point-set-topology

[opaque scroll]

That's exactly right

[ebon dock]

How does the mandelbrot set relate to the fibonacci sequence

[queen prism]

erm

is this a topology question

[low flame]

anyone able to explain this to me since I cant seem to get it

[mighty hull]

Is the interval [0,1] path connected? What is its boundary?

[low flame]

yes its path connected, and its boundry is 0,1 i think its the (1-t)x + ty function

oh i see the boundry isnt connected

is that the point?

[novel acorn]

It's neither connected nor path-connected

[low flame]

the interval [0,1] isnt connected in real numbers?

[novel acorn]

The boundary I meant

[low flame]

yea i kinda got the point of that now 😛 i saw it as i was answering hehe

[low flame]

Thank you so much you made me see it

[mighty hull]

For the interior, imagine two closed balls touching at exactly one point on their boundary.

In R^2 for example

Like

The interior is the disjoint union of two open balls

[low flame]

that makes sense, and then i gotta show the intersection isnt an iterior right

[low flame]

could you also have two disjoint balls, then connect them with a line?

[mighty hull]

Yes, that also works.

[mighty hull]

Highly recommend this book, btw: https://editorialdinosaurio.wordpress.com/wp-content/uploads/2012/03/counterexamples_in_topology_-_l-_steen__j-_seebach__1970__ww.pdf

(Support Dover and pay $10 for a real copy)

[low flame]

i actually have the pdf for this alreayd 🙂

i got an exam in less than a month and im just trying to keep afloat

ill look at it when i got time

when i looked at it last time i couldnt understand it

[mighty hull]

It's meant as a reference.

[jagged prairie]

I continued watching the video I was watching yesterday to review and here he says this that contradicts what I've been told here. Just what is going on?

[alpine nest]

What have you been told before and what's the contradiction?

Is it about the fact that in this example (1,3] is both open and closed?

If so, then: whether a set is closed and/or open depends on the space it's in. (1,3] is closed (and open) in the space X from this example, but it's neither closed nor open in the space R (the set of all real numbers with the usual metric)

[jagged prairie]

Oh it seems that that was my mistake. I mistook A being open / closed in X as it being open / closed in the space R. Thank you for clearing that up.

[jagged prairie]

So in the example, A is open in X because despite including 3, the epsilon ball B_1(x) is fully contained within X? If this is the case, why is A closed? Does A have any boundary points?

[low flame]

is the closure of A (0,inf) for x and for y its [-1,1]?

no the closure of A would include zero right, since its a limit point?

[alpine nest]

Did you sketch A?

[alpine nest]

Proving that A is closed (in the space X) is a good exercise, and how exactly you should go about it depends on what definitions/characterizations of closed sets you have available at this stage.

[low flame]

yes of course, thats not difficult

its all wiggly around the orgin, and it goes towards zero as it goes to infinity

[alpine nest]

Also I'm not sure what you mean by "for x" and "for y" in your description of the closure

[fringe thorn]

a set is closed if its complement is open. the complement of A in X is (4, inf), which you can check is open

[alpine nest]

Yeah, that's the fastest way, but relies on knowing this characterization of closed sets.

Although it's also a good exercise to prove closedness using various criteria (for example, a set is closed if and only if it contains all its limit points, or a set is closed if and only if for every convergent sequence of its elements, the limit is also its element)

Or indeed a set is closed if and only if it contains all its boundary points

My preferred three characterizations of closed sets, in no particular order, are "the complement is open", "is closed under the operation of taking limits of convergent sequences", "contains all its adherent points (i.e. if x is in X and for every epsilon the ball B(x,epsilon) has nonempty intersection with A, then x is in A)"

The last one is almost the same as "contains all its limit points", but the definition of adherent point is simpler (there isn't that clause about excluding the point itself from consideration).

[cedar jungle]

Is it possible that a subspace of X is homeomorphic to Y and a subspace of Y is homeomorphic to X, but X is not homeomorphic to Y? Why or why not?

[1/2, 1) is homeomorphic to x

And X is not homeomorphic to Y

[fierce lily]

The thing I am confused is: U_alpha is open convering of A, so it might contains some elements even outside of A. But U'_alpha intersects A must totally contained in A, why U_alpha=U_alpha' intersects A?

if that is typo, then how we can modify this part, can someone give me some ideas?

[cedar jungle]

Is R\Q separable?

[rustic carbon]

yes

[rustic carbon]

if U is open in A, then U = U' \cap A for some U' open in X. this is the definition of the subspace topology.

[fierce lily]

but my confusion is, U might contain some elements in X but outside of A, so I think we can not say U is open in A?

I mean U_alpha might not be subset of A, why it can be open in A

[rustic carbon]

no, U' is open in X. U is open in A by definition.

