point-set-topology

[real granite]

I mean, without getting into anything about geometric series, if you did what I suggested, you could have c=1 and d(Tx, Ty) < d(x,y) (note the strict inequality), in which case T is still shrinking the distances between points

[viral terrace]

yes, but a convex function on [0,1) does that, yet [0,1) itself isn't shrunk

it's still 1 unit long

and it's because it gets arbitrarily slower at shrinking

in the original definition, it would only go up to c and no less

without making the diameter of a set smaller, there wouldn't be the fixed point under iterations

[real granite]

What? An example might make things clearer

f:[0,1) -> [0,1) defined by f(x)=x^2 is convex, but I'm not sure what you mean by "[0,1) itself isn't shrunk"

[viral terrace]

so, $|x_2^2-x_1^2|< |x_2-x_1|$ for all $x_1 \neq x_2$. And your definition says it's a contraction

[gentle osprey]

Capy

[tender halo]

i mean it has a fixed point

doesnt have one on (0; 1) though

[viral terrace]

anyway, the map is surjective

and I would expect [0,1) to be mapped onto [0, a) with a<1

[real granite]

Yeah I suppose this makes sense

I presume the reason we can't let m tend to infty is because the RHS of V.2 would be zero?

[quick crane]

isn't it because V.2 is only true for $m < n$?

[gentle osprey]

L

[real granite]

Uh yes that is a good point

[sonic crane]

Are the topologies induced by any collection of metrics on a set always comparable?

[tender halo]

what do you mean by a topology induced by a collection of metrics?

[real granite]

I assume the idea is that you might have a set X equipped with various metrics d_i, and we want to know whether the topologies induced by each d_i are comparable

[sonic crane]

Yeah i figured that question was not well posed

[sonic crane]

This is what i meant yea

[tender halo]

oh ok, then not true

[sonic crane]

Ill show u why i was wondering this

This lemma

So i was wondering, does this mean if u have any two metrics, we can always talk about whether one topology is finer than the other?

[tender halo]

no, not true

take uhh idk

R

[sonic crane]

But maybe not because the condition that the d’ ball is contained in the other probably implies that the d and d’ metrics are related somehow

[tender halo]

add a point to it, its still metrizable

then take a nontrivial bijection on R + point

and take the induced metric

actually nvm, too hard (it works but probably not as intuitive)

take the metric topology on {0, 1, 1/2, 1/3, 1/4, ...}

then the same topology but on {1, 0, 1/2, 1/3, 1/4, ...} (same set but you exchanged 0 and 1) places

not comparable

also consider this

if it were true, then for any bijective map, either the map itself or its inverse would be continuous

which is obviously not true

[real granite]

Do continuity and continuity of inverse not need to be stated explicitly?

Oh actually I'm muddling homomorphism with homeomorphism. Isomorphisms are not a kind of homeomorphism

[viral terrace]

they give rise to homeomorphisms

since metric spaces have topology

[real granite]

Hmmm. Let me see if I've got this right.

1. f being an isometry is not sufficient for f to be a homeomorphism (f may not be bijective)
2. f being bijective is not sufficient for f to be a homeomorphism (as this doesn't tell you anything about continuity)
3. f being bijective is a necessary condition for f to be a homeomorphism (that is in the definition)
4. f being bijective and an isometry is sufficient to be a homeomorphism (the fact that it is an isometry means you can make a topological argument for continuity and continuity of inverse)

Is that right?

I assume that f being an isometry is not necessary for it to be a homeomorphism (I'm thinking that is f scales everything by a factor of 2).

[viral terrace]

Prove that every isometry is bijective.

Well, at least injective, not bijective.

But you can choose codomain

so, every isometry is homeomorphism of the domain to the range. the only problem if the codomain is large that the domain and it's not surjective

[grave solstice]

how does this follow from Theorem 4.17 and the definition directly?

The definition is

And the theorem is

the theorem works only when Z is simply connected, pathwise connected and locally pathwise connected, but the universal property is stated for Z just connected

so how is this direct. Idk if I'm missing something

[dim meadow]

I think the Z in the definition corresponds to Y in the theorem

@grave solstice

[grave solstice]

Ah, you are right

Thank you

[civic juniper]

I understand the reduction of the proposition to the totally disconnected case. So, this would mean we need to show that intersection of all open subgroups is trivial. But I do not understand how the latter part implies it. Can someone explain?

[plush folio]

Stupid question: [1, 2] is a complete metric space, and f(x) = x/2 is a contraction on [1, 2]. Why doesn't the Banach fixed point theorem work in this case?

[civic juniper]

It's not a contraction of the space in the sense that it doesn't map the space to itself.

You can rectify the problem by adding 1 to f(x). Then 2 is a fixed point.

[plush folio]

Right, f is not a map from [1, 2] to [1, 2] thanks 👍

[wispy veldt]

im not familiar with this topic much but here is my attemp, since the space is hausdorff is g is not in C then you can seperate g and the identity by open nbh V and U respectively, and we proved we can find a open subgroup in U, this subgroup doesnt contain g

just needs modification to get a compact nbh

which idk if thats possible to keep the seperation of g and e

okay i think we can since if we take a compact nbh K of e and intersect it with our U, we have a regular space K cap U

[civic juniper]

Nice. Question, why do we need a compact neighbourhood, isn't it suffice that for any nonidentity element g, we can find a open normal subgroup not containing g?

[civic juniper]

I mean compactness here.

[real granite]

Is my proof valid?

1. Assume $f:X\to Y$ is continuous. Given any $U\subseteq f(X)$ open in $f(X)$, because the topology on $f(X)$ is inherited from the topology on $Y$, $U$ is also open in $Y$. Then using our assumption, $f^{-1} (U)$ is open in $X$.

2. Assume now that $f:X\to f(X)$ is continuous. Given any $U\subseteq Y$ open in $Y$, $U\cap f(X)=V\subseteq f(X)$ is open in $f(X)$ (this is the definition of a subspace topology). Then, by assumption, $f^{-1}(U)= f^{-1}(V)$ is open in $X$.

[gentle osprey]

Douglas

[civic juniper]

Note, in 1, U being open in f(X) does not imply U is open in Y (Recall definition of subspace topology - seems you already know this).

[real granite]

How would you "patch" it?

[limpid fern]

The idea here is that preimage of U under f is precisely preimage of (U cap f[X])

oops I just realized that's the answer

but anyway if you're preimaging under f then it should make sense that only the stuff thats in U thats also in the range of X would have a preimage so to speak

if U was completely disjoint from range f then the preimage would be empty

[real granite]

Um, is that not a good thing?

I get that the preimage will just be the intersection with f(X) (you can "trim off" the stuff in $Y \setminus f(X)$), but how would you show that for arbitrary open sets in f(X), there exists an open set in Y that intersects f(X) to form it, i.e. $$\forall V \in \tau_{f(X)},\ \exists U \in \tau_Y,\ \text{s.t.}\ U\cap f(X)=V?$$

Then you can use that to argue $f^{-1}(U)=f^{-1}(U\cap f(X))=f^{-1}(V)$, and since $U$ is open in $Y$, $f^{-1}(U)$ is open in $X$, and so too is $f^{-1}(V)$. Because this works for any $V$ open in $f(X)$, we have that $f:X\to f(X)$ is continuous.

[gentle osprey]

Douglas

[real granite]

It's just the existence of open U in Y that is the problem

[prime elbow]

Then if A and B are nowhere in M. Now let any non-empty set U of M and U intersection ( A union B) is (U intersection A) union ( U intersection B).

Let since ( A and U) and (B and U) is not dense in U, there exists open set V_A in U such that V_A has empty intersection with (A intersection U) and V_B in U such that it has empty intersection with ( B intersection U)

So V_A intersection V_B has empty intersection with ( A and U) union ( B and U) , right?

So (A and U) union ( B and U) is not dense in U, and U is arbitrary

[real granite]

1. What is M?
2. We haven't studied density

Would any constant function f(x)=c work? f sends every set to the singleton {c}, which is closed and not open.

[limpid fern]

yes

[limpid fern]

Recall the definition of subspace topology. What you wrote down is the definition actually.

If a set V is open in f[X] then there is some U open in Y such that U cap f[X] = V

so it is necessarily true

[real granite]

Ah I see what you're saying now

This makes sense

[prime elbow]

M is metric space

[plush folio]

If you want to be explicit, you could write $U = (U \setminus f(X)) \cup (U \cap f(X))$. Then the preimage is $f^{-1}(U) = f^{-1}(U\setminus f(X)) \cup f^{-1}(U\cap f(X)) = f^{-1}(U \cap f(X))$, which is open

[gentle osprey]

sheddow

[wispy veldt]

how will you find a normal subgroup not containing g? i was mainly using what was proved in the theorem that each compact nbh contains a open subgroup

[grave solstice]

what does "connected cover" mean?

they are assuming that X is locally connected btw

[grave solstice]

I suppose that means that Y is connected?

[rancid umbra]

they likely mean that Y is connected, as you have pointed out.
a deck transformation that fixes a point must be the identity by the unique lift property

under these hypotheses, of course

[grave solstice]

thanks, just wanted to make sure

[grave solstice]

btw, Deck transformation and "automorphism" needn't be the same right? I think Deck transformations needn't be homeomorphisms?

[red yoke]

They are by definition

[rancid umbra]

it’s a homeomorphism that satisfies pf = p

[grave solstice]

oops yeah

I forgot what homeomorphism meant

[red yoke]

Deck transformations are the automorphisms of a cover over the base space

[grave solstice]

well idk what I did, I think I just saw homeomorphism in the definition and read it as just "continuous mapping"

[red yoke]

If the covering is connected any endomorphism is an automorphism anyway I think

[grave solstice]

that's what I thought. But it is not clear to me

I believe you can show that if f : Y-->Y is an endomorphism then f(Y) is open in Y

Suppose p : Y-->X is a cover and f : Y-->Y is an endomorphism of the cover. Let x in X and consider the preimages x_i of x. Say V is a neighborhood of x and U_i are disjoint nhoods of the x_i mapping homeomorphically to V with p^{-1}(V)= union U_i. Then I wonder if it is necessarily true that for each i, f(U_i)=U_j for some j

this would be true if you could assume that the U_i are connected, I think

[rancid umbra]

it is if V is connected i believe

then the decomposition into slices is unique, as each slice must be a component of the preimage of V

[grave solstice]

so if X were locally connected you would be able to choose such a V?

