point-set-topology

[prime elbow]

Is it correct?

I am taking function R to R

[plush folio]

Yep

[prime elbow]

Okay thank you

[prime elbow]

b part, we will prove that if they are comparable then they are equal.

Let T \subset T', let U be open set in T'.

Then X\U is closed and compact in T', since T' is compact.

Since T \subset T' implies X\U is compact in T, and X is Hausdorff therefore X\U is closed. Hence U is open in T.

Is it correct?

[tender halo]

that works yeah

an easier way to see it is consider the identity function on X

its continuous bijection from a compact space to a Hausdorff space therefore it is a homeomorphism

[prime elbow]

i see

thanks

[prime elbow]

{ x × y | y = 0 } and { x × y | x > 0 and y = 1/x} both are closed sets in R^2, right?

How can I deduce that they don't contain the limit point of the other from the diagram? I know they don't contain the limit point of the other but I want to know how I can get information from the diagram?

[iron bolt]

if they are closed, then they contain all their limit points. if they are furthermore disjoint, that means they don't contain any limit points of each other

idk what answer you're looking for with the diagram other than that the two lines never touch

[gritty widget]

How to approach this 🙂..I am clueless

[unreal stratus]

Presumably "aleph" should be aleph_0? Lol

[gritty widget]

I don't know 🙂 hebrew

How can I approach this

[alpine nest]

It's not really about hebrew, it's about notation convention. "cardinality aleph" almost certainly means "countably infinite" in this case

[plush folio]

What have you tried? If you take X to be the integers for example, what possible topologies can you come up with?

[opaque scroll]

I mean, the hint basically tells you the answer.

For any subset Y < X you have that {Ø, Y, X} forms a topology.

[opaque scroll]

I mean, they're just using Aleph as a variable. But unconventional symbol I guess

[alpine nest]

True, countability isn't really relevant here.

Weird choice of variable though

[opaque scroll]

I mean, when you think about cardinalities it's the symbol that comes to mind I guess

Slap on your favorite subscript

[versed pivot]

it's unconventional but I respect it
it wouldn't be weird e.g. to use the notation x_i to enumerate the rationals while using x as a variable name for a generic rational number

[opaque scroll]

You can almost drop the infinite part as well. Only Aleph = 1 should give trouble.

[scarlet turtle]

hopping off of this, there is a trivial upper bound of 2^(2^κ) distinct topologies on an infinite set of size κ. can we improve any of these bounds?

what about the number of distinct topologies up to homeomorphism?

[hidden abyss]

2^κ is always a lower bound in the first case, since {∅, Y, X} is a topology on X for any Y ∈ P(X)

In the countable case, the second has a lower bound of $2^\kappa$ too: take any decimal string in $[0,1)$ let $Y_0 =\emptyset$, and define $Y_i$ to be either $Y_{i-1} \cup {\text{a point}}$ or $Y_{i-1} \cup {\text{two points}}$ depending on whether $b_i$ is 0 or 1. The collection of $Y_i$s and the set itself form a topology, and distinct binary strings give nonhomeomorphic topologies

[gentle osprey]

Jussari

[gritty widget]

I got the first part that there are atleast 2^k distinct topologies..
But what about ..every infinite set has an uncountable number of distinct topologies on it..
I am actually studying topo for the first time..I am in first year of my undergraduate, if I am asking some really basic que, I am sorry

[haughty yew]

The power set of every infinite set is uncountable, so your statement follows

[gritty widget]

Ohh yessss..I got it 🥹

Thankyouu

[prime elbow]

d, I know how to show it by using continuous function.

But I want to use c part here.

Since X has at least two points say p and q, then d(p,q) ≠ 0.

Take t in [1,2]
Now let A = B(p, d(p,q)/t) and B is the set containing r such that d(p,r) > d(p,q)/t.

Since X is connected so X cannot be equal to A u B.

So there must be a point in x such that d(p,x) = d(p,q)/t.

Can I assign t -> x, such that d(p,x) = d(p,q)/t.

This mapping is injective, and t's are uncountable so X is uncountable

Is it correct?

[haughty yew]

wait, this is silly but I think this lemma is also false as stated. Just let U_1 = (-1, 1), U = (-200, -100) U (-2, 2) U (100, 200), and let f be the identity on (-2, 2) and let it be the doubling map on the rest of U.

It seems like both of these lemmas actually want to say this: for some restriction of f, there exists an extension of that restriction to all of R^n

and I’d guess that the flaws in the proofs come from the fact that they restricted the sets

[rancid umbra]

this is correct.

note: you need axiom of choice.

[prime elbow]

To choose x such that d(p,x) = d(p,q)/t?

[rancid umbra]

yes

[prime elbow]

Okay

[prime elbow]

If A and B are closed subset of metric space M. If A u B is connected then A and B are connected, right?

If A is disconnected then there exists a continuous function f:A -> {0,1} which is onto.

Since A is closed so A is closed in A u B.

Now By Tietze Extension we have continuous function g : A u B -> {0,1} which is onto. But it contradicts that A u B is connected. Right?

[haughty yew]

Consider A = {-1, 1} x R and B = R x {-1, 1}

[haughty yew]

the codomain of g may not be contained in {0, 1}

[prime elbow]

I have to think, A u B is connected or not

[prime elbow]

But why should the domain of g be contained in {0,1} ?

[haughty yew]

right, I meant “codomain”. I’ve edited it, thanks

[prime elbow]

I see your point

Thanks @haughty yew

[desert lodge]

How do I find the explicit form of this graph ?

[prime elbow]

Isn't imply if x is an isolated point then {x} is open?

[prime elbow]

Isn't the tan x function?

[desert lodge]

I was able to figure out the required function I need must look like this

[desert lodge]

the asymptotes are x = -1 and x = 1

[alpine nest]

It could be a rescaled and translated tangent, it could be something like x/(1-x^2)

[prime elbow]

Translate your graph

[alpine nest]

There are many functions that could have a graph of vaguely this shape

[desert lodge]

this is what I need to arrive

but how did you get that

[alpine nest]

Intuition, really, from having taught quite a few calculus courses

[desert lodge]

I see, I understood why the denominator

but numerator is a bit strange

ahh i see

numerator helps to tend + infty and -infty accordingly

[prime elbow]

Anyone?

[haughty yew]

Yes, if you’re asking “if x is an isolated point, then {x} is open”

[prime elbow]

Yes

Sometimes I am confused x is the isolated point of subset X of M or M, so that's why I asked

[haughty yew]

it is only open in the subspace topology, but for this exercise, it’s true

[solemn iris]

Trying to understand what this means, but can’t really picture it. Can someone explain to me better what a saturated set is?

[tiny obsidian]

you can define an equivalence relation on X by saying points are in the same equivalence class if and only if they map to the same point under p

then saturated sets are sets that are unions of equivalence classes i.e. union of preimages of points

(I would hope there's at least one example after what you've sent, so look at those too)

[hidden abyss]

I think the more intuitive definition is the reformulation in the next sentence: C is saturated iff C = p^{-1}(f(C))

[prime elbow]

Yes

[gritty widget]

Let X be an infinite set, I need to prove we have uncountable number of distinct topologies on X

Proof :

By cantors theorem P(X) is uncountable, now take any subset of X in P(X) , {X, null set, A} is a topology on X,
If I define a function g from P(X) to {topologies on X} by g(A) = {X, null set, A} , then definitely it will be an injection so codomain must be large as the domain, so it will be uncountable

Is this correct?

[hidden abyss]

Yeah that works

There's a slight technicality with ∅ and X being mapped to the same topology

Of course P(X) \ {X} is still uncountable so nothing changes

[gritty widget]

Ohh yesss ..I didn't notice it.. thankss

[hidden abyss]

according to this stackexchange answer that I'm not qualified to understand, there are in fact 2^(2^kappa) homeomorphism classes https://math.stackexchange.com/a/34841/877430

[red sage]

This is about R is finer than R with finite complement toplogy

I have question, Here to prove the above, they have taken an x in R and a basis element containing x of R_f

but what is a basis of R_f ?

[hidden abyss]

an obvious one is the whole topology

so unless the proof relies on a specific basis of R_f, any open set containing x should be fine

[prime elbow]

X is infinite, therefore any non-empty open set cannot be disjoint in finite complement topology so X is connected.

[queen prism]

"cannot be disjoint" what does that mean

[prime elbow]

That means if A and B are non-empty open sets then they cannot be disjoint

A intersection B ≠ \empty

[hidden abyss]

the argument is correct

but yeah saying an open set is/isn't disjoint doesn't make sense

[queen prism]

so say something like "any [two] non-empty open set[s] cannot be disjoint" or "any non-empty open set cannot be disjoint [from another]"

[prime elbow]

Yeah my mistake

[prime elbow]

I don't know how to use C connected here

[opaque scroll]

The boundary of A is the closure of A minus the interior of A. What happens if you intersect those with C?

[prime elbow]

I mean C has an element c such that every open set containing c intersect with A and X - A

[opaque scroll]

Okay, if you know that then you're already done

[prime elbow]

But how can I show the existence of such an element in C

[opaque scroll]

Like... You prefer to prove it in that way?

[prime elbow]

Yes

[opaque scroll]

Okay, well
C = C\cap A union C\cap(X - A)

Could these sets be closed in C?

[prime elbow]

No

[opaque scroll]

So then one of them has a closure that's bigger

Then take a point in that closure

[prime elbow]

Yes

[prime elbow]

say A has bigger closure cl A, let x in cl A\ A

[opaque scroll]

Then x is in the boundary, so badabing badaboom

[prime elbow]

Yes x is in the boundary but how x is in C then?

