it's also contained in an open ball contained in that set - can you say something similar about the order topology?

[devout sorrel]

open intervals we can take as a base, so that?

[iron bolt]

open intervals and open rays like (a,∞), yes

but in this specific case (c < b) you can actually show that c is always contained in an open interval

and I mean, once you have that c is contained in an open interval in B_0 you're basically done, right?

[devout sorrel]

oh

B_0 is open so c has (d, e) around it when c < b so (d, c] subset B_0

[iron bolt]

yep

there is only one more subtlety here

you know that B_0 is open in [a,b] with the subspace topology, but what we just said is true for the order topology only

has the text you're working with already shown that on subsets intervals in ordered sets the order topology and subset topology agree, or do you need to be careful with that here?

[devout sorrel]

the topologies are equal when convex

and intervals are convex, so it should work

thank you for guiding me through this lol i really appreciate it

[wise mist]

Let $ f: \mathbb R→ \mathbb R $ be a continuous function and let $S$ be a non-empty proper subset of $\mathbb R$ . Which one of the following statements is always true?

\ \

A. $\operatorname{int}f(S) \subseteq f(\operatorname{int}S)$

B. $f(\overline S) \subseteq \overline {f(S)} $

C. $f(\overline S) \supseteq \overline {f(S)} $

D. $\operatorname{int}f(S) \supseteq f(\operatorname{int}S)$

[gentle osprey]

yeshua

[rancid umbra]

have you tried with any examples?

[wise mist]

negated D

[rancid umbra]

? i mean examples of functions

and varying S

[wise mist]

yeah it was some counter example

$f=x^2, S=[-1,1], intS=(-1,1), f(S) = [0,1], f(intS)=[0,1), intf(S)=(0,1)$

[gentle osprey]

yeshua

[wise mist]

this ain't faulty right?

[wise mist]

Let $f:X\to Y$ be a continuous map between metric spaces. Then $f(X)$ is a complete subset of $Y$ if \

A. the space $X$ is compact\

B. the space $Y$ is compact\

C. the space $X$ is complete\

D. the space $Y$ is complete

[gentle osprey]

yeshua

[sonic crane]

Was so confused by this too, but yeah, its not claiming (d,c] is open in B0 which is what i thought it was saying at first

[gritty widget]

a, f(X) is compact implies it is complete

[sonic crane]

That whole proof for connectedness of R was honestly confusing

[gritty widget]

if you can show for open interval (0,1) then you can easily get insight for R, refer Topology without tears

[sonic crane]

The proof for that is the same isnt it

With the sup and all that

[gritty widget]

yes

[sonic crane]

I just find that proof confusing

[gritty widget]

but Topology without tears has better proof for it

[sonic crane]

Hmm ok. Ill chrck it

[wise mist]

i completely forgot the definition of complete set

[gritty widget]

every cauchy sequence is convergent sequence

[wise mist]

right

the converse is true, hmm now i can recall

[alpine nest]

In a metric space, a convergent sequence is always Cauchy.

A Cauchy sequence isn't guaranteed to be convergent, unless the space is complete.

I.e. "convergent -> Cauchy" holds always, "Cauchy -> convergent" holds if your space is complete

[wise mist]

correct

also, A continuous function on a compact set produces a compact image.

[gritty widget]

yes

[rapid lagoon]

I don't know exactly where this goes, because my question is a bit more general

But, first a motivating example. We say that the subspace topology of A \subset X is the coarsest topology that makes the inclusion continuous

The product topology is the coarsest topology that makes the projections continuous

Yet, the disjoint union topology is the finest topology that makes the natural inclusion continuous

When is it the coarsest, when is the finest?

As in, suppose we have some topological space $X$, and another set $Y$, with a natural mapping $f: X \rightarrow Y$. Would the topology that is induced by X be dependant on whether f is a surjection or an injection?

[gentle osprey]

dulg_n

[dire dove]

More or less yes

I mean not exactly

When you have a map (or a bunch of maps) out of a set you want the coarsest topology usually

Because the discrete topology trivially makes any map out of the set continuous

[mighty hull]

Sure, consider if Y is a one-point space.