[real granite]

Is it valid to say "obviously x_n converges to x under rho if and only if it converges to x under d, hence rho and d are equivalent, therefore they generate the same topologies"

Seeing the solution, I think the ball argument is a little better because it emphasises the point that openess is all about being about to fit balls into sets, but I presume my reasoning is alright

[rancid umbra]

yes, that’s correct

[real granite]

.

[rancid umbra]

why do you think that having the same convergent sequences means that the topologies are the same?

[real granite]

"hence rho and d are equivalent"

[rancid umbra]

hmm. i think that argument needs some justification

its easier and more direct to just say that the metrics are equivalent: p <= d <= 2p

drop the whole thing about the spaces having the same convergent sequences

[real granite]

iirc that is how equivalence was defined in my topology module

[rancid umbra]

the definition that i had in mind was that d1 and d2 are equivalent iff for each x, there are positive real numbers a and b such that for each y, a d1(x,y) <= d2(x,y) <= b d1(x,y)

but since that is the one you are given

your argument is logical

the only other thing that i would check for is somewhere in your class showing that two equivalent metrics determine the same topology

otherwise, an argument like the one given in the solution is needed for it to be complete, without skipping any details

[real granite]

When you say complete, do you mean complete as in metric space or complete as in full?

[rancid umbra]

i mean without skipping any details

[median sand]

If L is a topological field, does the Krull topology on G(L/K) coincide with the compact open topology?

[grave solstice]

are you assuming that the functions in G(L/K) are continuous?

I think the compact-open topology is only defined for subspaces of continuous functions X-->Y (in this case I assume X=Y=L), hence why I'm asking

but note that you can consider Q(sqrt 2)/Q and sqrt 2-->-sqrt 2 is not continuous

[median sand]

Oh right, I forgot COT requires continuous functions, nvm.

[grave solstice]

Your question does make sense if K=Q_p, for example (or more generally if K is Henselian) and L is algebraic over K. Then the functions in G(L/K) are isometries. But I'm not sure what happens in that case

[umbral panther]

I think you should have a continuous map from the Galois group to the mapping space. But the Galois group is compact, so an injective map to a Hausdorff space is a homeomorphisms onto its image

This is only interesting for an infinite extension, but then L is not locally compact and the compact open topology might be pathological

[prime elbow]

If f is continuous mapping from metric space M to N. Now suppose A is a separable subset of M then I need to prove it f(A) is separable.

Claim: since A is separable so there exists D in A such that D is countably dense in A.

Now f(D) is dense in f(A).

It is just my claim but I think we can prove it, right?

A metric space is separable which has countably dense set

[limpid fern]

it is true

Actually M does not even need to be a metric space

[prime elbow]

I need to prove f(D) is countable

[limpid fern]

it is automatic

[prime elbow]

Is it true in topology too?

[limpid fern]

yes

a space is seperable if it has a countable dense subset

[prime elbow]

How?

[limpid fern]

the cardinality of f[D] is at most the cardinality of D

[prime elbow]

Yes

[limpid fern]

so it follows that f[D] is countable

[prime elbow]

But what if I need an explicitly bijection function?

[limpid fern]

you don't

[prime elbow]

Actually I have no such background in cardinality argument

[limpid fern]

because you cant guarantee its a bijection

if I take the constant function for example

[prime elbow]

Yes

[limpid fern]

cardinality of f[D] is 1

[prime elbow]

Yes

Got it, thank you ❤️

[limpid fern]

np

[real granite]

Where he says "Define E... which is closed", the reason we know this is closed is because {0} is closed in the discrete topology, and since f is continuous by assumption, it's preimage should also be closed.

Is that correct?

[quick crane]

yes

[real granite]

I understand that f(X) is connected, but in theory if f was originally defined so as to be non-surjective, then Y might not be connected. Is that correct?

I.e. the image is always connected, but the whole codomain might not be.

[opaque zodiac]

Yes merely being the codomain of a continuous function with connected domain does not imply that you are connected

Consider the map from R to N that takes everything to 0 for instance

[real granite]

Note that compactness has not yet been defined, only sequential compactness.

Why is X being referred to as being sequentially compact but A in X being referred to as compact?

[inland thistle]

I just want to check a definition of local compactness, so for each point in x, and for each nighbourhoods of x, U, we can always find a compact subset C that contains U

is this correct?

[tender halo]

no

consider the nbhd that is the whole space

local compactness just tells you there is at least one compact nbhd

[real granite]

Also, does the existence of a convergent subsequence imply the original sequence is Cauchy?

I ask because to prove sequentially compact ==> complete, apparently you are meant to use the fact that if a Cauchy sequence has a convergent subsequence, then the original converges to the same limit. However, how are you supposed to use that fact unless the convergent subsequences in the sequentially compact space are necessarily Cauchy?

[limpid fern]

no it does not

[inland thistle]

so can we rephrase it like if there exists an open neighbourhood of x that is contained in some compact set for each x, then X is locally compact?

[limpid fern]

you can take (-1)^n and thats not cauchy

but it has a convergent subsequence

it just says IF you have a sequence its going to have a convergent subsequence. the original sequence may not converge

[real granite]

Oh actually yeah apparently you are just supposed to start with a Cauchy sequence and proceed from there

[tender halo]

for example yes

[real granite]

.

[alpine nest]

In the case of a metric space sequential compactness is equivalent to compactness (as defined in terms of open covers), so I'm not sure there's much value in defining a "sequentially compact metric space"

A compact metric space is one where each sequence has a convergent subsequence, or equivalently one where every open cover has a finite subcover.

In general topology the notions are no longer equivalent, but in metric topology "compact" and "sequentially compact" are synonyms.

[real granite]

Highlighted bit.

Suppose $C$ is closed in $X$. The induced topology is $$\tau_C={ U \cap C | U \in \tau_X }.$$

If $A$ is closed in $C$, then $X\setminus A$ is open in $C$, i.e. is an element of $\tau_C$. Thus $X\setminus A = U\cap C$ for some $U$ open in $X$.

Not sure where to go now, because $U$ is open in $X$ but $C$ is closed in $X$, so I'm not sure what you can say about their intersection (and hence whether the compliment of $A$ is open, i.e. $A$ itself is closed)

[gentle osprey]

Douglas

[real granite]

Actually, it's C\A that is open in C I think

Not X\A

Yeah I'm not sure what to do

[alpine nest]

Yes, if A is closed in C, then C\A is open in C, and your goal is to show that X\A is open in X.

Additional observation: since C is closed in X, we know that X\C is open in X

[real granite]

I'm not getting anywhere.

I've got $A=C\setminus (U\cap C)=C\setminus U$ (de Morgan's law). Then $X\setminus A=X\setminus(C\setminus U)$, which is what I'm stuck on.

[gentle osprey]

Douglas

[alpine nest]

C\A is open in C, therefore it's the intersection of some set V (which is open in X) with C.

Note that V must be disjoint with A

X\C is also open, and also disjoint with A

[real granite]

I'm really not getting it

Please just tell me the argument

[alpine nest]

X\A is the union of V and X\C (because C\A is a subset of V), which are both open

[left eagle]

Is there a way to do this without complex numbers? I can't think of a nice function that I can prove continuity in a way thats not super tedious

[winged viper]

What’s wrong with complex numbers?

[hexed steppe]

do you know any linear algebra

you should read up on it if not

its could be useful to have a strong foundation in linear algebra when you learn algebraic topology

[left eagle]

?? yes im not sure how that helps here

[hexed steppe]

as you seem to be preparing to do

[grave solstice]

why would you need linear algebra or complex numbers (for this particular problem I mean)?

[hexed steppe]

you can think of RP1 as being parametrized by angles in [0,2pi), identified with a subset of the plane: S^1

and now think of a 2x2 matrix acting on a coordinate via a linear map R^2 -> R^2

and then again you normalize so the resulting thing is back on the circle

[left eagle]

yes i was thinking of a similar thing. [0,1] is homeomorphic to S^1 but then proving continuity would be kind of a pain

[hexed steppe]

the action of any vector space isomorphism will do for your purposes

[grave solstice]

[hexed steppe]

the linearity will give continuity somehow

it is like

[robust drum]

I think maybe (x, y) mapsto (cos(2|arctan(y/x)|), sin(2|arctan(y/x))|) works? Idea is mapping (cos(theta), sin(theta)) to (cos(|2theta|), sin(|2theta|))

[hexed steppe]

going into the definitions of euclidean space

[robust drum]

This is just bc matrix multiplication is a multivariate polynomial in the coefficients of the matrix

[hexed steppe]

i think it should be possible

yeah

exactly

[robust drum]

But if you don’t want any linear algebra I think this works

[hexed steppe]

idk i cant visualize it

is it equivalent via some trig identity?

[robust drum]

This is a map S^1 -> S^1 which respects the identification

[robust drum]

Well if (x, y) is a point on the unit circle

Arctan(y/x) gives you the angle

[hexed steppe]

yes

[robust drum]

Then take the abs, times 2, and use cos and sin to pass back to (x, y)

This is taking the abs of an angle and multiplying by 2

Which I think works?

As a continuous surjection such that I think the preimage of every point should precisely be each pair (x, y) (-x, -y)

Oh you probably don’t even need to take abs bc arctan(y/x) gives abs angle

Or wait

You shouldn’t take abs

Arctan gives [-pi/2, pi/2] and then *2 gives the whole range of angles

So anyway the correct formula is (x, y) mapsto (cos(2arctan(y/x)), sin(2arctan(y/x)) and you need to do smt piecewise to fix x=0 (you can check continuity at x = 0 by limits of sequences probably)

Then you have a piecewise 2-1 function S^1 -> S^1 where the preimage of each point is precisely the equivalence class defining RP^2

So then by the definition of the quotient it induces a cont bijection RP^2 -> S^1

[prime elbow]

To show every closed set in metric space is countably intersection of open sets.