[opaque scroll]

Well, the assumption was that
C\cap clA was bigger than C\cap A

So you just pick x in the first set but not the second

[prime elbow]

Oh

I assumed Cl A is bigger than A

I missed C part

Got it thank you ❤️

Got it thank you ❤️

[desert lodge]

I want to prove comparability property for this,

Case 1: x^2 != y^2
1.1. x^2 < y^2
1.2. y^2 < x^2

Case 2: x^2 = y^2
2.1 x <y
2.2 y <x

[desert lodge]

are the cases chosen correctly?

[desert lodge]

Please help, how do I prove the comparability holds?

[ruby delta]

what about x=y?

[gentle osprey]

nux

[torn kayak]

ah didn't mean to delete the original message

[burnt tendon]

Suppose U is non-empty, clopen, and strictly smaller than X. What can you say about the complement, X\U?

[torn kayak]

that would also be clopen, right?

[burnt tendon]

Yes. Moreover, it would be non-empty and also not all of X

Hence, X can be partitioned as the disjoint union of open sets U and X\U.

So here’s the intuition: The existence of a non-empty, proper, clopen subset U gives us a partition of X into open sets. Thus, the space X is disconnected, for it can be separated by these open sets

[torn kayak]

ahh i see, that wasn't so hard

thank yoiu

[burnt tendon]

Note also that this condition, that X can be separated into disjoint open sets, is equivalent to the one given in your lecture

[torn kayak]

yeah i see now it was one equivalent statements given. i'll try to prove the other implication

[burnt tendon]

Some further food for thought: If you try to separate a connected space by two open sets, then inevitably you will miss points contained in the boundaries of the two sets

[civic ivy]

We know that second countable regular space is normal. What if we have a first countable regular space? Will it be always normal?

[fringe thorn]

some googling led to this counterexample

[solemn iris]

Not really getting why these small grey areas are the saturated sets…

[hidden abyss]

There is also the equivalent statement X is connected iff every locally constant map from X is constant

Might help with intuition

[desert lodge]

We have check comparability only when x! = y

[ruby delta]

That’s not the definition of comparability I know, but… sure

[desert lodge]

This is the definition I’m working with

I have a question,

Suppose two set A and B orders <, <‘ respectively.

Suppose there is an order preserving map f between A, B.

Is it true that, if A with < has supremum property, then <‘ also have?

My guess is,

Given a subset A0 of A,

If s is the supremum of A0, then f(s) is the supremum of f(A0)

[opaque scroll]

Consider B = Z x {0, 1} with
(m, 0) < (n, 1) and
(m, x) < (n, x) Iff m<n.

Let A be the subset of B containing Zx{0} and Nx{1}.

Then Zx{0} has a supremum in A, but not in B

[desert lodge]

would this be true,

If s is the smallest element of A0 then f(s) is the smallest element of f(A0)?

This is my orginal question,

This is Z_+ x [0,1] with dictonary ordering

Does this set have a supremum property?

does f preserve order?

[hidden abyss]

If f is assumed to be order preserving then yeah

Proof is basically s≤x for all x in A => f(s) ≤ f(x) for all f(x) in f(A)

[desert lodge]

yes

[desert lodge]

I thought this would also help to prove supremum property

[desert lodge]

^

[hidden abyss]

Supremum property is defined as every non-empty subset with an upper bound has a supremum?

Should be smallest element of upper bounds, no?

Or more precisely s is supremum of A if a≤s for every a ∈ A and s≤s' for every s' with this property

[desert lodge]

Ahh my bad

yes

[desert lodge]

If I were to take a subset here that looks like an interval then it does have a supremum

I guess

All the upperbounds are either above claimed supA or to the right of it

[hidden abyss]

But yeah that has supremum property

[desert lodge]

This is dictonary order for reference

[desert lodge]

But does this hold for every non subset A0?

[hidden abyss]

One way to show it is via casework: if (n,x) is an upper bound of A, then every element (m,y) of A has m≤n, i.e. the first coordinate is bounded

So you can choose m to be the maximal first coordinate in A, and prove that the supremum of A ∩ ({m} × [0,1)), if it exists, is the supremum of A

If it doesn't exist, the point (m+1, 0) is the supremum

Another way to do this is to construct an order preserving bijection between a set you know has the supremum property and $\mathbb{Z}_+ \times [0,1)$.

[gentle osprey]

Jussari

[hidden abyss]

Hint: || the set of nonnegative reals ||

[desert lodge]

Sorry I was away

yes,

f(n x t ) = n + t -1 does the job

Wouldn’t this takes us back to the original question, does having an order preserving map imply supremum property is preserved?

[hidden abyss]

I think it does for surjective maps, since you can take the supremum of the preimage and map that back into the codomain

[desert lodge]

I’m sure this helps to claim largest (smallest) element

but does it also help with supremum

How do I try to prove it?

[hidden abyss]

Let $f: A\to B$ be order-preserving and surjective and suppose $A$ has the supremum property. Let $S\subseteq B$ be a nonempty set with an upper bound $b$. Prove that any $a\in f^{-1}(b)$ is an upper bound of $f^{-1}(S)$. Then $s:= \sup f^{-1}(S)$ exists. Prove that $f(s)$ is the supremum of $S$

[gentle osprey]

Jussari

[hidden abyss]

I think that argument should work

[desert lodge]

order preserving by my definition is a bijection

[hidden abyss]

oh

I think the standard definition is x<=y in A implies f(x)<=f(y) in B

[alpine nest]

That's how I would have interpreted it as well

[solemn iris]

Would I be correct in claiming that $\mathcal{T}=\mathcal{P}(X)$?

[gentle osprey]

bluepianist

[alpine nest]

Doesn't have to be.

If I is a singleton, i.e. the family consists of just one topology T_0, then the topology you seek in this problem will just be T_0 again.

Also in any case, you want T to be refined by the T_alphas

So in the worst case it will be the indiscrete topology

[solemn iris]

oh so I want T such that T subset T_alpha instead of T_alpha subset T?

i'm a bit confused with the terminology cuz doesn't this mean that T is coarse

[alpine nest]

"coarse" doesn't mean much when applied to an individual topology

T is coarser than all of the T_alphas

And finer than any other topology that's coarser than all the T_alphas

[solemn iris]

ohh

got it

thanks

[dull lance]

f : X × Y → M
if f(x,y) is continuous when fixing x, and when fixing y.
then f is continuous?

[ruby delta]

consider f(x,y) = xy/(x^2+y^2), f(0,0) = 0

[dull lance]

thanks

[solemn iris]

Show that X is Hausdorff if and only if the diagonal {x × x | x ∈ X} is closed in X × X.

Hi can I get some hint? My first idea was taking the complement and showing it’s open but I’mm not sure how..

[prime elbow]

okay pick a point in complement

[solemn iris]

Yeah so there I can produce two disjoint sets for a pair (y,z) right

[prime elbow]

disjoint open sets

[solemn iris]

yeah sorry that

[prime elbow]

so what do you think now?

can you make open set of (y,z) such that it contains in complement?

[solemn iris]

oh hmm

like this?

wait no that looks wrong

[prime elbow]

why did you take set S ?

[solemn iris]

i was thinking of taking the union of these cartesian products

[prime elbow]

just take product U and V

this is open in X \times X

[prime elbow]

you want to show complement is union of all the elements of S?

that's fine

but you can also show if (x,y) in complement so there exists open set W contaning (x,y) such that W contained in complement

and here W = U \times V works

[ruby delta]

Lemma. A set S is closed in a topology T iff for every point x outside S, there exists an open neighbourhood of x that is disjoint from S

[solemn iris]

Ooo got it thank you notknow and hchan

[desert lodge]

I have a question, I’m going to modify (2) and (3) as follows

2. There exists an interval [a0, b), where b in B, and a0 is the smallest element (if any)

3. There exists an interval (a,b0] where a in B, and b0 is the largest element (if any)

Wouldn’t this changed B be a basis for a toplogy

[ruby delta]

There can be more than one basis for a topology

Exactly in the same way there can be more than one basis for a vector space

[rancid umbra]

that is not what they are asking

[desert lodge]

Would this be a basis for a topology?

ie satisfy the conditions for basis

[desert lodge]

If I didn’t formulate this properly, instead of taking infinite neighborhoods around a0, b0, why not just one ?

[rancid umbra]

a basis is a collection of subsets satisfying some properties.
the author is specifying which sets need to be included for B to be a basis.

first, if you only include one such set of the form [a0, b) or (a, b0], then you would first need to specify what a and b are.

second, this is indeed a basis for a topology on X, but the topology it generates is, in general, not the order topology.

for example, consider X = [0,1) and say that the set you include in your basis is [0,1/2).
in this topology, [0,1/3) is not open, as the only open neighborhoods of 0 are [0,1) and [0,1/2).

contrast this with the situation when you include all sets of the form [0,b) for b in [0,1):
[0,1/3) is open in the order topology on X, since it is a basis element

[digital thunder]

I think the modified topology wouldn’t be hausdorff even right?

[rancid umbra]

right, no non-zero element less than 1/2 can be separated from 0 by open sets

[desert lodge]

If I were to think for motivation for this definition of order topology

it’s because we need to approach the end points sort of however close through neigbordhoods!

So things like sequences may converge in these points

[rancid umbra]

exactly! the intuition carries over from studying subsets of the real line. one should be able to approach the end points with open sets as close as the ambeint space affords

[desert lodge]

Sorry, I might thinking too much

[desert lodge]

Thank you

[desert lodge]

Oh, the wording ‘ambient space affroads’ strikes.