[dire dove]

And a finer topology makes it easier in general to have continuous maps out of it

[gritty widget]

let (X, T_1) and (X,T_2), T_1 is finer than T_2 if and only if f: X -> X , x -> x domain with T_1 and co - domain with T_2, f is continuous

[dire dove]

So you care about the other extreme

When you have a map into the set the interesting extreme is the finest topology (because again, a coarser topology makes it easier for maps into your set to be continuous, i.e. they are less interesting)

[dire dove]

(and the indiscrete topology which is the coarsest makes everything continuous)

And indeed you can see that for both product and subspace you want a topology on a set from which you have a map coming out, i.e. you want the coarsest, while for the disjoint union you have the inclusions which are into it, and so you want the finest topology

[rapid lagoon]

Ah I see

Thqt actually makes a lot of sense

[mighty hull]

What you're looking at is the difference between the "initial topology" and the "final topology".

If we have a space X we can generally look at functions that map into X (i.e., where X is the codomain) or functions that map out of X (i.e., where X is the domain).

If we're looking at function that map out of X then the finest topology where those functions continuous is always the discrete topology, because every function out of X is continuous if X has the discrete topology and it's the finest possible topology.

So it's only interesting to ask about the coarsest topology such that blah blah blah.

And the converse is the case when we're looking at maps into X, but with the trivial topology instead of the discrete topology.

https://en.wikipedia.org/wiki/Initial_topology

https://en.wikipedia.org/wiki/Final_topology

[rapid lagoon]

I don't mind if you guys started talking in category theoretic language

Because I know that the discrete topology is the initial object, and indiscrete is the final

[rapid lagoon]

Hold on, is that even true?

For example, suppose $X \subseteq Y$, with $X$ being a topological space. Wouldn't the superset/ancestor topology on Y be the finest topology that makes the inclusion continuous?

[gentle osprey]

dulg_n

[rapid lagoon]

So here, we have a mapping going out of X, yet we require it to be the finest

Or is this example wrong?

[dire dove]

You want the topology on Y

And the map is into Y

[rapid lagoon]

Yes

[dapper kestrel]

Can I interrupt you guys to make a question or do I wait lol

[rapid lagoon]

dont ask to ask, just ask

[mighty hull]

It's not like anywhere you see X it will work, X is just a placeholder.

[dapper kestrel]

I mean it was more about being polite about your conversation lol

[rapid lagoon]

omg

[dire dove]

X is "the thing you want to put a topology on"

[mighty hull]

🙂

[dapper kestrel]

How can I prove that $f(\sigma) = 2 \sum_{j \in \mathbb{N}} \sigma_j 3^{-j}$ is continuous from $X = {{0, 1}}^\mathbb{N}$ to $[0, 1]$, $X$ with the product topology, $f(X)$ is the Cantor set (which I havent proven yet) and what I was thinking was given a (a, b) open interval in [0, 1] and say that I would get really close to a with finite indexes, finite zeros and ones and then the other indexes to be determined "could reach" to b and that would be an open set in the product topology, I dont know how to formalize that though.

[rapid lagoon]

Actually, no, my point still stands

[gentle osprey]

pablo

[mighty hull]

The discrete topology is finer than the topology you described.

Discrete topology = every subset is open

[dire dove]

In this case you want Y to be an initial object

[mighty hull]

oh no nvm

Uhh what I said originally was correct.

[dire dove]

I.e. among the ones with a map coming from X you want the initial one

[mighty hull]

Y is the space that doesn't have a topology, so we're looking at maps into Y.

[dire dove]

I guess the confusion is that among the objects with maps into them we want the initial one, that is one with maps from it

[mighty hull]

You know at some point you're going to have to ensure $\left|f(\alpha) - f(\beta)\right| < \varepsilon$, where $\alpha, \beta \in \left{0,1\right}^{\mathbb{N}$.

\\

To get some ideas, I might start by trying to estimate $\left|f(\alpha) - f(\beta)\right|$ for two arbitrary elements of $\left{0,1\right}^{\mathbb{N}$.