Let E be closed in M, then we can define function on M to R such that x maps to d(x,E).

Then the preimage of {0} is exactly E.

And we can write {0} as countably intersection of open sets but f^-1( A and B ) ≠ f^-1(A) and f^-1(B).

How to tackle this one?

[prime elbow]

Let x in metric space M and x_n be sequence in M. If f(x_n) converges to f(x) for every real valued continuous function then x_n converges to x.

Therefore I need to show that d(x_n, x) converges to 0.

Let f be the function such that y maps to d(x,y) it is a real valued continuous function.

Hence, d(x_n,x) converges to d(x,x)=0.

[civic juniper]

Is finding a normal subgroup necessary? I don't see. Doesn't finding just open subgroups suffice? Here is a writeup of a proof, inspired of course by your first comment:

[left eagle]

Any suggestions for part 2?

[rancid umbra]

use the uniqueness of the quotient space

along with the fact that the upper hemisphere with the equator is the same as the disk

@left eagle

[real granite]

I think I was working while too tired to be productive.

So the argument goes:

1. $A=C\setminus U$ (by assumption closed in $C$), so $A$ and $U$ are disjoint.

2. $C$ is closed in $X$ by assumption, so $X\setminus C$ is open in $X$.

3. $A\subseteq C$, so $X\setminus C$ must be disjoint with $A$.

How are we getting $X\setminus A=U\cup (X\setminus C)$ at the end?

[gentle osprey]

Douglas

[real granite]

This makes sense to me. I don't know how the argument about compliments works though

[alpine nest]

The argument about complements is essentially replicating the proof that closed sets in the relative topology on C are intersections of closed sets in the original topology on X with C

[real granite]

Let $f:X\to Y$ be continuous and $Y$ a subspace of $Z$.

Every set that is open in $Y$ has a pre-image that is open in $X$, i.e. $f ^{-1} (U) \in \tau_X$ if $U\in \tau_Y$.

If $V$ is open in $Z$, then $f^{-1} (V)=f ^{-1} (V\cap Y)$, and by definition of the subspace topology, $V\cap Y=U$ is open in $Y$. Thus $f ^{-1} (V)$ is open in $X$.

[gentle osprey]

Douglas

[real granite]

Is that a valid proof?

[tender halo]

yea

although I would just say that the inclusion h: Y -> Z is continuous and therefore the composition hf is continuous

[real granite]

C is apparently clopen. If Y=B u C, then C must be open (by definition of separation). Why is C closed?

Again using the definition of separation, B and C are disjoint, so is the idea that C=Y\B, and since B is open in Y, Y\B must be closed, hence C is closed?

Is that right?

[rich cliff]

I understand the proof given for (a), but I'm wondering if it's possible to argue directly (i.e. that P <=> Q)

For P => Q:

• if x ∈ closure(A) and U is an open set containing x, then X - U is a closed set not containing x. Not sure where to go from here

[rancid sand]

If x is not in X\U then X\U must not contain A otherwise X\U would be in the closure. So U intersects A. It's still basically by contradiction though.

[rich cliff]

Yeah that's the same I guess

[brittle jay]

Came across this problem, and I'm having difficulty in finding a way to use the given in a meaningful way.

Let $X$ be a topological space with the property that every convergent ultrafilter on $X$ is principal. Prove that the (arbitrary) intersection of open sets is open.

[gentle osprey]

Morrow

[real granite]

Is another way of doing this by re-defining the characteristic function of a set to be $$\chi_E (x) =\begin{cases} 1 & x\in E \ -1 & x\not\in E \end{cases}$$ and then saying that $X$ is connected if and only if whenever continuous $\varphi:X\to \bR$ with $\varphi(X)\subseteq {-1, 1}$ (equipped with the discrete topology), $\varphi$ is constant?

Since $F$ is nowhere zero, $\mathrm{sign} F=F/|F|$ is continuous (as $F$ and therefore $|F|$ is). Moreover, $\mathrm{sign} F(T) \subseteq {-1, 1}$. $T$ is clearly connected, so we must have $\mathrm{sign} F$ is constant.

[gentle osprey]

Douglas

[real granite]

Also for (iii), I said the following.

If the sign of $F$ on $T $ is 1, then $(x,y)\in T \implies f(x)-f(y)>0$, i.e. $x>y \implies f(x)>f(y)$ (notice that if $(x,y)$ was in the other half of the plane, we would flip $x$ and $y$, so $y>x \implies f(y)>f(x)$, so $f$ still preserves order).

If the sign of $F$ on $T$ is -1, then $(x,y)\in T \implies f(x)-f(y)<0$, i.e. $x>y \implies f(x)<f(y)$ (and for similar reasons to before, this applies to the other half of the plane too).

So $f$ either preserves or reverses the order of elements, hence $f$ is monotonic.

Does that work?

[gentle osprey]

Douglas

[quartz horizon]

Yes, so an alternative characterization of connectedness is “any locally constant continuous function is globally constant”

[real granite]

Okie thank you. What about (iii)?

[real granite]

.

[quartz horizon]

Just make clear where you’re using that f is a continuous injection

Otherwise this is actually really nice

I love this

This is now my favourite way to show continuous injection on R implies no ironic

[real granite]

Yes I did think my proof had a "plothole" of sorts

However, thinking about it some more, we only need f to be monotonic, it doesn't have to be strictly monotonic. So the fact that "the other half" is actually the points for which x≤y (rather than x<y) shouldn't matter I don't think

So you'd end up getting x>y ==> f(x) > f(y) in T, and x≤y ==> f(x) ≤ f(y) in R^2\T (I think), and therefore f is monotonic

But yeah, I'm not actually sure where injectivity comes into play

There is an issue with continuity of sign(F) at (0,0), and I'm not sure there's a way of getting around that

[quartz horizon]

You need to show the hypothesis of part ii holds

[real granite]

Uh... for injections, x≠y ==> f(x)≠f(y), so if f is an injection then F(x,y)=f(x)-f(y) is never zero on T (i.e. x>y so x≠y, hence f(x)≠f(y), hence f(x)-f(y)≠0).

[quartz horizon]

Yep

[real granite]

As f is continuous, F is continuous, and since F is nowhere zero on T, F/|F| is continuous on T

[quartz horizon]

Yep!

[real granite]

Hmmm, yeah that works. Because the case x=y lying in X\T is covered by the fact that f doesn't need to be strictly monotonic

Perfect

[granite crane]

does munkers is good start up for topology?

[limpid fern]

sure

[quick crane]

Looks fine. I personally found "Real Analysis: Modern Techniques and Their Applications" by G. Folland to be a good introduction to topology, among other things (measure theory, functional analysis, fourier analysis).

[granite crane]

But Folland looks much advance than munkers

[tender halo]

munkres is probably the best starting point for topology if you have enough mathematical maturity

and he doesnt need much of it

[manic seal]

I'll have you know that I'm very mathematically immature and I can read munkres just fine

Sorta

[granite crane]

Oh that sounds good. Btw Idk why I didn't get notification.

[tender halo]

you can reply to people without it pinging them

[ocean canyon]

I suppose John Lees ”Intro to Topological Manifolds” should work as well (atleast that is the book my university uses; it supposes you have taken analysis in the sense of ”baby rudin” before, so that you know something about ”basic topology”).

[granite crane]

That's cool i didn't know about it.
Thank you for this information!

[granite crane]

Oh I see. I check the first few pages and I found stuff understandable. I think I have studied such things like limit points and there corresponding theorem in analysis so I found initial pages easy. It was FUN i enjoyed

Btw I am studying Baby Rudin. I have completed Ch2, and I enjoyed it (at the same time it was painful too).

Edit : Idk why previous edit vanished lol

[ocean canyon]

Right, I have not taken topology (so take my recommendation with a grain of salt), I am registered for the course this fall. I did take differential geometry a year ago, where I accidentally bought this book instead of ”smooth manifolds” (also by J. Lee).

[granite crane]

Oh cool. I have his smooth manifolds book but not his "topological manifolds book". I thought "smooth manifold" is his first book lol but later released it's 2nd book

[ocean canyon]

Yeah I mean in some sense it might seem weird to take DG before topology since I believe it is generally agreed (alteast so I’ve read) that ”smooth manifolds” is more advanced than ”top. manifolds” (I am not sure that this is true). But one can learn DG with just knowing ”basic topology” and some analysis.

In a sense I think they might just be covering different things, I am not too sure that one is more advanced than the other, but I have not read much in either of them (we only used ”Smooth manifolds” as a reference, iirc).

[granite crane]

That's interesting

Oh I understood. At the moment my plan is to go though rudin and munkers (with topological manifolds as reference)

[real granite]

Am I being dumb? E=ø means f=1 right? Because E=ø means there are no elements of X for which f(x)=0, so must be identically equal to 1

I know he hasn't said anything wrong it's just that at the end he seems to reverse the order of the cases if that makes sense

[alpine nest]

Yes, if E (the preimage of {0}) is the emptyset, then f is constantly 1

[quick crane]

The topology section is less advanced, but I guess the other analytical sections are more advanced

[real granite]

Doesn't it make more sense to say if $X=B_r(x)$?

[gentle osprey]

Douglas

Douglas

[tender halo]

well

kinda not really?

the important part is that the ball covers the whole space, i.e. that X is inside the ball

which is precisely what is stated

of course that is the same as them being equal, but you be the judge which one is more natural

[rancid umbra]

they are equivalent

[tender halo]

i find the version presented a little cleaner

[alpine nest]

Its more consistent with the notion of bounded set in your space

[granite crane]

Oh. I haven't checked it yet, but I will check it later.

[sonic crane]

Why is {U1, U2} in the simplicial complex?