In some examples where the X is a not a linear conntium, getting infinitely close doesn’t make sense

[rancid umbra]

yea, exactly

[desert lodge]

Alright, Thanks a lot for helping me build intuition

Have a great day

[rancid umbra]

u2!

[brave geode]

if p: E → X is a covering map and X is locally path connected, E connected then the group action of Aut_{CovSp(X)}(p) on E is properly discontinuous by path lifting. can you say the same if you omit the locally path connected assumption?

hint only please

oh it's the usual connectedness trick

if f is an automorphism then the set of points e in E with f(e) = e is clopen

well, nice to know

[desert lodge]

Let X = R be the topological space with the standard topology. Let A = [0,1].

Is the subspace topology same as the order topology of A by the usual order relation ?

I verified that all the types of intervals described in the basis of the order topology is an open set in the subspace topology

[unreal stratus]

I think the subspace topology is always at least as fine as the order topology, so it's the other direction that is more subtle i guess

[desert lodge]

yes, I should show any open set in the subset topology is in the order topology right ?

[unreal stratus]

But you should be fine by considering a basis for the subspace topology. All the intervals of the form (a,b) where 0 < a < b < 1 are obviously open in either, so you just need to consider stuff of the form [0, a) (and symmetrically (b, 1])

[unreal stratus]

Yes

[unreal stratus]

but then these are open by definition again i guess lol

I believe basically this should be true because of [0,1] being an interval in R - i would assume that this works whenever you have an interval as the subset

[desert lodge]

How do I prove that the only open sets of the subspace topology of [0,1], is either (a,b), or [0, b) or (a, 1] ?

[unreal stratus]

Well that isn't true, but what I described is a basis

[desert lodge]

i mean

the union of these

is the topology yes

[unreal stratus]

Well just by considering the basis for R and intersecting with [0,1]

[prime elbow]

b part, each non empty set M has non-empty open subset that disjoint from A.

Since int cl A = empty implies int A is empty so any non-empty open set must be in X \A, so every non-empty open set must be disjoint from A.

Correct?

[tender halo]

the topologies are the same if the subset is convex

[desert lodge]

right, so subspace topology on [0,1], where X = R, is same as the order topology right?

[young stone]

yes

[digital thunder]

As beaver said, you have to be sure they are convex, which in the case of [0,1] is true, but not true in general

[rancid umbra]

no. consider A = Z, M = R, U = (-1,1).
U is not disjoint from A

[prime elbow]

Thanks for pointing out

[prime elbow]

okay, let there exists U such that it is non-empty and every non-empty open set of U intersect with A. We will show that U \subset cl A.

let x in U, then there exists open ball B(x,r) \subset U, so it must be intersect with A, and since every non-empty open set of U intersect with A, therefore B(x, t) intersect with A , for all t<= r. So x in cl A, implies U \cl A, but U is open contradicts that int cl A = empty.

is it correct? @rancid umbra

[rancid umbra]

is your definition of nowhere dense that int cl (A) is empty?
if it is, then this is fine.
remember, you also need the reverse direction

[desert lodge]

\subsection*{$X = [0,1) \cup {2}$, with usual order}
For instance, $(0,1)$ is an open set in the order topology. Take $a = 0$, $b = 2$. We have $(0,1) = (0,2)$ in the order topology, since all elements between $0$ and $2$ in $X$ lie in $[0,1)$.

Can you check if this is correct ?

[gentle osprey]

Dubs

[desert lodge]

I'm a bit skeptic about saying (0,1) = (0,2)

(0,1) is not a basis element I suppose, since 1 doesn't belong to X

[rancid umbra]

in this setting, it is correct to say (0,1) = (0,2) for the reason you stated.
you are correct that it is not a basis element

[desert lodge]

thank you

[prime elbow]

yes nowhere dense means int cl A is empty, yes reverse direction is easy, we can go b to c, trivial and c to a, let int cl A is non-empty open set, then there exist open ball B(x,r) \subset int cl A such that B(x,r) intersection A is empty, but B(x,r) \subset int cl A implies x in cl A, then B(x,r) intersect A, but it contradicts that B(x,r) intersection A is empty

which definition you using for nowhere dense set?

[rancid umbra]

none in particular. i just wanted clarification on the one you were using

[prime elbow]

Okay thank you

[prime elbow]

i want to verify intersection of two connected space is connected or not? in R it is, but in general space, any hint?

[rancid umbra]

look in R^2 for two connected sets whose intersection has two components

hint: ||both sets are homeomorphic to a disk||

[digital thunder]

I would start by assuming the intersection is not connected, and find a contradiction

Actually there's two possible contradictions, either you show there's a separation of one of the connected spaces. Or you show that in the separation of the intersection, one of the open sets is actually empty.

[prime elbow]

you mean intersection of connected space is connected ?

[digital thunder]

Yes it is, but easiest to prove via contradiction

[rancid umbra]

that is not true

[digital thunder]

wait, yeah you are right

[rancid umbra]

in R it is, but not in general

see the hints i gave above

[prime elbow]

is it true that X is connected iff every continuous map X -> {0,1} is constant map, {0,1} with discrete topology, right?

[opaque scroll]

I wonder, if the union of two connected spaces is simply connected. Would their intersection be connected?

[prime elbow]

simply connected?

i think no

[opaque scroll]

If one replaced connected with path-connected it seems it should be true

[prime elbow]

we know that if A and B is connected and they intersect each other then A u B is connected, is converse true?

if A and B connected and A u B is connected, is their intersection non-empty?

[digital thunder]

(-1,0) U [0, 1]

[prime elbow]

yes

[rancid umbra]

do you consider the empty set to be connected?

[prime elbow]

so you mean if one of them path connected then A n B is path connected?

[prime elbow]

i consider

[opaque scroll]

I don't, but I'm happy to throw in the assumption that they intersect

[rancid umbra]

i see. if you don’t, then taking an annulus and then its open center as the two sets, or vivian’s example work

but i think that you can modify the annulus example:

take a closed annulus as one set, and an open disk with the top left quarter boundary and bottom right quarter boundary.

then the union is simply connected but the intersection is the union of the two quarter boundaries

which is not connected

[opaque scroll]

Yeah, just realized that myself.

Probably you want to assume the sets are open (or closed maybe)

[rancid umbra]

right

[opaque scroll]

Anyway, thx

[rancid umbra]

np. what motivated that one?

[opaque scroll]

The counterexamples to notknows question

The most natural ones are not simply connected

[rancid umbra]

right

that’s kind of interesting

they also don’t really leverage being open or closed

[digital thunder]

The intersection of the closure of A with the closure of B is non-empty though I think

[rancid umbra]

yea, otherwise cl(A U B) = cl(A) U cl(B) is disconnected (as long as A and B start off non-empty)

[prime elbow]

can i get hint how to show in second countable metric space X, every open covering of X has countable subcovering?

[rancid umbra]

in a metric space, second countable and separable are equivalent

[prime elbow]

yes i know this

[rancid umbra]

wait, i may have misread the question

[young stone]

Pick an open cover. Write each element in the open cover as a union of basis elements, use this to choose a countable subcover

[prime elbow]

but how to choose them?

[young stone]

For each of the (countably many) basis elements choose an open set containing it in the open cover

[prime elbow]

let $X = \bigcup U_i$, let's say $U_i = \bigcup_{j=1}^{\infty} V_j$

[gentle osprey]

Notknow🙇

[prime elbow]

now how we do choose countable U_i s?

[young stone]

.

Is it not clear?

You still have to show that this covers X but I leave that to you

[prime elbow]

sorry, i don't get this part

[young stone]

I'll rephrase it, Take the countable basis, for each element in the countable basis take an element in the open cover that contains it (if there is no such element don't pick anything). we claim that this is a countable subcover of the cover we started with, it's clear that it's countable since we are picking atmost one open set for each basis element

It remains to show that it covers X

[prime elbow]

i see

[prime elbow]

it comes from basis definition that union of basis element is X, right?

[young stone]

Yes but we are leaving out some basis elements so we should be careful

[young stone]

Like this is not a correct proof that it covers X, since there maybe a basis element thats not contained in any element of the cover

[prime elbow]

yes

[frosty prism]

Would anyone recommend a background about vectors of spheres?

[ruby delta]

it's just basic linear algebra

there's nothing special about the sphere

start with a LI set of n vectors. then normalise every vector so that you get a LI set of n vectors on the sphere

so it boilds down to understanding that there always exists a LI set of n vectors

[frosty prism]

so vectors of d-dim sphere are exactly the (d+1)-dim euclidean space vectors whose norm is 1, right?

[ruby delta]

well, I dont know if dimension of a topological manifold has been defined for you yet

because if it hasn't then the sphere does not yet have a well-defined dimension, because it's not a vector space

[frosty prism]

the proof says "consider vectors on the d-dim sphere"

does that imply dimension is defined?

[ruby delta]

yes it is defined but I don't know if the definition has been introduced to you yet

but for the sake of this proof you can ignore that, you are correct

[frosty prism]

so for the proof, vectors of a sphere constitute a vector space, right?

[ruby delta]

no, that is my point, it's not a vector space

just... they're linearly independent in the bigger vector space

and that's good enough

[digital thunder]

Can anyone let me know if I did this correctly? And I'm new here so forgive me if this isn't the best way to share proofs (let me know how I should do it next time). Also is there an easier way to prove?

[frosty prism]

thanks 👍

[brave geode]

the ideas are there :)

• at least one upper bound (i assume that's a typo)
• Take a closed [...] c ∈ C
• elaborate on why Ω has the finite intersection property

[digital thunder]

Ok thank you that helps!