[gentle osprey]

Cufflink
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[dapper kestrel]

Sorry for the late reply but wouldnt I need to define a metric in the product topology in order to say that alpha and beta are near?

(I was in class :P)

[dapper kestrel]

Or just taking an open set in product topology would sufice

[mighty hull]

No, f(α) and f(β) are real numbers in [0,1]

[dapper kestrel]

I was thinking about the epsilon-delta definition

[mighty hull]

Yes. You can't use the metric on both sides, but if the codomain is a metric space then the idea is:

Given ε > 0, find an open set U such that for all α,β in U, we have |f(α) - f(β)| < ε

Then think about what it means for α,β to be in the same open set.

I understand prompts like this can be annoying because you don't see exactly where it will lead, but when you're dealing with novel/unfamiliar problems you have to strike out in any direction where you can plausibly get traction.

I'm willing to tell you exactly what to look at, if you want. Just let me know. But the "hard part" of any novel problem is figuring out what to look at and what perspective to take while looking at it.

[dapper kestrel]

Not at all, I understand. I appreciate a lot your help and I think I got this problem, when I get home Ill redact it properly :)

[rain nova]

$(X_1, T_1)$ and $(X_2, T_2)$ be topological spaces and let $f:;X_1\xrightarrow{} X_2$ be a bijection. Then $f$ is homeomorphism \emph{if and only if} $f$ sends $O_1\in T_1$ to $f(O_1)\in T_2$.
\
\emph{$(\longrightarrow)$ has been proved by myself so I omit here}
\
For the $(\longleftarrow)$ direction, we only need to prove $f$ and $f^{-1}$ are continuous. Consider $f^{-1}$, let $U\in T_1$ be arbitrary open set. Since $f(U)\in T_2$ and $f=(f^{-1})^{-1}$ it will eliminate the effect of $f^{-1}$ and enable us to get the inverse image of $U$, which implies that $f^{-1}$ is continuous. Now consider $f$. Since $U\in T_1$ correspond to an \emph{unique} $f(U)\in T_2$ and by bijective,$f^{-1}$, $f=(f^{-1})^{-1}$ has the similar effect which let every $f(U)\in T_2$ correspond to an \emph{unique} $U\in T_1$, which implies $f$ is continuous.

I would like to ask is this argument valid? Thanks for the opinion!

[gentle osprey]

Tanxs

[round oyster]

Let $(E,T)$ and $(F,T')$ be two toplogical spaces, and let $f : E \rightarrow F$ be a continious map, show that the image by $f$ of a dense subset $A$ of $E$ is a dense subset of $f(E)$.

[gentle osprey]

MadAbdo

[iron bolt]

that's not a question, that's the statement of an exercise

what did you try? how was "dense" defined in the text you're working with, and what does the exercise look like if you expand out that definition? do you see where a proof could go from there?

[real star]

once you know the trick it is really easy

[real star]

if you use this defintion, I suggest you try using statement 5.

[iron bolt]

I was thinking of statement 6, but it's basically the same

1 too

[real star]

they all are actually

LOL

[craggy cedar]

Not sure if you’ve solved it by now, but think about the literal definitions of continuity and density

[quick crane]

this is trivial with nets too

[iron bolt]

😂

gotta love it when somebody asks a question, and the entire chat just decides that no matter how you do it it is trivial

they'll surely love to hear that

[round oyster]

nets ?

i wrote a proof, yet not sure of it

would you like to hear it?

[real star]

sure

[real star]

go ahead

[round oyster]

$\bar{A} = E \implies f(E) \subset \bar{f(A)}$ \
take $x \in \bar{f(A)}^{c}$
\begin{align*}
&\implies \exists U_{x} \in V(x), \quad U_{x} \cap f(A) = \phi \
f^{-1}(U_{x} \cap f(A) ) \subset f^{-1}(U_{x}) \cap
f^{-1}(f(A)) \
\implies f^{-1}(U_{x}) \cap A = \phi \
\end{align*}
We have $f^{-1}(U_{x})$ is open of $E$ and it's neighborhood
of $f^{-1}(x)$.
[
f^{-1}(x) \notin (\bar{A} = E) \implies
x \notin f(E) \implies f(E) \subset \bar{f(A)}
]

[gentle osprey]

MadAbdo

[round oyster]

does this work?