They have a non empty intersection

For context, i am learning about topological data analysis and this section was talking about the nerve of an open cover

I can show the definition they gave of abstract simplicial complex and the nerve if u want

[real granite]

Maybe this is small, but isn't he being a bit sloppy when he says "so we may assume $Y=f(X)$"?

E.g. define $f:[0,1] \to \bR$ by $f(x)=x+1$ (obviously this is continuous). $X=[0,1]$ is closed and bounded, so it is compact (Heine-Borel). According to the theorem, $f([0,1])=[1,2]$ is compact, which is true, but the whole codomain is not compact: $\bR$ is not compact.

[gentle osprey]

Douglas

[real granite]

Like uh, the codomain will be compact if the function is surjective, that is true, but we have no reason to assume the function is actually surjective, so we can only talk about the image

[sonic crane]

The theorem is proving that the image set is compact not the codomain

[real granite]

Yeah but in the statement of the theorem, he doesn't say f is surjective, so assuming Y=f(X) does not seem entirely sound

[sonic crane]

It is valid because youre proving something about the image set not all of Y

Its just like, ok u have a cts function from X to Y, but the theorem is talking about how f(X) is compact so instead lets just look at X -> f(X)

Thats all its doing

f is surjective on X->f(X)

[real granite]

Yes, I understand that $Y\setminus f(X)$ won't make a different to $f(X)$ being compact, but it is an abuse of notation to use $Y$ to denote the image, because $Y\neq f(X)$

[gentle osprey]

Douglas

[real granite]

in general

[tender halo]

its not an abuse of notation

its just WLOG

[sonic crane]

Yeah its like, Y might not be f(X) exactly, but for the purposes of the proof we can assume so

[real granite]

hmmm

okie

[sonic crane]

Bruh my bad lol, they NEED a non empty intersection for it to form a simplex . I got it backwards

[rich cliff]

Why is the fiber over p homeomorphic to F?

[unreal stratus]

Restrict phi.

The diagram says that phi sends the fibre of p to {p} x F

@rich cliff

[limpid fern]

Is the product of 2 ccc spaces ccc?

This doesn't feel like a question that I can answer

[opaque zodiac]

No?

I think this can fail

Errrr

[brave geode]

i'm pretty sure it's true

[opaque zodiac]

Ah I'm thinking of arbitrary products

[brave geode]

hey! are you here?

[opaque zodiac]

I think 'arbitrary products of CCC spaces are CCC' is independent of ZFC

[limpid fern]

yeah

hello

Yeah okay I must have been asleep in my set theory class because this is MA(omega1)

if it holds for finite i think it holds for arbitrary, no?

[brave geode]

i don't know enough set theory to answer the arbitrary case but the proof i have in mind for finite doesn't extend

take a disjoint collection in the product topology

let's find a fundamental open set contained in each open set in the collection

now disjointness in the product space says that when we project these pairwise, there has to be at least one component in the product where the projections are disjoint

does that follow so far

[limpid fern]

hmm

sorry one moment

[limpid fern]

this is not immediately obvious to me sorry

[brave geode]

if all of them had something in their intersection we could take the points' cartesian product

[limpid fern]

sure

[brave geode]

ok now say for each pair we pick a component whose projection to that component leaves those two disjoint

[limpid fern]

okay no im still confused sorry

[brave geode]

on which point?

[limpid fern]

i actually did not see how the previous point implies this again

can you show me an example for the case of 2 products

[limpid fern]

right okay

sure

[brave geode]

do you use dark mode

[limpid fern]

yeah but it's no big deal

also I thought that product of 2 ccc spaces is ccc is not provable from zfc?

so are we assuming additional axioms here

[brave geode]

no

it is

[limpid fern]

huh okay

[brave geode]

continuing on this

whoops the proof doesn't work

[limpid fern]

?

[limpid fern]

We proved last week in my set theory class if finite products are CCC then arbitrary products are CCC

But it seems suspicious to me that arbitrary products are ccc

Intuitively

[limpid fern]

This is using the delta system lemma

[opaque zodiac]

wait yeah that sounds familiar

what the fuck am I thinking of?

hm

Right I think I remember this now

You do get that 'finite products of ccc spaces are ccc' implies 'arbitrary products of ccc spaces are ccc' but finite products of ccc spaces can consistently fail to be ccc

[real granite]

If $d(y,x_n) \leq \delta_{x_n}$, doesn't that mean the distance between the points depends on $x_n$? Isn't that pointwise continuity then?

[gentle osprey]

Douglas

Douglas

[granite crane]

thats a cool theorem I recently done it in analysis.

Btw if you don't mind can I ask which book are you using

[real granite]

If $V\in \tau_Y$, then $Y\setminus V=Y\setminus (U\cap Y), U\in \tau_X$. Then $Y\setminus V = Y\setminus (U\cap Y)=Y\setminus U \cup Y\setminus Y=Y\setminus U$, which must be finite as $X\setminus U$ is finite. Is that the correct reasoning for why $Y\subseteq X$ must also have the finite compliment topology?

When he says "since ${U_i}$ covers $X$", I assume he means "covers $Y$"?

[gentle osprey]

Douglas

[real granite]

lecture notes

[granite crane]

oh got it thank you

[real granite]

Also, he only proves the subspaces Y in X are compact, he doesn't prove X itself is compact. If X is infinite, then the cofinite topology does not coincide with the discrete topology

So it's not clear why X would be compact

There could be infinitely many subspaces, so idt you could say "finite union of compact subsets is compact"

[alpine nest]

X is a subspace of itself

The key argument is that the cofinite topology is always compact.

If X has cofinite topology, then every subspace of X (including X itself) has cofinite topology, therefore is compact.

[real granite]

This is true.

Is what I said above correct?

[real granite]

Here

[alpine nest]

Yep

[real granite]

Thx

[limpid fern]

my set theory professor told me its consistent with ZFC

but not provable from it

[manic seal]

Im reading about the fundamental group right now and iv'e just gotten introduced to the concatenation operation then this equation pops up

with k being a continuous map. Can someone give me something i that can google so I can understand why this is the case better

i barely know any group theory btw

nvm, ig they were kinda right

[robust drum]

This is just post composition

If you actually look at what happens to some t under both maps you’ll see it’s the same

[manic seal]

If A is a retract of a larger set X with retraction r: X -> A and given an continuous map g : [0,1] -> A does there then exist a continuous map f : [0,1] -> X s.t. r o f = g?

[manic seal]

ok ig i just need to prove that there exist a continuous inclusion l : A -> X

[worthy olive]

What does it mean for a quotient function π to have an inverse

(By π I mean the function that goes from a set to its equivalence classes)

[gaunt linden]

Umm... that there's a function in the other direction that composes to the identity on both sides?

[worthy olive]

But how can a function be in another direction

I was thinking like the equivalence classes of X are A={a,b,c}and some other stuff, which aren’t the same under =, but then you can add another term like d into A

And that’s different than the original A

So how could a function have an inverse, if one of the elements can be 2 things

[gaunt linden]

So that puts very tight conditions on when the projection function can have an inverse.

[manic seal]

pretty lame projection if it does have an inverse imo

[worthy olive]

Also my book is saying that the inverse is open

I’m a little slow when it comes to open maps

[manic seal]

the preimage is open (assuming that the set taken the preimage of is open)

if thats what youre thinking of

[worthy olive]

Yeah but how can π be continuous?

[manic seal]

because you have here constructed a topology on X/~ s.t. pi is continuous

do you remember the definition we use in topology for a continuous map?

[worthy olive]

But how can pi be continuous?

[quick crane]

The definition can be rephrased as the topology on $X / \sim$ is the smallest topology that makes $\pi : X \to X / \sim$ continuous.

[gentle osprey]

L

[umbral panther]

Every function f has an associated function called f^-1, but it is not the inverse and it does not simply swap domain and range, but instead takes sets in the target to sets in the source

[limpid fern]

suppose U is open in X/~. what's the preimage of U under pi?

is it open? yes. by construction

[gilded steeple]

hi, can anyone help me proving the first assertion of this proposition? I've proved the bijectivity of the two functions and the fact that they're inverses of one another; the continuity of g is known. is my attempt headed in the right direction or is it straight-up wrong?

[plush folio]

How did you show that g and (φ o s) are inverses of each other?

[gilded steeple]

given the decomposition in red (where φ is the canonical surjection, g is the bijection of X\R onto f(X), and ψ is the canonical injection of f(X) into Y i.e. the identity mapping of Y in this case), f=gφ. then g=φs for s is a section of f.

I sorted it out by now though. I didn't realize the section was continuous by hypothesis, which implies that the inverse of g is continuous, i.e. that g is a homeomorphism. sorry for the trouble

[plush folio]

Is the term "sequentially closed" not commonly used? I can't find it anywhere in Munkres

It seems pretty useful to know that sequentially closed implies closed in first countable spaces

Okay, I found the relevant theorem relating closed and sequentially closed sets, but without the terminology, ditto for continuity vs sequential continuity. Huge hole in Munkres, I really like those terms

[tender halo]

its not commonly used because non first countable spaces are not common

[plush folio]

Hmm, makes sense. I still like it though, there's so many theorems so just putting a name to it makes it somehow easier to remember

[tender halo]

it is nice

in general i think people should know about sequential and Frechet-Uryhson spaces

and how for example sequential and epsilon delta definitions of continuity are the same precisely because R is a sequential space

[plush folio]

aha, sequential spaces are the weakest spaces where sequentially closed implies closed? That seems useful, I assume almost all interesting spaces are sequential

[tender halo]

they are just spaces where sequentially closed implies closed

no "weakest"

Frechet-Uryhson spaces are where sequential closure is the same as closure

[dawn lynx]

I think they just mean that sequentially closed is the weakest condition you can put on a space so that "sequentialy closed --> closed"

Whereas "metric" would be a stronger condition to put on a space to ensure this condition

[tender halo]

uhh i guess thats true

the condition itself is in fact the weakest condition you can put on a space to make it respect that condition

i assume almost all interesting spaces are sequential

not at all!