[brave geode]

np

[prime elbow]

I got it, thank you ❤️

[gritty widget]

Can anyone help me understand why I'_c without f^(-1)(0) is Fσ?

[fickle tendon]

is there a name for topological spaces with a basis consisting of sets, where for any two sets they are either disjoint or one is included in the other?

[alpine nest]

Not sure but 1. you can get such a basis in a Cantor space; 2. My gut tells me this would generally require your space to be hugely disconnected.

[fickle tendon]

yes, I can imagine too lol

[tender halo]

Alexandrov spaces qualify but im not sure if there are any others

[fickle tendon]

since I'm constructing this from the top down in a sense any open subspace would be disconnected or indiscrete

[fickle tendon]

hmm, I'll look into em, thank you

nvm, I just realised that my space would have every open set also closed, rather boring

wait is it?

[opaque scroll]

I would guess that for Hausdorff spaces, this would be equivalent to being totally separated

[unreal stratus]

I am tempted to try to prove this lol

[fickle tendon]

In a covering, the fiber of some x in the covered space as a subspace has the discrete topology, right?

[echo folio]

Hey I'm really confused on this proof showing that [a,b] is connected.

I dont understand why [a,c) is contained in U

I'm good with everything before

[hidden abyss]

That's by definition of c, it's the greatest lower bound for V ∩ [a,b]

If there was a a≤x <c in V, then x<c = inf (V ∩ [a,b]) which is impossible

[echo folio]

Okay follow up question

Like why is x in V

I'm trying to draw a picture to understand but I think its just confusing me more

I get that c is the greatest lower bound for V \cap [a,b]

But i dont see why $[a,c) \not\subseteq U$ means that everything "past" it is contained in $V$

[gentle osprey]

cola_drinker12

[echo folio]

if that makes sense

[hidden abyss]

By assumption $[a,b] \subseteq U \cup V$, so if $x \notin U$, then $x\in V$

[gentle osprey]

Jussari

[hidden abyss]

Are you talking about x ∈ [a,c) or x>c?

[echo folio]

the x in [a,c) thats not in U

[hidden abyss]

Yeah it would have to be in V then because U and V cover the interval

[echo folio]

ok thanks

i still dont understand but I think I just need to sit and think about it for like 5-10 minutes

this should be more than enough

thank you

[hidden abyss]

Happy to help

[echo folio]

ok yes I get it

GOAT

[craggy portal]

can anyone help me understand saifert van kampen

[paper wedge]

just ask

[gritty widget]

Ok nvm I figured it out

[rugged dagger]

I was reading a proof on "Let X be a topological Hausdorff space, then X is locally compact if and only if every point x in X has a compact surrounding"

As for the <- implication (first one is trivial) I'm gonna explain the proof until the point I don't understand how it follows:

Let x be an element of X, and U subset of X an open surrounding of X. Let K be the compact surrounding of x, then L=K\U is a closed set in a compact, then L is compact, because it is Hausdorff, there exist sets W and W' such that their intersection is empty, and x is an element of W, and L is a subset of W', then V=W intersec interior(L) is an open surrounding of x

And then it continues, but my doubt is, how is V a surrounding of x if L=K\U, and K and U contained x, therefore L cannot contain x, and therefore interior(L) cannot contain x (and obviously its intersection with something else can't either)?

Thanks

(As on how the proof would end, then it says the adherence of V is closed, and because K is closed, as it is a compact in a hausdorff, then because V is a subset of K, then the adherence is also a subset of K, therefore the adherence of V is a compact contained in U, and therefore X is locally compact)

[heady skiff]

Why is V an open subset of |K_2|? Are we viewing |K_2| under the subspace topology endowed by |K|?

[native arrow]

Yeah I can try seifert van kampen theorem is about how to compute the fundamental group of union of two spaces if you know the fundamental group of each of them. Hope it's clear. Can you specify what's your doubt?

[native arrow]

And there's a condition that the intersection should be connected.

[fickle tendon]

in my case the basis also consists of clopen sets, so we have T0 <-> totally separated <-> Hausdorff

[urban zinc]

if it's compact hausdorff, then that sounds like a stone space

(my beloved...)

[fickle tendon]

yeah I am looking into that rn lol

compactness, that is

I've got a hunch for some equivalent condition but having trouble proving it aueguahg

[fickle tendon]

my savior 🙏 (The condition makes it so that the space is an inverse sequential limit of compact spaces)

nvm the maps aren't closed
I'll never get this proven I have gotten this proven

qwq \o/

[warped bone]

Guys quick question, if i want to prove that if we suppose that A is an open set, then i can find, for all x in A, an open set U such that x is in U and U is subset of A,can i just pick A=U?

[paper wedge]

proper

[cloud kindle]

No, because U is open, and you dont know what A is yet. This also boils down the the statement, for all x in A, x is in A which tells you nothing

You need to show that for all x in A there is an open U containing x which is properly contained in A

[warped bone]

Since A is open i can write it as Union of open sets, so these U must exists for each x in A and they must be subsets of A?

[cloud kindle]

Im confused about what youre trying to do. It seems like youre both assuming A is open and trying to prove that its open

[warped bone]

Let (X, τ) be a topological space and A a subset of X. Prove that A is open if and only if for every x in A there exists an open set U such that x is in U and U is a subset of A.

[prime elbow]

Yes

[cloud kindle]

Ok in that case yes I appologise, if you assume A is open you can just take A to be a neighbourhood of all its points

[warped bone]

Thanks guys

[analog kelp]

Hello

Can I talk about icosahedron based sphere point distribution here?

I have a method for obtaining the nearest node on a unit sphere for an arbitrary spherical coordinate, the node set is well distributed longitudinally and latitudinally, but I'm trying to improve it by warping the set into a subdivided icosahedron set which has much better equidistance-ness between nodes. Anyone interested or am I at the wrong channel?

The general method for obtaining something like this would be to pre-generate the ico-sphere node set and sort out the nearest node, but with my theoretical method it should be possible to obtain the nearest node without any storage.

[craggy portal]

like wedge sums

sorry for late response

[rancid umbra]

wdym

is your doubt van kampen with wedge sums?

[tender halo]

i mean you just kinda look at the properties of the regular closure operator i guess

as to what problems it solves - sometimes its just more convenient, although not that much different from just specifying closed sets

i would say the most important alternative way to specify a topology is neighborhood bases

[brittle rapids]

i don't think there's a nice way to define compactness using kuratowski closure

you can say that for every family of sets with the finite intersection property, there exists a point that is a limit point of all of them

[red sage]

I’m trying to prove an Interior of set A is open.

Definition: Let A subset of X, Int A is the union of all open sets contained in A.

My answer: Let {U_a : a in I} be an collection of open sets in contained in A. Since arbitary union of open sets in open in X, Int A is open in X.

My question: Is it possible to make all open sets contained in A into an arbitary collection? How do I make sure I indexed all sets with the indexing set I

sorry if this question is a bit too much 🙂

[brittle rapids]

why are you indexing the collection with points in A?

[red sage]

I’ve changed it

[brittle rapids]

alright

you don't need to index every collection

you can just say {U ⊆ X | U is contained in A and open}

[tender halo]

whats the interval doing there lol

I is just some index set

[red sage]

right

[tender halo]

no relation with the unit interval I

[red sage]

I understand that

it may countable or countable

[red sage]

This is a bit helpful

is it because in the defintion of topology doesn’t specify an indexed collection?

[brittle rapids]

in general you can specify a collection of subsets, or elements, without needing to use a function

"let S be the subset of real numbers that are roots of integer polynomials"

[red sage]

Alright, my misconception was this: when we say A arbitary collection of sets, then I thought

A = {u_a : a in I}

Is it possible to always index an arbitary collection of sets?

[brittle rapids]

what is an indexing?

[red sage]

I don’t have a specific definition in hand. But it’s a function which defined between I to the target set A

[brittle rapids]

precisely

I is the "index set"

any such indexing function determines a collection

via its image

[red sage]

T = {U ⊆ X | U is contained in A and open}

Suppose we define T as such

How do I find an I an indexing set

[brittle rapids]

don't use the same letter as the subset we're defining this as

well according to our definition we need to find a function from a set, to P(X), such that the image of this function is exactly T, right?

P(X) powerset

[red sage]

right

[brittle rapids]

one such function is just the inclusion map T -> P(X)

so the answer is yes

[red sage]

so our index set is same as our target set T

[brittle rapids]

sure

well we're targeting P(X)

but the image will just be T

[red sage]

by target i mean the image set of the indexing function

[brittle rapids]

you want to be careful and think of the target as the codomain instead of the image

if i consider paths in R^2 i really mean certain functions from the unit interval I to R^2

i don't restrict these functions to their images in their definitions

[red sage]

Makes sense

In general, I may just say an “arbitary collection of sets”

I have a one more question,

I was trying to show intA is the largest open set contained in A.

My idea is to set U as the largest open contained in A, and show U = int A.

Clearly, IntA C U and A C Int A followed.

Question: How do I be sure there exists some U that is the largest open set contained in A. Ie prove the existence of such a U

[brittle rapids]

the empty set is open

your definition of int A shows that the union is not the empty union

[red sage]

isn’t empty set the smallest open set contained in A?

[red sage]

Yes, because A is trivally contained in A?

[brittle rapids]

because the empty set is trivially contained in A

i think the proof would be better phrased if you directly showed that int A was the largest

it would look like this: showing that int A is open, and showing that any other open set contained in A is also contained in int A

[red sage]

any other open set contained in A clearly belongs to IntA since it’s the union of such open sets contained in A

If I can show given any open set U contained in A, then U is contained in Int A. Doesn’t that mean int A is the largest?