[quick crane]

yeah suppose $y = f(x) \in f(E)$. Take a net $x_{\alpha} \in A$ with $x_{\alpha} \to x$. Then $f(x_{\alpha}) \in f(A)$ and $f(x_{\alpha}) \to f(x)$.

[gentle osprey]

L

[real star]

I would've taken x in f(E), take any open neighborhood U of x and take the preimage which is open.
f^(-1)(U) intersects A (by density), so f(U) intersects f(A)

I just have trouble following along this lol because my proof is simpler

[round oyster]

oh this works nicely

thanks for your help, maybe some one will review my proof later

[round oyster]

net? is that a sequence ?

[quick crane]

a sequence is a net, but nets include things other than sequences. For metric spaces, sequences are enough to characterize the closed sets, but for general topological spaces, you need nets to characterize the closed sets in a similar way that sequences do for metric spaces.

[round oyster]

oh, from my french lecture, i think they call these "nets" sequences, because they were defined along other toplogy definitions, as if a sequence converge to x_n in A, and x_n -> x, => x in closure of A, and things like that

[real star]

lol

https://en.wikipedia.org/wiki/Net_(mathematics) this has an example on R^R where sequences aren't enough to describe a closure

[quick crane]

from "Real Analysis" by G. Folland.

[round oyster]

Oh wait, i thought the sequences definition works all the time?

ist not a necessary condition?

[real star]

sequences is enough for metric spaces as L said, in non-metric spaces you need to generalize to nets

[plush folio]

More generally in sequential spaces, which includes first countable spaces

[iron bolt]

sequential spaces are cool because they're the coreflective hull of the extended natural numbers

that gives them some neat categorical properties, and also makes them seem really fundamental in a way

[alpine nest]

The cultured person's uwu.

[alpine nest]

Sadly the weak or weak* topology is rarely sequential

[alpine nest]

(but also you're often specifically interested in weak or weak* convergence of sequences, i.e. sequential closure matters more than topological closure)

[rain nova]

I would like to ask is $\emph{identify function}$ on X continous automatically or it depends on the topology assigned to the set X?

[gentle osprey]

Tanxs

[shrewd night]

For exercise 27, what are your ideas on showing for all x in P, x is a limit point of P

I've proved everything else here

[faint gyro]

https://math.stackexchange.com/questions/1961656/example-of-an-identity-function-thats-not-continuous depends on the topology

does this theorem have a specific name? or is this just another theorem in a long list of theorems in topology

[red yoke]

Complete metric spaces form a reflective subcategory of the category of isometries

[faint gyro]

lol i think i'll keep this theorem nameless then

[alpine nest]

Well, the concise phrasing of the theorem is that "every metric space has a completion" (https://en.wikipedia.org/wiki/Complete_metric_space#Completion), but it doesn't have any better "name" than that.

[shrewd night]

Hello?

[mighty hull]

hi

[alpine nest]

There's some discussion of that problem here: https://math.stackexchange.com/questions/1386387/the-set-of-all-condensation-points-is-perfect

[round oyster]

Ist possible to prove that
$\mathring{A} \subset A$ without using the definition, there exist an open for which is in A and contain a point x

[gentle osprey]

MadAbdo

[mighty hull]

Is that the interior of A?

[alpine nest]

To prove anything about the interior you need to have some definition of the interior, and invoke that definition somehow (directly or indirectly)

[mighty hull]

Yeah. The interior is the union of all the open subsets of A.

[alpine nest]

Another definition of the interior is that it's the union of all open subsets of A, and from this definition the inclusion follows trivially

[mighty hull]

The empty set is open, so every set has at least one open subset.