[unreal stratus]

Depends what you mean by interesting lol

[tender halo]

in beta N all sets are sequentially closed for example

[tender halo]

i guess so, for some only metric spaces are interesting

[tender halo]

the cantor cube of weight c (i.e. the space 2^c) has a countable subspace which has no countable base at any point

actually hmm

hmmmm

i am not sure if you can get anywhere in that space with sequences

[tender halo]

nvm no it doesnt work

[red folio]

are the half-open interval and open ray topologies second countable?

[umbral panther]

Not both

[left eagle]

This seems like a homeomorphism but im second guessing myself?? Is it?

also how would that help me deduce S^n is path connected?

[manic seal]

homeomorphic spaces have the same topological properties

and a product of path connected spaces is also path connected

you can try to argue that if one of the two spaces isnt path connected then the product space isnt either

[manic seal]

I.e. show that this is an if and only if statement (for a finite product)

[rancid umbra]

just compose with the projection onto S^n. this composition surjective onto S^n, so all you have to show is that R^{n+1}\{0} is path connected

[rancid umbra]

yes, it is a homeomorphism. it’s the same homeomorphism that allows for “polar” coordinates

[manic seal]

when we speak of "the fundamental group of a top. space" does this imply that the fundamental group for each point in the given space is isomorphic to one another?

[gentle girder]

yes but it's not canonical, the isomorphism is gonna be given by a choice of path between basepoints

but since the groups are at least isomorphic, then you can speak of "pi_1(X)" as shorthand for "pick a basepoint x_0, and then take pi_1(X, x_0)"

[gentle girder]

in particular, if you pick a path $\gamma$ such that $\gamma(0) = x_0$ and $\gamma(1) = x_1$ then you get an isomorphism $\pi_1(X, x_0) \to \pi_1(X,x_1)$ given by $\alpha \mapsto \gamma * \alpha * \bar\gamma$

[gentle osprey]

microwavable coordinate chart

[gentle girder]

but a different choice of gamma might induce a different isomorphism

[manic seal]

ty for the very informative answer

yeah i've read about that path composition thing earlier

makes sense

[gaunt linden]

In particluar if you set x0=x1, this gives you exactly he inner automorphisms of the fundamental groip.

And if the space is not path-connected then there doesn't need to be any isomorphisms between different base points. (But usually path-connectedness is assumed before we say "the" fundamental group.)

[real granite]

sirs and maams and comrades

pls wld sm1 recommend a good textbook (or some other learning resrouce) to learn about the lower limit topology

i dont need to know in huge amount but its come up in the some past papers but isnt in the lecture notes

[red yoke]

It's just a common counterexample

I'm sure the stuff in Munkres suffices

I've never seen it appear outside point set topology

[real granite]

Okie. Relevant to lower limit topology, Munkres gives the example of $f:\bR\to\bR_L,\ f(x)=x$ and says that $f$ is not continuous because $f^{-1} ([a,b) )$ is open in the range but not the domain (this makes sense). Can this argument be generalised to $f: X_{\mathcal{T}} \to X_{\mathcal{T}'}$ where $\mathcal{T}\subset\mathcal{T}'$?

[gentle osprey]

Douglas

[real granite]

My inclination was to take $U\in \mathcal{T}'\setminus\mathcal{T}$, but I think maybe it depends on the function

[gentle girder]

yes, take the set U that's in T' but not in T

[gentle osprey]

Douglas

[gentle girder]

yep

[real granite]

Oh ok

Sweet

[tender halo]

counterexamples in topology

[real granite]

this doesnt appear to have lower limit

Munkres doesn't actually have much

I think lower limit is mentioned 6 times

[umbral panther]

That’s enough

[tender halo]

it is no. 51

[real granite]

ah ok

different name

ty

[real granite]

For some reason there aren't solutions for a past paper I did, so I would very much appreciate it if someone could help me "mark" it

1. Fine.

2. (i) No. This space isn't Hausdorff so it cannot be metrizable. Consider $x=1$ and $y=3$. These are both odd, so the only open neighbourhood of $x$ is the whole space, and likewise for $y$, thus there are no non-empty disjoint neighbourhoods for which $x \in U$ and $y \in V$.
   (ii) Probably not but I'm not sure.

3. Suppose $x=2k$ is an even integer. Then the open neighbourhoods of $x$ are the subsets of even integers containing $x$. For all $n>1$, $x_n=1/n$ is not an integer (specifically not an even integer), so $x_n$ cannot be an element of the open neighbourhoods of $x=2k$. Thus $x_n$ does not converge to any even integer.
   Conversely, if $x$ is a non-even integer real number, then the only neighbourhood of $x$ is the whole real line, and then for any $n$, we have $x_n \in \bR$, so $x_n$ converges to every non-even integer real.

4. I struggled with this.

5. (i) The subspace topology is ${ {1,3} \cap U | U \in \tau } = {1,3} \cup \emptyset$. So the only separation of ${1,3}$ is ${1,3}\cup\emptyset$ (as ${1}, {3}$ are not open), which is trivial, so ${1,3}$ must be connected.
   (ii) ${2}, {4}$ are open, so there is a non-trivial separation. Hence ${2,4}$ is disconnected.

6. Consider $\tau_{\mathcal{E}}={\mathcal{E}\cap U | U \in \tau }$. $\bigcup_{n=0} ^{\infty} \qty( {-2n} \cup {2n})$ is an open cover of $\mathcal{E}$, but it has no finite subcover any finite subcover would contain only finitely many elements. Thus $\mathcal{E}$ is not compact.

[gentle osprey]

Douglas

[real granite]

The ones I need help with are 2ii and 4

But any feedback on the others would be appreciated

[tender halo]

2ii sure, its a finite discrete space

4i just uhh take a constant function idk

4ii take the identity

others are like fine

[real granite]

Huh... so my initial guess that I didn't write down was actually right.

Suppose f(x)=c. If c is an even integer, then any open set containing c will have open pre-image (i.e. all of R), and any open set that does not contain c will have empty pre-image (also open). If c is not an open integer, then the only open set containing c is R, so the pre-image is either empty (open set does not contain c) or all of R (does contain c).

Is that correct?

I guess you don't really need to split it into two cases depending on the value of c

[tender halo]

any constant function between two topological spaces is continuous

nothing to do with their topologies

[real granite]

For the reasons I gave (R replaced with general space X)?

[tender halo]

yea

[real granite]

Amazin

For the others

That I didn't need much hjelp with

When you say "like fine", does that mean you have improvements?

[alpine nest]

Fine like the discrete topology

[real granite]

There was also a Section B

[gentle osprey]

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

[real granite]

If someone could take a look at my answers and give feedback that would be much appreciated

Please ping

Need help with 7iv and 8iv specifically

Otherwise I feel alright

[rancid umbra]

hint for 7iv: || |sqrt(x) - sqrt(y)| |sqrt(x) + sqrt(y)| = |x - y| for non-negative x,y||
hint for 8iv: || What happens if you poke a hole in R^3? What happens if you poke a hole in R? maybe refer to part (iii) of this problem||

@real granite

[quick crane]

Hint for 7iv: Use the mean value theorem.

[real granite]

First makes perfect sense, silly oversight from me.

Second: $\bR^3 \setminus{0}$ is path connected but $\bR\setminus{0}$ is not. How does this lead to our conclusion? Aren't we dealing with different spaces by removing the origin?

[gentle osprey]

Douglas

[real granite]

Uh so... |f(x)-f(y)|≤ sup|f'(x)| |x-y| so we can choose sup|f'(x)|=c in [0,1)
Is that right?

[rancid umbra]

yes, but you have a way of transporting R^3 \ {0} to R \ {0}

[real granite]

I'm not following you. If R^3-{0} is path-connected but R-{0} is not then surely there is no homeomorphism?

But how does this help you with R^3 and R

[rancid umbra]

suppose there were a homeomorphism f : R^3 --> R

wlog taking 0 to 0 (otherwise, just post-compose with a translation)

[rancid umbra]

think about how poking holes in these two spaces would affect f. does it affect continuity? injectivity? surjectivity?

[real granite]

Assume f:R^3 to R is a homeomorpshim. If 0 is mapped to 0 then you would expect f:R^3-{0} to R-{0} (i.e. same f with domain restriction) to also be a homeomorphism, but this is not the case. By contradiction, original f must not have been a homeomorphism.

Is that the argument?

[rancid umbra]

yes, because of this.

more precisely, this argument shows, with some slight modifications, that there can't be any continuous injection from R^n to R for n > 1

[rancid umbra]

you can actually compute the max of f'(x) = 1/(2 sqrt(x + 4)) on [0,infty) since it is decreasing

[quick crane]

yeah, you should compute $\sup_{x \geq 0}|f'(x)|$ explicitly.

[gentle osprey]

L

[red folio]

is a basis for a topological space (X, T) a local basis for every x in X?

[umbral panther]

A basis probably contains open sets that don’t contain x, whereas in a local basis at x, every open set should contain x

[rich cliff]

A space is Urysohn if distinct points can be separated by closed neighborhoods. How do I show that Urysohn spaces are Hausdorff?

For points x and y in a Urysohn space, there exists closed nbhds U and V of them respectively. Now I wanted to just use their complements, but they're not necessarily disjoint

It could be wikipedia is inconsistent with their usage of the term neighborhood

But if you interpet "closed nbhd of x" as a closed set containing x, then is the original statement I made wrong?

I see

Then wikipedia is wrong

OH

[umbral panther]

I think you’re misreading Wikipedia

[rich cliff]

A closed nbhd is a closet set that contains an open set that contains your point

Ok

I see

Thanks TTeppa and bw

[rich cliff]

hausdorff <-> can separate two points with nbhds
completely hausdorff <-> can separate two points with a continuous function

regular <-> can separate a point and a closed set with nbhds
completely regular <-> can separate a point and a closed set with a continuous function

normal <-> can separate two closed sets with nbhds
perfectly normal <-> can precisely separate two closed sets with a continuous function
completely normal <-> every subspace is normal

Why did they mess up the flow by calling them perfectly normal spaces and not completely normal

Then "completely" could be associated with "separating with a function"

[red yoke]

Because normal spaces already separate closed subsets with a function

[rich cliff]

Oh

[red yoke]

Though I agree something like hereditarily normal is a better name

[rich cliff]

Me too

[real granite]

I think this was asked here a while ago

But can't diam(ø) be any real number?