[brittle rapids]

it does

kind of a tautology right

[red sage]

yes

[brittle rapids]

it's a pretty common construction in math

you'll see the same with closed sets

the key property being that the arbitrary intersection of closed sets is closed

[red sage]

formally, given a collection of sets with certain property.

The union of the collection gives the largest sets with the property

And intersection gives the smallest right?

[brittle rapids]

you need to make sure that the union of sets with property P also has property P

likewise with intersection

in math it's much easier for intersections to preserve some property that it is for unions to preserve some property

think intersection of linear subspaces

[red sage]

yep, intersection of a collection of topologies is a topology

But union is not

[brittle rapids]

there you go

[red sage]

Can I instead take the identity map as well?

[arctic mural]

I can't prove that in a first countable space adherent points of a set are also limits of sequences over that set.

I mean the converse of this theorem

[hexed steppe]

what have you tried

[prime elbow]

i want to show that if every function M -> R is continuous then every subset of M is open.

M is metric space with metric d.
fix y in M.
Now define f : M -> R such that f(x) = 1 when x \neq y, f(y) = 0.

so this is continuous, now take inverse image of (-1/2, 1/2) it will be open in M, hence {y} is open.

is it correct?

[brittle rapids]

it's correct but M doesn't need to be a metric space

you can even be more direct about it; let χ_U be the characteristic function on U

[prime elbow]

it is given that M is metric space

[prime elbow]

i see

thank you

[dry seal]

https://topology.pi-base.org/spaces/S000015

Is there any reason we use ω here instead of the usual notation for natural numbers?

[hidden abyss]

ω is the first infinite ordinal

[tender halo]

to a set theorist \omega is the usual notation for natural numbers

[ruby delta]

it's a good self-report, whenever someone does that you stay the hell away

[dry seal]

I know what ω means I just find it odd in this context because in this case we are literally just taking it as a set and don't really care about its relationship with other ordinals

Unless I'm missing something?

Also I get that it's the same to a set theorist but like this is a topology database

[tender halo]

not really, to me personally thats more natural than N

because we are only talking about set theoretic properties (being countable) instead of arithmetic

[dry seal]

Ironic choice of wording lol

But I see your point I guess

It's just the first time I've worked with these things

[tender halo]

also point set topology is a branch of set theory

at least at higher levels

[dry seal]

My university offers a course in set theory but it's only given once every other year and I missed it while on exchange

So I'm def behind on this side of things

[glad laurel]

might be a stupid question, i'm sort of new to set theory, but would this be how you'd write like, "the set of elements in B where x != y and (x,y) != (y,x)"?
{(x,y) ∈ B | x≠y, (y,x) ∉ B}

(where B is the cartesian product of some set A with itself)

basically, i want only the unique pairs of different values, so not (0,0) or having both (3,4) and (4,3) in the set

[plush folio]

No, (x, y) != (y, x) isn't the same as (y, x) \notin B, and neither formulation does what you want. If you want to consider (3, 4) and (4, 3) as equivalent, then you probably want a set, not a pair. So { {x, y} in B | x != y }. If you actually want pairs, then you need to decide whether you want (3, 4) or (4, 3) in your set; you can for example choose the pair with the smallest x-value

[glad laurel]

huh, ok, didnt think just saying 'this is a set' would work

thanks

[plush folio]

Yeah, {x, y} is literally the same set as {y, x}, so just writing { {x, y} in B | x != y} is sufficient, you won't have any duplicates

btw, this question might belong better in #discrete-math

[glad laurel]

kk

[rugged dagger]

Someone?

[tender halo]

whats an adherence and whats a surrounding

[rugged dagger]

Is there a different name in english? Fuck

Let me try to search it

[rugged dagger]

Well surrounding is basically "a surrounding of x is a subset A of X such that it contains x and an open set that also contains x in X"

And then the adherence of a subset A is such that for every open surrounding U of A, U intersec A is non-empty

What's the names of each?

[tender halo]

the second one doesnt make sense, you mustve misplaced a variable name

first one is called a neighborhood

second is closure

[rugged dagger]

Yeah I defined the second a bit loosely so that it still could be understood what it was

[rugged dagger]

Yeah those

I would've said that the closure of a subset A in X is basically the elements x of X such that for every open neighborhood U of x the intersection between A and U is non-empty

If I'm not mistaken

But whatever

[dawn coyote]

can anyone explain to me why this proof doesn't work?

[tough hamlet]

on my soul this looks completely fine

[hexed steppe]

there are some typos in the first paragraph?

that is hardly a “big issue” though

[dawn coyote]

hmm

i see

[prime elbow]

I am thinking about the question that, if f: X -> Y is a continuous and bijective function, where X and Y have the same topology then does it imply that f^-1 is continuous?

I don't want an answer

f: X -> X

[tender halo]

i guess they mean whether all bijective continuous endomorphisms are homemorphisms

[hidden abyss]

There's an answer in this stackexhange thread ||https://math.stackexchange.com/questions/20913/are-continuous-self-bijections-of-connected-spaces-homeomorphisms||

[prime elbow]

Yes

[unreal stratus]

I'm sure this is there, but just to say that you can simply endow the same set with two topologies (one finer than the other) such that both are connected

For example, consider R with the indiscrete topology or with the usual topology

The identity map is obviously bijective, and continuous if you make it go from the finer one to the less fine one

But not a homeo or the topologies would coincide

[prime elbow]

I know, therefore I assumed both have same topology

[prime elbow]

Connected? I don't get it this point

[unreal stratus]

But then like what does that mean if they have different sets

To me it just sounds like you are considering the identity map from a space to itself

which is clearly a homeomorphism

[prime elbow]

I mean f : X -> X which is continuous and bijective and same topology on X, is f homeomorphism?

[unreal stratus]

Ah okay sorry that's cool

That is a good question

[opaque scroll]

||Let X be the the disjoint union of Z with discrete topology and Z with the trivial topology.

You can make a bijection by moving some of the discrete points over to the trivial part||

[unreal stratus]

I guess here X was meant to be connected

(That was not stated here but was stated earlier)

But nice example

[opaque scroll]

Well just add in a generic point to make any space connected

[unreal stratus]

This is a good point lol

[opaque scroll]

You can make connected Hausdorff examples though, as someone posted above

[unreal stratus]

Pog

[ebon galleon]

champ

[desert lodge]

\textbf{Theorem.} Let ( X ) be a topological space, and ( A \subseteq X ). Let ( F ) denote the set of all limit points of ( A ) in ( X ). Then ( F ) is a closed set in ( X ).

[gentle osprey]

Dubs

Dubs

[desert lodge]

Can you check if my proof is correct?

[prime elbow]

Idea is correct, if y ≠ x, maybe U intersect with A {y}, x can be in U intersection A

[desert lodge]

I can’t really figure out how to proceed

if y ≠ x then if y belongs to A, then we have U intersect A-{y} is empty, because U doesn’t contain any element of A different from x.

[opaque scroll]

I guess take a two point space {x, y} with the trivial topology and take A = {x}. Then the only limit point is y, but {y} is not closed

[desert lodge]

right

[desert lodge]

I think it says X is a hausdorff space

[solemn willow]

Construct a set $A\subset[0,1]\times[0,1]$ such that $A$ contains at most one point on each horizontal and each vertical line but boundary $A=[0,1]\times[0,1]$. A hint is provided that says it suffices to ensure that $A$ contains points in each quarter of the square, sixteenth, etc.

Not really sure how to start with this beyond looking for some very particular subset of vectors with just rational entries. Any help would be appreciated

[gentle osprey]

cole

[rancid umbra]

i think you have some typos

[solemn willow]

Its word for word from the textbook

I’ll send a picture as well

[tender halo]

we need to find a bijection from I to I such that the graph is dense in I^2

we can leave irrational points where they are, rational points are enough

[rancid umbra]

sorry, seemed weird to me to phrase it as bd A = I^2, i would’ve just said the closure

[urban zinc]

that's different, no?

bd A = cl A minus int A

[urban zinc]

What definition(s) of boundary do you know so far?

[rancid umbra]

usually it is, but in this case we have
I^2 = bd A ⊆ cl A ⊆ I^2

which is why we can do what bussy beaver suggests

[young stone]

saying cl(A) = I^2 is not the same as saying bd A = I^2

[urban zinc]

yeah, in this case it works since the restrictions on A guarantee the interior is empty but you have to argue that

[rancid umbra]

sorry, is what i wrote here wrong?

[tender halo]

its enough to find a function Q -> Q that sends open sets to dense sets

[urban zinc]

that says that if bd A = I^2 then cl A = I^2 but not the other way around

[solemn willow]

Set of all the points $x$ that satisfy:
If $B$ is any open rectangle with $x\in{B}$, then $B$ contains points of both $A$ and $R^n-A$

[gentle osprey]

cole

[solemn willow]

sorry for editing lol the textbook defines it somewhat informally

this is calc on manifolds by spivak btw

[urban zinc]

ah okay so we want to check that every point x in [0,1] x [0,1] satisfies these two properties: property 1) every B containing x contains a point in A and 2) every B containing x contains a point not in A

one of these properties just follows from how A is described in the question

which one?

[solemn willow]

every B containing x contains a point not in A since the horizontal and vertical is unique for each point in A?

[urban zinc]

yea!

so now we gotta figure out how to satisfy property 1

the problem claims that it's sufficient to show A contains a point in every 1/4 of the square, 1/16 of the square, etc.

why is this?