[round oyster]

yes

also i have another question, about $A \subset \bar{A}$, following from this result, it implies that every point $x$ of $A$, the set of neighborhoods of $x$ is not empty, which means that every point of our set X, has an open subset that contains it

[gentle osprey]

MadAbdo

[faint gyro]

yea, X contains every point x in X and X is open

[alpine nest]

You're skipping too many words so I'm not entirely clear on what you mean.

[mighty hull]

This is why we demand the empty set and the whole space be open.

So that every subset of a topological space has at least one an open subset and one open superset.

That also means every subset of a topological space has at least one closed subset and one closed superset, so we can talk about

1. The interior, i.e., the union of open subsets of A
2. The closure, i.e., the intersection of closed supersets of A

[round oyster]

because if you consider any arbitrary point, $x$ of $A$ there exist an open neighborhood of $x$, because since $A \subset \bar{A}$ means all points of $A$ are already in the closure, so this means atleast one neighborhood of each point of $A$ must exist

[gentle osprey]

MadAbdo

[alpine nest]

Every element of A has an open neighborhood, but that neighborhood might not be a subset of A

[round oyster]

This makes it clear, i always not consider that the empty set, the set it self are open

[mighty hull]

Every point in A is contained in some open set in X, it's true, but that open set need not be a subset of A

[round oyster]

yes sure, those points cannot be in the interior, but they can be in the closure, from the definition of closure we only consider the set of neighborhoods V(x), not the once that are contained in A , since X \in V(x), always X \cap A != phi

[gaunt drift]

yes, in closure we check for all neighborhoods of x if they intersect A for x to be in the closure, while the interior is the union of all open subsets of A

[shrewd night]

I read through it and I have a hard time understanding that

[mighty hull]

What ideas do you have for proving every point is a limit point?

[shrewd night]

My first thoughts are this. I'm thinking if we have to show if there is a neighborhood with no points in P, you'd get a contradiction

[mighty hull]

Yeah, always a good idea to assume the opposite is the case and see what breaks down.

If x ∈ P is not a limit point of P (i.e., it is an isolated point) what does that mean?

[shrewd night]

||Then there would be a neighborhood D of x where D int P = x||

[shrewd night]

||Is it right to say that as E is uncountable and P^c int E is at most countable, could we say that P^c would be at most countable and then P would be uncountable||

Not that it could get anywhere

[mighty hull]

I'd go a more direct route.

Like you said, if x is isolated then you have some open neighborhood V of x such that V∩P = {x}. V contains uncountably many points of E, directly from the definition.

Let U = V\{x}. What can you say about neighborhoods of points in U?

Keep in mind that U∩P is empty.

[shrewd night]

||That they also must have a neighborhood which when intersected with P must also be empty||

Surely that has to be the case

[mighty hull]

Maybe, but I'd still stick w/ the immediate definitions in hand.

Can a point in U have a neighborhood that contains uncountably many points in E?

(Write out what it means for U∩P to be empty. P is the set of points such that XYZ. Therefore no point in U has property XYZ.)

[shrewd night]

Oh yes of course

[shrewd night]

I'm just trying to work through Rudin's book

[mighty hull]

Yep.

[shrewd night]

This chapter has just been pain

[mighty hull]

Just saying, for a lot of these problems it's a matter of unwinding the definition and putting it in front of your face.

Thinking you have it in your mind's eye isn't enough.

[shrewd night]

That's what happened when I did the other part of this question

Problem*

[mighty hull]

The idea here is you'll be able to cover V with a countable collection of countable sets, but V is uncountable.

(I'll leave you to fill in the rest.)

[rain nova]

Thanks

[mighty hull]

Or another way to think about it is that: each point y of V has an open neighborhood N_y that only contains countably many elements of E. By definition, there's a basic open neighborhood of y contained in N_y.

But there are only countably many basic open neighborhoods, so...

[shrewd night]

How I did it was that I went back to $W$. Every point $y$ of $V$ is not in $P$ so it must then be in $W$, which also means $V$ is a subset of $W$. Keep in mind $W \cap E $ has at most countable points of $E$. This means also that $U \subset W \cup {x}$.