Or uh

negative infty

I'm looking at a proof of Cantor's Lemma (if we have a countably infinite collection of nested closed sets in a complete metric space with diameter tending to zero, then their intersection is the singleton), and we use the fact that diam(A)≤diam(A_n) to argue diam(A)=0, and then say A is either the singleton or ø

And I'm just wondering if that is actually true, that A=ø is a possibility

[quartz horizon]

Well it is true that diam(emptyset) = 0

With a suitable definition of diameter I guess

[real granite]

The way my lecture notes define it is as
diam(A)=sup{d(x,y) | x,y in A}

[quartz horizon]

Yeah well it’s like

You have to decide what sup(emptyset) means

[tender halo]

usually supremum of an empty set is negative infinity

but in metric spaces its a little inconvenient

honestly it doesnt matter much, you can make arguments in favor of both definitions, just pick the one you like more and stick to it

[alpine nest]

I think the diameter of the emptyset being 0 is more convenient

[tender halo]

also third option is to explicitly exclude empty set from the domain of diam

[alpine nest]

That's another, although if we agree that the diameter of the emptyset is 0, then it's true that you can cover the emptyset using a ball of any radius greater than 0, and we preserve the property that the diameter of a union of two sets is less than or equal to the sum of their diameters.

But either way it doesn't matter very much because you're rarely particularly invested in the notion of diameter when your set is empty

[real granite]

yeah like convention seems to be thing(ø)=0, e.g. span(ø)=0

or the group generated by ø is {e} iirc

so i think sup(ø)=0 fits better with other stuff

[tender halo]

sup(ø) is almost always -inf

or like

⊥

whatever you happen to have

span of empty set is different, because its the smallest substructure generated by an empty set and has little to do with size

[real granite]

i mean

its kinda related

through dimension of a vector space

that kinda tells you smth about size right

[alpine nest]

But also you could define diameter as the infimum of all nonnegative r such that your set is contained in a ball of radius r

Which is equivalent to the sup(d(x,y)) definition for nonempty sets

[real granite]

Maybe better in #real-complex-analysis, but what are the assumptions about the domain and its compactness?

E.g. continuously differentiable ==> Lipschitz on a compact subset of R (or R^n for that matter), but in general a contraction doesn't need to have a compact domain does it?

[alpine nest]

In general you'd need a space where differentiability is a thing.

The Lipschitz condition is generally definable for a map between two metric spaces, but differentiability is only a thing for certain kinds of spaces.

The continuous differentiability criterion cites compactness, because compactness of the domain implies that the derivative is bounded.

A more general version would be that boundedness of the derivative implies the Lipschitz property

(if your function is continuously differentiable on a compact set, then the derivative is a continuos function on a compact set, so it's bounded)

[real granite]

Yes this makes sense

[alpine nest]

The general definition of the Lipschitz condition is this: if $(X,d_X)$ and $(Y,d_Y)$ are metric spaces, then a map $f:X\to Y$ satisfies the Lipschitz condition, if there exists an $L\geq 0$ such that for all$s,t\in X$ we have $d_Y(f(s),f(t)) \leq L\cdot d_X(s,t)$.

[gentle osprey]

Outsider

[alpine nest]

It implies uniform continuity, but isn't equivalent to it.

For functions from R to R, bounded derivative implies the Lipschitz property, but also isn't equivalent to it.

[real granite]

Yeah, and f being Lipschitz doesn't generally imply f is a contraction (unless f is X to X)

[unreal stratus]

particularly if L > 1 lol

[plush folio]

Your argument for why A not connected => A not path connected doesn't seem quite complete. Why does U and V being disjoint imply that gamma must take values outside A?

[alpine nest]

Oh yeah, I missed that you mentioned/asked about contraction.

A Lipschitz map (from a space into itself) with L < 1 is a contraction.

Actually that might be the definition of a contraction

[real granite]

Well I'm assuming A is a subset of some broader ambient space

Like R^n

So it might be something like $A=B_1 ((2,0)) \cup B_1 ((-2,0)) \subseteq \bR^2=X$

[gentle osprey]

Douglas

[real granite]

You could replace the balls with disjoint open neighbourhoods

So it works for general topological spaces

[plush folio]

But [0, 1] and (1, 2] are disjoint, but you could still have a path from 0 to 2. You need to use the fact that U and V are open

[unreal stratus]

You can have stuff where d(f(x),f(y))/d(x,y) gets arbitrarily close to 1 (but is always <1 )

wait

no

ignore

i'm silly

[alpine nest]

So that would be with contraction defined as a map such that d(f(x),f(y)) < d(x,y) for all x,y?

[unreal stratus]

ur right

[alpine nest]

That kind of contraction wouldn't be sufficient for the Banach Fixed Point theorem though

[unreal stratus]

contraction = 1-e lipshitz for some e > 0

[unreal stratus]

Ig this is sometimes called a shrinking map

[alpine nest]

Yeah, as usual in topology there's like 15 almost identical but subtly differing notions.

[unreal stratus]

thouogh ofc "for all x != y"

[alpine nest]

Fair point

[unreal stratus]

Lol idk if this is usual for topology

i would argue this isn't rly topology though aha

well ig borderline

metric space theory

so ig it counts

[alpine nest]

What makes me think this about topology is mostly the axioms of separation and the many flavours of compactness

Statements dreamed up by the utterly deranged

Salvaged only by the conclusion that metric spaces are in fact perfectly normal.

[real granite]

Yeah I did actually ask about switching the inequalities (so you had a strict inequality between the distances but 0≤c≤1) a few days ago, and while reading a proof of the fixed point theorem I realised we used c<1 to do some stuff with geometric series

[quartz horizon]

Aww it looks like a bow of some kind

[alpine nest]

To be clear, I like (point set) topology quite a lot

[alpine nest]

yeah, for the Banach Fixed Point theorem you need contraction defined as "Lipschitz map with the constant less than 1" (and indeed the typical proof will invoke the convergence of a geometric series where the constant is the ratio)

[alpine nest]

Neither Lipschitz property with L=1, nor the "weak contractivity" as defined in this post, will be enough for the Banach Fixed Point Theorem to hold.

[alpine nest]

I don't really get the disdain some people (not necessarily you) have for metric topology; admittedly it makes a lot of stuff much easier than in general topology (sometimes downright trivial compared to general topology), but on the other hand a lot of very important and widely used spaces are metrizable.

[real granite]

Something like $\gamma^{-1}(U)\cup\gamma^{-1}(V)=[0,1]$ and $\gamma^{-1}(U)\cap\gamma^{-1}(V)=\emptyset$, but this is impossible as $\gamma^{-1}(U), \gamma^{-1}(V)$ are both open by continuity of $\gamma$

[gentle osprey]

Douglas

[alpine nest]

Heck, the weak* topology is ostensibly not metrizable, but the unit ball (over a separable Banach space) is metrizable (and compact), and that's often the only set on which you do actually consider the weak* topology.

Since that's where the probability measures live.

[real granite]

I haven't studied the weak* topology

[alpine nest]

Yeah, I was more rambling into the void.

[real granite]

Okie

Do any of the separation axioms imply metrizability

[alpine nest]

You do need at least T6 and Hausdorff (since all metric spaces are T6 and Hausdorff), but it's probably not equivalent

[real granite]

Or actually, a more useful question (or rather a question with a more useful answer) for this module I'm studying, what are sufficient conditions for metrizability? I know that being finite discrete is sufficient because all such topologies are metrizable by the discrete metric

[alpine nest]

https://en.wikipedia.org/wiki/Metrizable_space#Metrization_theorems

[real granite]

But that's quite specific

[alpine nest]

Actually not bad, Hausdorff and second countable isn't that hard to verify

Oh wait, you also need regularity

Which might be less easy

[plush folio]

Yep, this works 👍 To be clear, it is impossible because [0, 1] is connected but gamma^{-1}(U) and gamma^{-1}(V) would be a separation of [0, 1]

[real granite]

Nice

Thanking you

[alpine nest]

I think the Urysohn metrization theorem (and variations) is the main criterion

In specific cases you can sometimes show that a space is metrizable by explicitly constructing the metric.

Actually that's what the Urysohn metrization theorem does, for a specific value of "explicitly"

But you basically construct a sequence of real-valued functions on your space and define the metric using those functions.

It's a surprisingly easy proof

[plush folio]

By the way, it seems you're working with the definition that $A \subseteq X$ is path connected if every two points in A can be connected by a path $[0, 1] \to X$ that takes values only in A. I think it is easier to work with the definition that A is path connected if it is path connected in the subspace topology. Then you don't have to worry about whether the range is contained in A

[gentle osprey]

sheddow

[tender halo]

I always forget what Uryhson metrization theorem and Heine Borel are

cuz they look like lemmas to me

[alpine nest]

Fair, especially because the main tool in the Urysohn metirzation theorem arguably is the Urysohn lemma.

The rest is basically just applying the lemma for your countable base of neighborhoods (obtaining a sequence of continuous functions that let you distinguish between points) and summing the functions up into a metric.

[tender halo]

you dont really need it

uhh

i guess you do need it to get tychonoffness

my proof is that its regular + second countable -> tychonoff + second countable -> embeds into Hilberts cube -> metrizable

[alpine nest]

Yep

[tender halo]

Bing's metrization theorem proof with the hedgehog is probably my favorite out of them

[alpine nest]

Urysohn's lemma is to give you the continuous functions for Tychonoff

(and second-countability is needed so that countably many functions suffice)

What I said about summing up the continuous functions is basically the same result as embedding into the Hilbert cube.