[solemn willow]

i suppose when you divide it into sufficiently small squares one can ensure that any possible open rectangle contains at least one of these squares and thus a point in A

ig that seems like something that needs proving itself

[urban zinc]

yes!

[urban zinc]

how would you go about proving it

[solemn willow]

here are my rough thoughts: letting s denote the length of the rectangle and w the width, you can keep dividing the square until you reach the point where the length of the square, L, is leq to s/2 and w/2. with a 1d argument we can show on a number line that on the interval [0,s] for any 0\leq\epsilon\leq{s/2}, we have that \epsilon+L\leq{s}. doing the same for w/2 shows that a square of length L can fit inside the rectangle

[urban zinc]

yup that's the right idea!

alright so now next part of the problem, how do you construct an A with the property of having points in every quarter, every sixteenth, etc.?

[solemn willow]

sorry went to grab food. i think the general idea is to iterate through all $n^2$ pieces and appropriately increment a rational to ensure uniqueness? something like
$A^r={\begin{bmatrix}\frac{r}{n^2}+\frac{c}{n}\\frac{c}{n^2}+\frac{r}{n}\end{bmatrix}:0\leq{c}\leq{n-1}}$ where r and c can be thought of as the row and column respectively. and then $A=A^0\cup{A^1}\cup\dots\cup{A^{n-1}}$

okay that was a disaster hold on

[gentle osprey]

cole

[solemn willow]

there

[urban zinc]

yeah! that's exactly right, how do you know this ensures uniqueness though?

[solemn willow]

if we regard a single row we know that none of $A^r$ share the same horizontal since the first entry as a function of $c$ is bijective in [0,1] and same principle for the vertical component

[gentle osprey]

cole

[solemn willow]

and then for the rest of the rows i can argue the vertical by construction since it differs from the corresponding entry in the previous row by r/n^2, while the horizontals are trivially different from that of the previous row(s)

wait now im confusing myself

no nvm this should be fine?

[urban zinc]

hmm this is okay for a single A^r, but how do you know that A^r and A^s don't have duplicates in the same row/column for r not equal to s?

[haughty jungle]

am I bugging or can you just like exactly follow the hint and like take the point in the middle of each quarter square recursively

[solemn willow]

but then you have shared verticals no?

[haughty jungle]

oh yes

I was bugging

🥀

[solemn willow]

suppose $A^r$ shares the x entry with $A^{r+a}$, then $\frac{r+a}{n^2}+\frac{c_1}{n}=\frac{r}{n^2}+\frac{c_2}{n}$ which implies that $\frac{a}{n}=c_2-c_1$ but $a\in[0,n-r-1]$ and so the fraction is never a natural number which is required since $c_i\in{N}$.

now suppose $A^r$ shares the y entry with $A^{r+a}$, then by the same procedure we arrive at $na=c_1-c_2$. The trivial case of $a=0\Rightarrow{c_1=c_2}$ can be ignored. This would imply that $c_1-c_2\geq{n}$ which is nonsense

[gentle osprey]

cole

[urban zinc]

oh wait but then are you taking the limit as n -> inf? I'm confused

or some sort of union

[solemn willow]

i was just doing it for arbitrary n

[urban zinc]

you need the same set to work for all n though

so you could try taking the union of all the sets for all n?

then you'd need to check that you don't introduce any duplicate points in any row/column but after that you'd be done

[solemn willow]

if im not mistaken i think my most recent proof above does this. was to demonstrate that each $A^i$ in the union $A=A^0\cup\dots\cup{A^{n-1}}$ contains points that have unique horizontals/verticals and thus the union also has no duplicates

[gentle osprey]

cole

[solemn willow]

oh i see where the confusion is let me fix it up

[solemn willow]

suppose there exists $v_1\in{A^r}$ that shares the x entry with $v_2\in{A^{r+a}}$, then $\frac{r+a}{n^2}+\frac{c_1}{n}=\frac{r}{n^2}+\frac{c_2}{n}$ for some $c_1$,$c_2$ which implies that $\frac{a}{n}=c_2-c_1$ but $a\in[0,n-r-1]$ and so the fraction is never a natural number which is required since $c_i\in{N}$.

now suppose there exists $v_1\in{A^r}$ that shares the y entry with $v_2\in{A^{r+a}}$, then by the same procedure we arrive at $na=c_1-c_2$. The trivial case of $a=0\Rightarrow{c_1=c_2}$ can be ignored. This would imply that $c_1-c_2\geq{n}$ which is nonsense

[gentle osprey]

cole

[urban zinc]

this assumes the n is the same for both points

I'm saying you could take the union of all these sets for all n

and then you'd have to check for collision if there's different values of n

[solemn willow]

OH i see what you mean okay thank you

[urban zinc]

You're welcome! :)

Btw here's another approach you could take here: if you don't care about the specific formula for the points in A, you can invoke the axiom of choice to just say "pick one point in each quadrant not in the same row/column as any of the points chosen up until then, then pick one point in each sixteenth not in the same row/column as any points chosen up until then, etc." Since at each step, you only have finitely many points to "avoid collisions with" and infinitely many points to choose from, you're guaranteed that you can choose such a point for each of the 4^n squares for all n, and then you're done.

Concrete formulas are always nice to have though

[solemn willow]

ah this is interesting… I have yet to be formally introduced to the axiom of choice in university but I keep seeing it sporadically pop up

probably a sign to do some reading beforehand lol

[urban zinc]

It's probably not that important to understand explicitly until later parts of your mathematical career. Often one uses it implicitly without comment in proofs anyways. (In fact, it was only enshrined as an axiom after mathematicians realized with a bit of surprise that it didn't follow from the other axioms.)

It's interesting to read about though.

[sly geyser]

what's the difference between a cluster point and an adherent point

(pls ping)

[cloud kindle]

For limit points of a subset S of X we require every neighbourhood of a point x contains a point, a, in S which is different from x

For adherent points we just require their intersection to be non empty, I.e. a=x is allowed

[sly geyser]

so it's only different if the adherent point is an element of S

so adherent points are just elements of S plus limit points right?

[cloud kindle]

I’m not sure I know what you mean. All limit points are certainly adherent points, that much is true.

Essentially for a limit point we require the intersection of any neighbourhood of x and S to contain at least 2 points, for adherent points we only require it to contain one point

[sly geyser]

two points?

[cloud kindle]

Ignore that I think i was being silly, but the difference is for limit point the intersection must contain a point which isn’t x, but we don’t assume that for adherent points

[fierce lily]

just a quick question that I want to confirm: (2)->(1) actually can conclude that T=T', is it correct( i think it is)

[tiny obsidian]

No, from (2) you can only conclude that T subset T'

[fierce lily]

I am actually consider it as: collection B' is also the basis for T by def that every open sets U in T and x in U have x subset of elements of B' subset of U, so we write it as union of all elements of B'=T, so T=T' because all union of elements of basis equals topology.

[tiny obsidian]

you're not guaranteed B' is a basis for T

not all of the definition of a basis is satisfied

sure x in B' subset U for all U in T and x in U, but to be a basis you also need B' open

[fierce lily]

ok, i just noticed it. really appreciate for pointing out

[steel wing]

Did anybody watch Michael Penn's video about how R^2\Q^2 is path connected

[cloud kindle]

I have not, I think I can believe that though, something something density you can take weird snaking paths to avoid a lattice of Q

Sounds like an interesting problem though, quite funky

[grave solstice]

let p,q be distinct points in R^2 setminus Q^2. As long as one of the coordinates is irrational you are fine. Move from q first in the x coordinate in a straight line until you are "close" to the x coordinate of p, but end up at an irrational x-coordinate. Do the same with the y-coordinate. Repeat this process ad-infinitum, you should obtain a legit curve

actually for most points p,q the segment joining them won't have any points from Q^2, so this is hardly surprising

[hexed steppe]

can you just use this fact and consider two line segments to handle the general case

[rancid umbra]

the argument i like is as follows:
the set of lines through a point which contain a rational coordinate is countable, so uncountably many of them lie in R^2 - Q^2.

you can do this for two points and the lines have to intersect somewhere (if they are parallel, just pick a different line through one of them)

[hexed steppe]

yeah

[grave solstice]

do you even need to mention uncountability? You just need two distinct lines with no points in Q^2 (through every point)

[rancid umbra]

its a nice way to ensure that such a line exists, but im not sure how else you would do that atm

[hexed steppe]

can you show that for (x,y) not in Q^2, a line L passing through (x,y) contains a point in Q^2 iff L - (x,y) contains a point in Q^2 other than (0,0)?

if so then it’s easy to just construct the two lines

hmm no that is false

for trivial reasons

like if L is horizontal and (x,y) = (0, pi)

[rancid umbra]

i have a hunch that you can do something with arcs

[hexed steppe]

does this argument work for any countable subset of R^2

[rancid umbra]

yes, it also works for R^n - C for any countable subset C of R^n

it fails for sets of measure zero though

e.g., R x {0}

[hexed steppe]

what if they have zero hausdorff dimension

[steel wing]

Why is nobody watching the video lol

[hexed steppe]

why should we

does he discuss these things

[steel wing]

I mean all of the arguments so far have been incorrect... The video gives an easy solution

[rancid umbra]

i haven’t worked with this enough to have anything insightful to say

[rancid umbra]

the arguments for what? constructing a path?

[rancid umbra]

i thought this was the argument in the video

[steel wing]

https://www.youtube.com/watch?v=EYLbV7iTjpE

[rancid umbra]

he makes the same argument as this one

at around 11 or 12 minutes into the video

[hexed steppe]

lol wut

c squared gave a complete proof

[hexed steppe]

me neither

[rancid umbra]

is this thing guaranteed to be continuous? like, what if you trace out a path that wiggles like sin(1/x)?