Then take both sides with an intersection of $E$, so $U \cap E \subset {x} \cup (W \cap E)$. Now $W \cap E$ has at most countable $E$, meaning $U \cap E$ has at most countable $E$. However, $x$ was in $U$ and was a condensation point, which provides a contradiction. @mighty hull

[gentle osprey]

s1yuan

[mighty hull]

That's a little too big hah

[gentle osprey]

s1yuan

[dapper kestrel]

How can I justify writing the open neighborhood of an element as a basis element

If I have an open neighborhood of x then I have a basis element that also contains x and is fully contained by the neighborhood?

Does that make sense?

[mighty hull]

Yes, that is what I mean. This is just true in general: if U is an open set containing x then there exists a basic open set B with x ∈ B ⊆ U.

That's sometimes the definition of a base for a topology.

[dapper kestrel]

Yeah it makes sense, thanks. I didnt get to writing the proof yesterday but today I will 😅

[dapper kestrel]

Talking about the exercise I sent yesterday :P

[shrewd night]

That's exactly the definition of the base in Rudin's book

[mighty hull]

Yeah, the other (equivalent) definition is that a collection of open sets form a base if every open set can be written as the union of basic open sets.

If an open set V can be written as the union of basic open sets then any element of V is in at least one of those basic open sets. Conversely, if every x in V is contained in some basic open set B(x) ⊆ V then V is the union of all all such B(x).

[shrewd night]

Yeah that's a theorem Rudin used

I'm assuming you've read all of his books

[mighty hull]

Nah. Baby Rudin was my undergrad analysis textbook. I've skimmed parts of RCA but never worked through it start to finish.

[heady skiff]

For 2b, can we just say that if $f \circ \pi$ is continuous and $U \in \mathcal{T}_Z$ is arbitrary, $(f \circ \pi)^{-1}(U) = \pi^{-1}\bigl(f^{-1}(U)\bigl)$ is open in $X$, so by the definition of the quotient topology $f^{-1}(U)$ is open and thus $f$ is continuous

[gentle osprey]

okeyokay

[shrewd night]

Fair enough

[shrewd night]

I've also had a look at RCA and FA

Those textbooks are so demanding

[mighty hull]

Yes

[gentle osprey]

person2709505

[swift sluice]

Wait nevermind I mixed up the x- and y-axes 🤦‍♂️ please disregard

[fringe thorn]

yes

[fierce lily]

if n=3 with x=(-1,0,0), y=(1,0,0), then it seems that the length of f(t) would be 2, which is much larger than 1, can anyone explain why it happens here?

[mighty hull]

Why do you say its norm is always 2? Show the calculation.

What happens when t=0 for example?

[dapper kestrel]

Finally got to writing the proof, is this correct? The writing could be improved as I'm not used to writing proofs in english but I think it's fine

[fierce lily]

no, I mean there exists norm that is 2, which is greater than 1, so I am confused what is happenning

[fierce lily]

ex:x=(-1,0,0), y=(1,0,0)

[fierce lily]

I am sure that when we list f(t), its norm is always smaller than 1, but it seems that this two points can let length of line greater than 1 such that f(t)=(1-t)(-1,0,0)+t(1,0,0) as t belongs to [0,1]

as norm(f(t)<=1, but distance between x and y is 2

[dapper kestrel]

f(t) gives you points that connect x to y, and what this is saying that the norm is bounded by 1 is that every point that the function gives you is in the unit ball, the lenght of the path that those points take could be 2 but that is not the norm of f

it would be something like the integral of the norm I think but i'm not sure

[mighty hull]

The norm of f(t) is the distance from the origin to f(t)

[fierce lily]

The thing that makes me confused is, I can apply f(t) to connect points x=(-1,0,0) and y=(1,0,0), but |f(t)|<=(1-t)||f(x)||+t||f(y)||<=(1-t)*1+t=1

[fierce lily]

so do you mean the norm of f(t) measures the distance from origin to each pair of points instead of measuring the line itself?

then it seems that ||f(t)||<=1 but the closed ball is of radius 2?