Except you take the metric from the Hilbert cube directly, without invoking the cube as an auxiliary object.

It's all pretty much the same ideas just worded differently

[tender halo]

kinda, although any tychonoff space of weight m is embeddable into the Tychonoff cube I^m

[alpine nest]

Explicitly referring to the Hilbert cube is probably more convenient

[tender halo]

so i like to think of them as separate steps

you just get lucky that Tychonoff cube is metrizable iff m = \omega

[alpine nest]

Good times

[unreal stratus]

Ig I just think of it as analysis usually lol

Metrisable being different to metric

But yeah it doesn't matter this is just terminology and I'm not a point set topology person aha

[real granite]

Where is 6/21 coming from and why do we need it?

Can't we just reason as follows?

1. By the mean value theorem, |f(x)-f(y)|≤sup|f'(x)| |x-y|.
2. |f'(x)|≤20/21<1 by the triangle inequality (and setting x=1), so |f(x)-f(y)|≤20|x-y|/21 (note I'm not saying sup=20/21, sup may be less than 20/21).
3. Since f is [0,1] to [0,1] (we know the image is in [0,1] because 20|x-y|/21≤20/21) and f satisfies the Lipschitz condition, it is a contraction.

[alpine nest]

How do you justify (3), exactly?

Specifically, how do you justify that the range of f is contained in [0,1]?

The purpose of this line is specifically to justify that if x is in [0,1], then f(x) is also in [0,1]

Although admittedly you could get the upper bound from the MVT

You definitely need to establish the lower bound directly, though.

[real granite]

Hmmm yes this a problem. What I've written in brackets in (3) is sufficient to argue that the relative distance between x and y is less than or equal to 1, but that doesn't mean x and y lie in [0,1]

[alpine nest]

Anyway, the 6/21 is coming from the estimate $\frac{1}{21}(x^7-3x^4+x+4)\leq \frac{1}{21}(x^7+x+4) \leq \frac{1}{21}(1^7+1+4) = \frac{6}{21}$

[gentle osprey]

Outsider

[alpine nest]

For $x\leq 1$

[gentle osprey]

Outsider

[real granite]

Uh...
Would it be valid to argue $$0\leq \frac{1}{21} (-3x^4+4)\leq f(x) \leq |f(x)| \leq \frac{20}{21}?$$ Like, $6/21$ is surplus to requirements in some sense, right?

[alpine nest]

How do you know that |f(x)| is less than or equal to 20/21?

[real granite]

Uh yes

Good point

I am muddling my f and f'

Hmmm, I suppose you could do triangle inequality on |f(x)| and get |f(x)|≤(1+3+1+4)/21=9/21=3/7

That would work right?

In place of 6/21

[alpine nest]

Yep

[real granite]

Okie I think I'm understanding this better now

[tacit basin]

the interior of an infinite set is never empty in the euclidean topology on R^N right

[rancid umbra]

think about some really sparsely distributed subsets of R

[tacit basin]

Oh, countable union of disjoint singletons

?

Empty interior I think

[rancid umbra]

yes, thats one counter example, but even something uncountable like R - Q works

since it contains no intervals

[tacit basin]

Right ty

[rancid umbra]

the point is that its not that its not just a disjoint union of singletons of some cardinality, there is something special about the geometry/topology of R^n that allows for counter-examples like this to occur

[tender halo]

any countably infinite set has empty interior in R

[tacit basin]

Thanks

[thin tide]

$$\bigcup_{n \geq 1} \left{(x,y) \in \bR^2 : \left(x-\frac{1}{n}\right)^2 + y^2 = \frac{1}{n^2}\right}$$

[gentle osprey]

chipotle

[thin tide]

Do "hawaiian earring" and "earring space" both refer to this?

cool thanks

[austere dirge]

if the stone cech compactification is essentially an initial object in the category of maps from X -> K where K is compact hausdorff, what are our final objects? i thought that the final objects would be the one point compactification when it exists, but i couldn't manage to prove this.

[opaque zodiac]

I don't think it will always have one?

[rich cliff]

A space $X$ is normal iff for every pair $Y, Z \subset X$ such that $Y \subset Z$, $Y$ is closed, and $Z$ is open, there exists an open set $U$ and a closed set $V$ such that $Y \subset U \subset V \subset Z$.

[gentle osprey]

plexcty

[rich cliff]

I have proved the forward direction, but for the reverse I am stuck

If $A, B \subset X$ are closed sets then $A \subset B^c$ furnishes an open set $R_1$ and a closed set $R_2$ such that $A \subset R_1 \subset R_2 \subset B^c$, and likewise $B \subset A^c$ furnishes an open $S_1$ and closed $S_2$ such that $B \subset S_1 \subset S_2 \subset A^c$.

[gentle osprey]

plexcty

[rich cliff]

So now my candidates for open nbhds were $R_1, S_1$ or $R_2^c, S_2^c$, but not sure how to show disjointness in either case, or if it even holds

[gentle osprey]

plexcty

[rich cliff]

I couldn't read all of that but im pretty sure that was the forward direction

[rancid umbra]

let Y and Z’ be closed and disjoint. set Z = X \ Z’

as Y is a subset of Z then due to the condition, there is an open U and a closed V’ such that
Y subset U subset V’ subset Z

now V = X \ V’ is an open nbhd of Z’ disjoint from U

[rich cliff]

@rancid umbra holy shit I suck

Thanks

[umbral panther]

The final object is the point

To get the final object to be the one point compactification, you need to make two requirements. One is that X maps homeomorphically onto its image. The other is that its image is dense

[austere dirge]

this makes sense

thanks

[tender halo]

compactifications of a tychonoff space form a complete (upper) semilattice, with the order given \alpha X >= \gamma X when there is a continuous map f: \alpha X -> \gamma X such that f | X = id

the top of this semilattice is the CS compactification

the bottom exists iff X is locally compact, and then it is the one-point compactification

[real granite]

We haven't actually studied bases in this module but I assume you could say something like "the lower limit topology is the topology whose basis sets are those of the form [a,b), where a<b"?

[gritty widget]

That is the usual definition of lower limit topology. What is the question?

[real granite]

That's it

It's literally just define the LLT

But since we haven't done basis sets it's a bit of a weird question

[keen seal]

oh

[real granite]

Without making reference to bases, I suppose you could say something like "the LLT is the topology where sets of the form [a,b), a<b, are regarded as open", but I'm not sure that's actually a formal definition

[gritty widget]

How do you know what the LLT is if you haven't defined it?

[tender halo]

its the minimal topology generated by those half open intervals yes

[gritty widget]

there are more open sets than those ones

[keen seal]

I assume through sequences I guess

[real granite]

Uh... so the lecturer used the LLT in the lectures but didn't write anything in the lecture notes, which means a student's familiarity with it can vary depending on the textbooks they have used (if any)

Someone here suggested Counterexamples in Topology, which I found useful

[gritty widget]

How did the lecturer define the LLT in the lectures?

[tender halo]

i am a little bit confused how you havent defined bases yet

uts like

second definition

that the course usually has

first is "topology", second is "basis of a topology"

[real granite]

Ye, this is why I think it's a weird question to ask, without bases then you can give factual statements about LLT but not really a formal definition

[keen seal]

I don't think so

When I took topology we started with Metric spaces again, even though we saw them in an Analysis course

[tender halo]

blergh

[keen seal]

I don't think it's that uncommon to start with metric spaces, changing statements to be in terms of balls

[real granite]

Not just yet, bases aren't covered at all in the lecture notes. It's possible they were mentioned in the lectures and I have forgotten, but I don't think they were

[gritty widget]

If you know what bases are, you can reword the usual definition of LLT without using the word base

[tender halo]

i maintain that starting with metric spaces is pedagogically harmful

still really weird

[keen seal]

ok

[gritty widget]

Why?

[quartz horizon]

Wait really?

[real granite]

The first time I heard bases used in reference to topology (rather than algebra) was after the lectures had finished and I asked a tutor for my quantum mechanics for some topology textbook recommendations, and he said Pure and Applied Topology (or smth like that), and there bases are defined midway through

[quartz horizon]

I am also curious why

[real granite]

Our course started with metric spaces and then presented topological spacies as generalised metric spaces

No bases ever

[keen seal]

yikes

[real granite]

Rip

[tender halo]

i mean i just think it really encourages wrong intuition

[quartz horizon]

Really? I’ve often used metric space intuition

[keen seal]

If you think so

[real granite]

Are non-Hausdorff topological spaces used in physics much?
(I would ask "at all" but realistically there is a probably an application of any and every part of maths in physics if you look hard enough)

[tender halo]

not really

[quartz horizon]

Hmm, I have not yet come across one

[tender halo]

non Hausdorffness is usually some kind of algebra

[wispy veldt]

in what sense? understanding the strenghts of a metric structure is important

[gritty widget]

I don't think it's good to consider topological spaces as "generalized metric spaces", because metric spaces have extra structure. Generalization of certain properties of metric spaces would seem more adequate.

[real granite]

Stuff like sequentially closed does not imply closed in general topologies

[wispy veldt]

and as mentioned its natural to generalize concepts from a metric space to a topological space

[real granite]

I assume this is what Beaver has in mind

[wispy veldt]

often as "this used to work when we had a metric structure but now it doesnt"

[keen seal]

I'd say probably in an area which uses topological groups

[gritty widget]

If you replace sequences by nets then it works for general topological spaces

[wispy veldt]

yeah thats fair but i think its in general important to understand and grasp the information you gain from your structure

[real granite]

Well yes I think when I said "generalised metric spaces" that was shorthand for what you've said here

[gritty widget]

In a topological group T0 implies hausdorff iirc

[wispy veldt]

what happens when we have a metric? what do we gain if we have a norm instead ? or a inner product?