[hexed steppe]

this is the only vaguely relevant thing ik of

[grave solstice]

yeah, you can imagine like an infinite staircase. But I think literally anything works as long as the distances get smaller (like, you could spiral around, or spiral then staircase, then spiral in the other direction, etc)

[grave solstice]

the correct analogy here would be wiggling like xsin(1/x), so it's fine

[steel wing]

@rancid umbra's argument is correct because it's the same argument essentially

[rancid umbra]

mhm, right, i see

[hexed steppe]

ok more relevant: if dim A < 1 then by fubini A_x is empty for a.e. x

[hexed steppe]

that is not the reason it’s correct…

michael penn is not the arbiter of mathematical rigor

[rancid umbra]

lol

[rancid umbra]

oh that’s neat

[hexed steppe]

im guessing it’s hard to improve over that though

actually fubini might not work

[rancid umbra]

we’re out of my realm of knowledge now haha

[urban zinc]

geometric measure theory

our last few classes of fourier analysis were dedicated to this stuff

it was fire

[hexed steppe]

i guess the point is to project A onto the line and use that the 1d lebesgue measure is zero

not exactly fubini

[timber slate]

@narrow tide found panopto yet?
consider natural bijection of open and closed sets of a topology

[gaunt linden]

That is a quite neat argument, especially since it works for any countable set of forbidden points. But in the particular case of R² \ Q², we can also avoid the appeal to countability simply by:

For every point in R² \ Q², depending on which coordinate is irrational, either the horizontal line through it or the vertical line through it is entirely in R² \ Q².
This gives you, for each of p and q, an axis-parallel line through it. Connect each of those lines to the line x+y=pi (which is itself disjoint from Q², and intersects every axis-parallel line in the plane).

[cloud kindle]

General topology exam in an hour, I’m sick as a dog, the entirety of my studying has been rereading the notes and skimming last years exam before deciding it looked fine. This can only go well.

[wispy veldt]

https://cdn.discordapp.com/attachments/1116324549835890738/1368081610787913789/image0.gif

[unreal stratus]

Good luck

[fringe thorn]

I hope the exam went well

[cloud kindle]

It was the right call, 50 marker defining the zariski topolgy and showing it’s not Hausdorff lol

People who didn’t take alggeo are rather upset but so it goes

It was honestly harder if you had taken alggeo because the notation he used to define things was impenetrably bad

I definitely could’ve done far better if I’d studied but I think I have a B given how ill I’ve been i think I’ll take that

[dry seal]

Idk what y'all covered in class but if I was asked to define something that can only be defined using shit from a class that isn't even a prerequisite on an exam I'm writing an email as soon as I get home lol

[novel acorn]

Well I don't think that's what he said

He said the professor defined what the zariski topology is

And you were supposed to prove something using the definition given in the exam

You didn't need to take an alg geo class to understand it

[dry seal]

I read it as the question asked you to define the zariski topology and then also prove it is not Hausdorff

Unless I'm missing something

I also have no idea if 50 marks is supposed to be 50% because if it is then what is even that exam

[versed pivot]

it's pretty common to have "restate the definition of something we defined in class" as a question, or first part of a question to help the student get started

"give definition for this word none of you have heard before" would be too nonsensical to have on an exam for me to think this is the correct interpretation

[dry seal]

Usually those definition type questions are about defining something general to the subject rather than a specific example so I see what you mean

[dry seal]

Would be interesting to get this clarified though

[brittle rapids]

got a q

trying to show that if p: G -> H is an open surjective continuous homomorphism between top groups and ker(p) is discrete then i can find a nbd U of e_H such that p|U is injective

it feels very true but just getting a nbd of e_H that doesn't intersect ker(p) \ e_H is way too weak for me

and i can't use anything like local compactness or path connected so i'm struggling

hint only please

[opaque scroll]

So I guess you mean neighborhood of e_G.
Anyway, notice that
f(x) = f(y), then xy^-1 is in the kernel. So if you can find neighborhood such that xy^-1 is always sufficiently close to the identity you would be done by discreteness of the kernel.

[cloud kindle]

He in the worst way possible gave a purely set theoretic description of the vanishing locus of an ideal of R^n[x] (like a genuinely impenetrable definition it wasn’t good) then described a set based on that, then said prove there is a topology on R^n whose close sets are precisely the elements of the set he defined

Then just a bunch of other questions about properties of the zariski topology

You don’t need any alggeo to do it but there are currently a lot of students very upset about it

The rest of the exam was a question with like 8 true/false give a proof or counter example about properties of continuous maps, then a question about the dyadic rationals

[dry seal]

This sounds like a master's course which makes this totally insane to me

Like what is the instructor doing

[cloud kindle]

The dyadic rationals was the worst question I genuinely couldn’t make sense of what i was being asked to do, he defined them in there the notation he used was very unclear

[cloud kindle]

I mean it’s a final year course at a pretty decent uni it’s not going to be super easy. Also the guy that set the exam didn’t actually teach the course, nor is he a topologist so it’s all a bit funky

[dry seal]

If it's a good uni then why don't they have a topologist teaching topology 😐

Like when I learned basic topology 2 years ago it was still taught by an actual algebraic topologist and my institution isn't even in the top 100 rn

Also it's totally insane to me that you'd have someone not teaching the course make the exam outside of qualifying exams

It's borderline impossible for instructors to accurately predict how much and what material they'll make it through by the time you get to the exam so they're really the only ones qualified to make it imo

Seems to me like they're more concerned with making it hard than making it instructive

[brittle rapids]

V that doesn't intersect and then symmetric U^2 ⊆ V

thanks

[cloud kindle]

The guy who usually teaches it was on sabbatical. It’s also UG topology any mathematician could teach it it’s rather basic

[cloud kindle]

This was only because the guy who was supposed to teach the course was ill all year and couldn’t make it in, so we had various people sub in, but he had to be the person to set the exam because it’s “his course”

[dry seal]

That's fair

[still vortex]

How to show that ]0,1[^N is not sigma-compact?

[tender halo]

remind me what the brackets mean

do you just mean R^N

[opaque scroll]

The brackets mean open interval

So (0, 1)^N in less french notation

[tender halo]

ok so R^N hmm

i think bct helps right

[still vortex]

What's bct?

[tender halo]

if it is sigma compact then every compact set in the union is nowhere dense

[opaque scroll]

I'm thinking, let
X = (0, 1)^N and let fi: X -> (0, 1) be the projections.

For any compact subspace C, fi(C) is compact so not all of (0,1)

Now let Cn be a countable family of compact sets. And pick xi outside of fi(Ci). Then (xi) is not in any of the Cn

[tender halo]

oh thats also a good solution

[opaque scroll]

I guess the same works for any infinite product of non-compact sets

[tender halo]

so their union cannot be the whole space because R^N is completely metrizable

[arctic mural]

What is a continuous open map in number 12?

[unreal stratus]

A map which is continuous and open, where "open" means that it sends open sets to open sets

[arctic mural]

So, an embedding?

I mean. Wouldn't it imply that X is homeomorphic to f(X)

[hidden abyss]

Not necessarily, it need not be injective

[unreal stratus]

Or surjective for that matter

[arctic mural]

Ok

[queen prism]

well it would be surjective onto its image but yea

[unreal stratus]

Sure yeah I just mean like

There is no reason to think this should be injective but not surjective

[gritty widget]

Both images and preimages of open sets are sent to open sets

[candid jewel]

hey, can anybody tell me about the basis of a point in a topology? i know about basis in a global sense that every open set can be written as the union of the basis elements, now what is "basis of a point"?

[haughty jungle]

it's a collection of open sets containing the point such that any neighborhood of the point contains a member of the collection

[candid jewel]

so like can it be said as the smallest neighbourhood of x? as every open neighbourhood will at least contain one of the basis elements of x??

or is it like the basis of neighbourhoods of x?

[prime elbow]

in R, with euclidean topology you can't get smallest neighborhood of x

because every neighborhood around 0, say (-e, e) you will get always smaller neighborhood (-e/2, e/2)

[candid jewel]

yeahh, got it

[prime elbow]

maybe that's true

[candid jewel]

hmm, i was learning about he first countability axiom, and i was confused about the basis of the element

[toxic prism]

hi guys, is there a topology reading group going on right now that i can join ?

[pallid comet]

is it true that if A and B are topological spaces such that A is homeomorphic to a proper subset of B and B is homeomorphic to a proper subset of A, then A and B are homeomorphic?

[plush folio]

No, consider [0, 1] and R for example

[unreal stratus]

I wonder what a "minimal" counterexample would be

e.g. a countable one

[alpine nest]

Are Q and Q \cap [0,1] homeomorphic?

[tender halo]

surprisingly yes

[alpine nest]

Neat

Is the homeomorphism able to be reasonably intuitively described?

[tender halo]

all countable metric spaces with no isolated points are homeomorphic

(to Q)

[tender halo]

yeaa you can like fiddle with half open segments of rationals

[alpine nest]

Yeah, I'm beginning to see that

[tender halo]

so Q^n is homemomorphic to Q for any n from 1 to omega included

also p adic topopology on Q is homemomorphic to Q

[opaque scroll]

Consider A = N with the topology whose open sets are [0, 2n], and B the subspace N\{0}.

A has an open point, B does not and n |-> n+2 is an embedding

Otherwise I guess you can do some kind of Eilenberg Mazur swindle.