[dapper kestrel]

It's not to each pair of points, but by triangle inequality you have $||(1-t)x + ty|| \leq ||(1-t)x|| + ||ty||$ and that is bounded by 1 because $x, y \in B^n$

[gentle osprey]

pablo

[dapper kestrel]

If $||x|| \leq 2$ and $||y|| \leq 2$ your bound would also be $(1-t)||x|| + t||y|| \leq 2(1-t) + 2t = 2$

[gentle osprey]

pablo

[fierce lily]

ok, so f(t) connects (-1,0,0) and (1,0,0) right?

and I am confused about the distance between this two point can be 2, but |f(t)|<=1 by triangle inequality, that's the thing I feel confused

because f(t) expressed as a line as that connects two points, but its norm seems does not match the length of line connecting two points

[dapper kestrel]

f(t) is not a line but the points that live in that line

and every point in that line is inside the unit ball

[fierce lily]

then what does norm of f(t) measure?

[dapper kestrel]

is the norm of those points

it's <= 1 because those points are in the unit ball

[fierce lily]

I realize that f(t) is the expression of a collection of points lying on this line, so it is the norm of these points on line

[dapper kestrel]

Yeah and since there is a path that connects two arbitrary points in the unit ball we can say that it is path connected

[mighty hull]

This is what's happening: https://www.desmos.com/calculator/qwuxkhje31

The norm ||f(t)|| is the length of the purple line.

You can drag x and y around if you want

[fierce lily]

if X is first countable, then each a in X has neighborhood U contains at least one of element of U, is that correct

for def part, it writes there is a countable collection B of neighborhood of x such that each neighborhood of x contains at least one element of B, so does it mean each neighborhood of x contains points other than {x}?

[fringe thorn]

not necessarily

consider when X is discrete

then {x} itself is a countable basis at x, because every nbhd of x contains x, by def

and {x} is a neighbourhood of x, that obviously only contains x

[fierce lily]

so when it says each nbhd of x contains at least one of the elements of B, it just means contains at least one of elements of this nbhd?

[fierce lily]

I mean, each nbhd of x must contain at least one element of itself because nbhd is obviously non-empty, so do I misunderstand sth?

[fringe thorn]

no it should contain all of the elements in at least one of the nbhds in B

[fringe thorn]

I fear I don't understand

[fierce lily]

what does all of the elements refer to

[fierce lily]

so there is a single point x here, and I am confused about all of elements here?

[fringe thorn]

this is the kind of picture you can imagine

U is an arbitrary nbhd of x, and we can always find at least one element A in B such that A is contained in U

in the case of a discrete space, {x} is the only element in B needed to make B a countable basis at x

because any open nbhd U is obviously going to contain {x} in it; think of the picture!

[fierce lily]

i think you mean each nbhd must contain at least one nbhd, namely an element from B.

[fringe thorn]

that’s correct

it’s just that in a discrete space, we can take B = {{x}} to be our countable basis at x

also, just a heads up, I’m heading off to bed now!

so I’m not gonna be able to respond to anything from here

[sonic crane]

Tbh im having a hard time visualizing quotient topologies

Why, for example, identifying the boundary of a circle as a single point suddenly makes it into the sphere

Or the rectangle identifications making the torus

[red yoke]

If it helps you can play with some paper or play-doh

[fringe thorn]

https://upload.wikimedia.org/wikipedia/commons/4/44/Disk_to_Sphere_using_Quotient_Space.gif

[fringe thorn]

https://upload.wikimedia.org/wikipedia/commons/6/60/Torus_from_rectangle.gif

do these GIFs help?

in the first case, we gotta somehow force the entire boundary of the disk into a single point

the only way we can do that is by pushing the interior of the disk downwards, and then moving the boundary up to a point

this forms a sphere, as you can see in the GIF

[mighty hull]

This one you can do w/ a physical piece of paper.

[fringe thorn]

in the case of the square, we identify the left and right edges with each other, glue them together to get a cylinder, and then identify the top and bottom of the cylinder (which are just the top and bottom edges of the square), and glue those together to get a torus