[alpine nest]

This doesn't uniquely define the topology, since the discrete topology satisfies that property as well

[keen seal]

yes

[wispy veldt]

having a sense of what a space does is good

[alpine nest]

You would need to speak of the smallest topology in which those sets are open

[keen seal]

{0} being closed is equivalent to Regular

[real granite]

Yes I've heard this but our course doesn't go into that

[real granite]

Ye, I think bases is easiest

[gritty widget]

I don't understand this reply then

[real granite]

Anyway I need to get back to grinding

Thx for the help tho

[alpine nest]

Eh, I think saying "the topology generated by <family of sets>" (which is exactly "the smallest topology in which those sets are open") is fairly standard and fairly good.

It frees you from worrying whether your family of sets is in fact a base

[keen seal]

Why?

[alpine nest]

Which an arbitrary family might not be

[gritty widget]

You could be more explicit and say: "the opens are unions of sets of form [a,b)"

[wispy veldt]

well a fair amount of physics is done on seperable hilbert spaces, so you have a lot more than just hausdorffness

i would be suprised if there is a area that uses them too

[gritty widget]

The question was if non-Hausdorff topological spaces are used in physics. And you answered probably in an area which uses topological groups. but for topological groups T0 equivalent to Hausdorff and there's not usually much practical difference between working with T0 or non-T0 spaces

[quartz horizon]

I mean I don’t even think I’ve met non-metrizable spaces

In physics that is

[wispy veldt]

topological groups are hausdorff

[keen seal]

Not necessarily

[wispy veldt]

if they are not you can just work with the quotient

so they are essentially hausdorff

[keen seal]

I mean the closure of <1> can be non trivial

[tender halo]

it is important, but also its the subject of metric geometry, not point set topology, a lot of "obvious" facts from metric spaces become really wrong really fast

[wispy veldt]

and thats why we work with G/closure<1>

[keen seal]

I mean yeah

[wispy veldt]

technically there is non-hausdorff top groups but for most purposes they are hausdorff

[keen seal]

wdym?

[wispy veldt]

edited

[keen seal]

ah I see

[wispy veldt]

due to this for reference to the others

[alpine nest]

In my area of mathematics even non-metrizable spaces are rare.

[keen seal]

I mean if you're working with a pro-V topology on some category /pseudocategory V, you might care about Cl(<1>) because it has implications on separation.

[alpine nest]

Some of the tools do use a space which is technicaly non-metrizable, but we only work on a subspace of it, which is metrizable.

[wispy veldt]

i suppose the space of distributions is a example of a non-metrizable space

with the topology we want to work with

but thats about it

[quartz horizon]

Even then, it’s not like topology is only useful to consider non-metrizable spaces

Topology lets you do operations on spaces in a way that metrics just can’t

Like, the square and the torus are both metrizable, but without a 3D embedding it’s unclear how to relate the metric on the square to the metric on the torus

But you can relate their topologies quite nicely, without any embedding

[gritty widget]

If you mean limits and colimits you could if you used (symmetric) Lawvere metrics instead

[keen seal]

wait really?

[quartz horizon]

Oh?

[gritty widget]

yes, the issue with usual metrics is that

[quartz horizon]

Hmm, even then, I know that a colimit in metric spaces can differ from the colimit in topological spaces

[gritty widget]

they can be seen as categories enriched over the monoidal poset [0,+infinity) with + as tensor product

[quartz horizon]

But I am curious to hear what you mean

Yes I know

[gritty widget]

and [0,+infinity) is not complete

[quartz horizon]

Ok so what does lawvere do

[gritty widget]

this is the reason why metric spaces are not complete

[gritty widget]

lawvere metric spaces allow metrics to take +infinity value and allow nontrivial distances to be 0

[real granite]

when you say "more", do you mean separation of axioms T_n where n≥2?

[quartz horizon]

Yes I know

But what I am interested in is how this gives rise to categorical constructions

[gritty widget]

they also allow non-symmetry but thats not as important

[quartz horizon]

That I have not seen

Mhm I know

[gritty widget]

When V is a complete cocomplete monoidal closed category, the category of V-enriched categories is also complete and cocomplete, or something like that

[keen seal]

I think separable and Hausdorff don't have any dependence (i.e. you can have one without the other, both or none)

[gritty widget]

the category of metric spaces is a subcategory of the category of [0,+infinity]-enriched categories

[keen seal]

Separable means having a dense countable subset.

[quartz horizon]

I see, that’s a cool result

[gritty widget]

if you took the whole category, instead of just the classical metric spaces, you'd get a complete and cocomplete category by this

[quartz horizon]

And you’re saying [0, infty] is such a monoidal category?

[wispy veldt]

separation is one property you gain, but also structure is what im emphasising, normed spaces by themselves have a much nicer structure than metric spaces, and a hilbert spaces gives you a geometric structure on top of that, and most situations where you are "non-hausdorff" are spaces that don't have a very interesting analytic structure to work with, its more so topological in nature.

[quartz horizon]

So it’s monoidal closed, complete and cocomplete

[keen seal]

Shoutout to all my projective and injective limits kachow

[gritty widget]

yeah with respect to +

[quartz horizon]

I see! Thanks for this, I wasn’t aware

Another reason to support lawvere metric spaces, I suppose

[gritty widget]

And if you consider only the symmetric ones it will still be complete and cocomplete iirc

[wispy veldt]

and ofcourse seperability gives you all the quality of life stuff similar to linear algebra

[keen seal]

lmao

We got QOL update on Hilbert spaces before gta 6

Man I do always assume separability

I ain't gonna lie

it's a massive QOL boost

[wispy veldt]

indeed

[keen seal]

There was like a theorem I remember in like Functional Analysis class

which held for all Hilbert Spaces

but the proof for non separable was just so much more annoying

[wispy veldt]

people used to define hilbert spaces as seperable for a reason

but its fun to study what happens too

similar to seeing what happens in a non-hausdorff space

gotta love the chaos

[real granite]

2(ii)
Discrete metric works right?

Because we can always put 0<r≤1, so B(r, x) (x in Z) is just B(r,x)={x}, so every singleton containing an integer is open and hence every union thereof is open

[gritty widget]

Yeah

[gritty widget]

Not sure what the first part of your message means

[alpine nest]

More succintly: Z as a subspace has the discrete topology, which is always metrizable.

[real granite]

Well if r>1, then the open ball B(r,x) covers the entire space

And if r=0, then the ball is empty

[gritty widget]

It's not possible to have B(r,x) entire space when r>1 and only {x} when r<=1

But the idea is close to something that works

[alpine nest]

That is exactly what happens in the discrete metric

[gritty widget]

No, you have B(r,x) entire space when r>=1

rather than only when r>1

[alpine nest]

B(r,x) refers to open balls here

[real granite]

Ye

d(x,y)<r

Strict inequality

[gritty widget]

Sorry you're right

[real granite]

Np

[alpine nest]

"Ball" meaning "open ball" is the convention I'm used to so I didn't question it

But I suppose some authors might use other conventions

[real granite]

yeah I've not come across closed balls are the default

[gritty widget]

Me neither, my brain just wasn't thinking properly

[keen seal]

wait what the heck

people use balls to refer to closed balls, that's so cursed

[gritty widget]

I've usually heard it called disks

For closed balls

[alpine nest]

I say "closed balls"

[keen seal]

why not say just closed balls 😭

[keen seal]

exactly

Point set topology is awesome

[gritty widget]

Idk, Some authors use disk

[alpine nest]

It is!

[keen seal]

even in analysis, I just find things easier to prove using the topological definition

than sometimes using the theorems 💀

[alpine nest]

What do you mean by "topological definition"? Avoiding the concept of metric?

[keen seal]

Yeah, I meant like the general topological definition

[alpine nest]

I was about to point out that metric is a topological definition

[keen seal]

Like proving continuity in terms of opens, instead of epsilon deltas

[alpine nest]

But yeah, the general-topological point of view sometimes helps.

Although metric gives you the notions of uniform continuity and of completeness

[keen seal]

yeah of course

[alpine nest]

Which can be somewhat emulated using the general topological language, but it's usually more trouble than it's worth

[keen seal]

I'm just saying it as a way to speedup some proofs

after all that's why there are so many equivalent definitions for the same thing 😛

[alpine nest]

So many almost equivalent definitions

[keen seal]

I mean like equivalent if property, and we have that property established

like Hausdorff or separability

[gritty widget]

Why?

[alpine nest]

It's an alternative way of defining the topology, using a metric rather than concepts of open set/closed set/closure operator/what have you

And it's not a side effect, the primary purpose of the metric is to induce a topology

[quartz horizon]

To be fair it does depend what you’re using the metric space for

There are more to metric spaces than continuity

[alpine nest]

Do people study metric spaces without considering the topology induced by the metric?

[quartz horizon]

I think this is common in applied maths yeah

Tons of discrete metric spaces e.g.

One considers “word distance” for codes

Sometimes you really do just want an actual notion of distance

[alpine nest]

Fair point

[gritty widget]

That doesn't mean it doesn't have more purpose than just the topology

[quartz horizon]

I think in pure maths your statement may be more true

[alpine nest]

Either way, it's the kind of semantic quibbles over which we could argue for days without any reasonable conclusion.

Either way, in the context of real analysis "metric" is definitely something I would consider a topological notion

[quartz horizon]

Mhm, I think that’s fairer

[real granite]

Same topology as here. Am I right in saying that both are connected?

i. The open sets are ø, {1} and (0,1], so there is no way of writing (0,1] as a union of non-empty disjoint open sets.

ii. The open sets are ø, ( (0,1) u (2,3) ) n Z = ø, ( (0,1) u (2,3) ) n R = (0,1) u (2,3). These are all trivial, so there is no non-trivial separation, i.e. (0,1) u (2,3) is connected

[gritty widget]

(0,1] is not open

[alpine nest]

It is, but {1} is closed, so (0,1) is open

(0,1) and {1] are both clopen sets in (0,1] with your topology

[real granite]

Uh, it's open in the subspace topology

Isn't that what the question means

[gritty widget]

oh i see

[alpine nest]

Yep, we're still talking that weird topology introduced above

Wait, I might be wrong about {1} being closed.

Yeah, I'm wrong