Like countable disjoint union of sierpinski space and countable disjoint union plus one extra point

[tender halo]

i mean you can do the same thing again but with rationals

Q + 1 and Q are not homemomorphic but contain each other

bonus points for being all metric and nice

[plush folio]

Are there any simple properties that would make it true btw? Like if A and B were compact for example?

[tender halo]

not really

I and I + 1 are both compact and contain each other

[opaque scroll]

If you replace homeomorphic with isometric, then ccompact should probably do it

[plush folio]

I see it seems like a property that is rare even for algebraic structures; AFAIK you can find two non-isomorphic groups containing each other, and I think even two fields?

[unreal stratus]

Ah yeah very nice

I like the swindle a lot lol

I can't remember what the context was but I saw smth v similar to this recently lol

Like not just a swindle but a similar one to this

[unreal stratus]

For fields yes, because alg closure of C(t) is C I think, or maybe ((t))?

So you can do C and C(t)

[opaque scroll]

I mean a swindle works anytime you have a structure with a non-isomorphic substructure pretty much

[unreal stratus]

It is kind of funny cause hm

In derived stuff like subobjects often don't make the best sense, or like aren't always useful

Hmm

Besides summands or whatever

[opaque scroll]

Triangulated categories don't really have monomorphism

[unreal stratus]

Yeah exactly

I have a funny thing where I show a map is injective on homotopy groups

And I initially was like huh does that make it have vanishing fiber hence be an equivalence

[opaque scroll]

Same in Set or in a semisimple category

[unreal stratus]

Why Set?

Maybe I am silly

[opaque scroll]

Why monomorphisms split? Well the inclusion of the empty set doesn't

[unreal stratus]

No I mean like

We didn't say monos split

So I was confused like assumed you were saying Set doesn't have a good notion of mono aha

[opaque scroll]

Well saying you can't do this swindle there

[unreal stratus]

Oh I mean ig you can sometimes just

I was more remarking that subobjects in general aren't as important

And only split ones rly make sense and sure then it is quite different

Ok nice I get ur angle now sorry

[opaque scroll]

Anyway, vector spaces would be an algebraic structure where you don't have this A and B relationship thingy

[unreal stratus]

Yee

[brave geode]

sanity check

if X is a space and B_X the borel sigma algebra, then does X being lindelof imply that the presheaf of measurable functions U → ℝ ∪ {∞} is a sheaf?

i am almost certainly sure that the answer is no and we require hereditarily lindelof

if the answer is yes then i'd only like affirmation (no hints)

[heady skiff]

Can somebody please explain the highlighted line? In particular, I thought that |K_1| is open (since it is the intersection of Euclidean space - s (which is open) with |K|), so that |K_1| \subset X_1^o makes sense. On the other hand, why is |K_2| - |K_1| an open subset of |K_2| if |K_1| is open?

[chrome trout]

If $X$ is compact hausdorff and $G$ acts continuously on $X$, and $R$ is a $G$-invariant closed equivalence relation, is $X/R$ hausdorff?

[gentle osprey]

qwerty

[chrome trout]

It suffices to show that $\pi:X\to X/R$ is an open map, but I am unable to see this

[gentle osprey]

qwerty

[iron bolt]

I think G doesn't matter for that question at all? not even the fact that it exists

because every equivalence relation is invariant with respect to the trivial action of any group

so you're just asking whether X/R is Hausdorff for every compact Hausdorff space X and closed equivalence relation R

[brittle rapids]

it doesn't sound true

actually quasicompactness might do something

hm

[unreal stratus]

What do you mean by closed? Closed as a subset of X x X?

[unreal stratus]

Pretty sure this is true right

[brittle rapids]

equivalently every equivalence class is closed i think

[unreal stratus]

Like use the fact compact Hausdorff spaces are ||normal||

[iron bolt]

if I knew that I would've said so

idk

[brittle rapids]

you need the saturation of your two disjoint open sets to also be disjoint

[unreal stratus]

True

[brittle rapids]

(referring to hot potato's hint)

[unreal stratus]

But there's also more to do

[unreal stratus]

Yeah it is, though it's been a while and I've forgotten the proof, lol

[brittle rapids]

it suffices to separate points not on the same equivalence class, by saturated open nbds, so i'm not sure compact => normal will be helpful (?)

maybe you do it in XxX tho

oh no need for anything fancy

[chrome trout]

Ah yes, it seems the statement is true. $X$ compact hausdorff and $R$ closed implies $X\to X/R$ is closed, and the image of a normal space under continuous closed map is normal.

[gentle osprey]

qwerty

[unreal stratus]

Oh nice lol

Is it easy to see this map is closed

I am glad suggesting normality wasn't bogus aha

[rugged dagger]

I had a very fun exercise in an exam it wrote like this:

Let Y be Hausdorff, and X a topological space, then if we have the following equivalence relation: x~x' <-> f(x)=f(x') for every continuous function that go from X to Y. Show that X/~ is Hausdorff.

There was also two indications:

Let Xi be a topological space for every i from I

Then prod(Xi) is Hausdorff (for every i in I) if and only if every Xi is Hausdorff

And another one that said, let X be a topological space, then X is Hausdorff if and only if there exists a hausdorff space Y and a continuous and injective function from X to Y

[plush folio]

Wouldn't ||X/~ be a singleton? Let f be a constant function, then f(x) = f(x') and therefore x ~ x' for all x, x' in X||?

[rugged dagger]

No, x~x' if and only if every continuous function is such that f(x) = f(x'), then if the ONLY continuous function was the constant one, it would be a singleton, but if you have more than one continuous function you need that for every function f(x) = f(x')

As an example, in R, if you have let's say the constant function and e^x

(From R to R)

No x would be equivalent to a x'

As e^x is not injective

*is injective

[plush folio]

Ah, I see. Interesting problem

[rugged dagger]

For there to actually be some x~x' that isn't itself, it must be two spaces such that every function X to Y is non injective

Basically

[rugged dagger]

Indeed

[unreal stratus]

Yeah there's a notion of "Hausdorffication" where you do this for all choices of Y as well [with X fixed]

Presumably this exercise is a variant on that

[rugged dagger]

Hm to solve the problem you had to define a function F where each component of the imagine is the image of every possible continuous function

Is it similar?

[rugged dagger]

So you would have injectivity

(And go from X/~ to prod(Y) where there's one Y for every continuous function there exists)

The other "indications" had to be proven but uh, the main exercise was the top one

[rugged dagger]

Any example using that idea of the hausforffication?

[tender halo]

there is a similar idea Tychonoff spaces, if you take a quotient of a Hausdorff space by all elements that have the same image under functions into R, then the quotient is Tychonoff

moreover, it has the same ring of continuous functions, and the mapping C(X/~) -> С(X): f |-> f \circ q where q is the quotient function is an isomorphism

[rugged dagger]

You don't take the quotient of a hausdorff space here tho

X isn't Hausdorff

But I understand the similarity

[heady skiff]

Anyone?

[severe swan]

Is there an error with the proof here?

what if we want to prove that every open subspace of a compact space is compact? can we take the closure of an open subset A, apply the proof as above, then conclude that as Ua ⊇ cl(A) ⊇ A, the statement is true?

[tender halo]

the union doesnt contain cl A

well it can but it doesnt have to

[heady skiff]

Any point set gods?

[vestal moth]

where do I go for basic set theory?

Undergrad level

[queen prism]

#proofs-and-logic

[unreal stratus]

Well that last bit isn't the implication - it's an "and". Probably just |K2| is closed in |K| so the complement is open, then intersect with |K1|

[warped bone]

If i want to show that S¹={(x¹)²+(x²)²=1} is a closed in R², i can show that it's complementary is open, since the complementary of S¹ is the union of {open balls with r<1} and {open balls with r>1} , the complementary of S¹ is a union of open sets, and i can conclude that S¹ is closed

[warped bone]

It says in my textbook that i can prove the same using hausdorff properties and/or cotinuous functions

Any advice on how to do so?

[ruby delta]

the union of {open balls with r<1} and {open balls with r>1} is just R^2

[rancid umbra]

i feel like you have the right idea but aren’t phrasing it correctly

if you could describe in words what the picture of R^2 - S^1 looks like, how would you do it?

like what parts of that picture are there?

[warped bone]

R² without the circumference

[rancid umbra]

without the unit circle, yes

ehrm

[rancid umbra]

is that how you would describe it?

[warped bone]

Yes?

[rancid umbra]

can you try to add some more detail?

there are two very distinct parts of R^2 - S^1

[warped bone]

The inner disc and the area on the outside of the circumference

My initial thought was to consider both of them as union of open balls but now that i think of it is not right at all

[rancid umbra]

right. the inner disc. it’s the set of all points whose length is less than 1

that’s just an open ball of radius 1 centered at the origin

[warped bone]

And that is open

[rancid umbra]

yes

[warped bone]

Now i need to prove that the outside is open too

[rancid umbra]

in a similar way, how would you describe the outer region

[warped bone]

Lenght >1

[rancid umbra]

right

so you need to be able to find an open ball around any point with norm larger than 1 that stays in the outer region

drawing a picture should help

[ruby delta]

actually, you made me realise it's much easier to just prove that ||the norm function is continuous|| to do the question

[rancid umbra]

i was going to mention that as well

don’t even need the norm function (it involves a sqrt), just (x,y) |—> x^2 + y^2

[hexed steppe]

you can also consider a closed ball of radius R>1 minus an open ball of radius r<1, and take the intersection of these closed annuli over R>1>r

[dry seal]

https://en.wikipedia.org/wiki/Arens–Fort_space