[fierce lily]

but U_alpha is also every open covering of A, this means we even can let U_alpha cover many points that does not belong to A. I know that subsapce topology on A is U' intersects A as U' is open in X, but it must be subset of A

so since U_alpha is possible to contain points that does not belong to itself, why we can define it as subspace topology of A

[fierce lily]

I think U_alpha can only be open in X, so it should be wrong. but we can write {U_alpha} union X-A that covers X. In that case, because X is compact, we have finite subcover of X. Then we can just discard the set X-A as the subcollection of subcover contains X-A, and the resulting subcollection of cover is finite for A

The above argument seems correct to me.

[chilly topaz]

Consider a set that as an empty interior, a non empty boundary, and it contains all of its boundary (so like the subset {(x, y) in R^2 | x^2 + y^2 = 1) . . . Is this set Clopen?

Because if yes because it technically contains all it’s interior points (which are none), then the example this is a non trivial Clopen set of R^2, which doesn’t make sense since R^2 is connected,

And if no, then how does it not qualify as a Clopen set which it contains all interior points (is open) and it is it’s closure (it’s boundary union it’s interior aka is closed),

Oh and I also checked to make sure open set (belonging to topology) means is contains only interior points (there might be something wrong with this?):

And likewise for closed sets:

[queen prism]

those are some interesting notes...

[chilly topaz]

But because it has an empty interior it technically contains its interior points?

We use the same logic for empty set right? That it is open because it contains it’s interior points (which is none),

Oh,

OH

So open sets ONLY contain interior points?

[wise mist]

,rotate

[gentle osprey]

Couldn't find an attached image in the last 10 messages.

Nov 6, 2024, 2:10 AM

[prime elbow]

i think converse is true but i am not sure, can anyone verify it?

[prime elbow]

How can I show that in order topology [a,b] is a closed set? Can I say it is a complement of open sets by studying cases ?

Order topology on X

[wise mist]

do i give a counter or is there a way to prove

[low flame]

obvioulsy the image of A is the ray or projection onto x, which is homemorphic to (0,inf) which is path connected.

to show that X is connected can i say that A is dense in X. Let X = U cup V

A = A cap X = (A cap U) cup (A cap V) Let V be empty, and then A is dense in X, and the space X has an everywhere dense subset A and is thus connected?

[wise mist]

$f_n(x) = \frac{x^n}{n}, \quad f_n'(x) = \frac{nx^{n-1}}{n} = x^{n-1}, \quad |f_n|Y \to 0, \quad |Df_n|{\infty} = 1$

[gentle osprey]

noctambulant

[wise mist]

is the counter enough to state the desired ?

[cedar pebble]

alternatively you can take f_n(x)=sin(nx)/n which goes to 0 in the \infty-norm, but f'_n(x)=cos(nx) does not

but yeah any sequence of functions which goes to 0 in the \infty-norm, but whose derivatives do not, suffices as a counterexample

[wise mist]

so that proves it right ?

[cedar pebble]

yeah

[wise mist]

alr

is it hard to generate continous fucntion from (a,b) \to R for which F^- is also continous

[cedar pebble]

well it's hard for me

I'm bad at these types of problems

[fickle tendon]

Using topology and geometry by Bredon, in definition 9.5 he uses the equality with three lines (which i only know from modular arithmetic). It's not introduced anywhere and isn't in the list of used symbols

Does anyone have a clue what it is?

[prime elbow]

my guess is f(x) = 1 for all x in C

[queen prism]

identically equal

it’s a form of emphasis

[prime elbow]

can i say in arbitrary topological space X times Y, (x_n, y_n) converges to (x,y) if and only if x_n converges to x and y_n converges to y.
i know it is true in metric space, but i think it is also true in topological space

[hot cove]

It's true here also as product of open sets is a basis for product Topology

[prime elbow]

yes, thank you

there is a question characterize the continuous function from R(co-countable topology) to R(usual topology), now i know constant functions are continuous, and identity function is not, but no idea how to characterize it?

[alpine nest]

It might be easier to think in terms of closed sets, since in the co-countable topology a set is closed if and only if it's countable (or all of R).

So for a function to be continuous, the preimage of every closed set has to be either countable or all of R.

[prime elbow]

yes

[prime elbow]

i got it but i don't get it how to write, can i write simply this one ?

[alpine nest]

Well, I wouldn't consider this to be a characterization, but it might be a good starting point to developing one.

[prime elbow]

okay

[alpine nest]

Disclaimer, I don't actually know the answer, but that would be my line of thinking

I suspect it might indeed be just the constant functions

Yeah, it will be, unless I'm misthinking.

[prime elbow]

this one is nice starting point

[mighty hull]

Let S be an uncountable set w/ the co-countable topology. You have the following properties:

1. If V ⊆ S is open then V is uncountable
2. If V ⊆ S is open and B ⊆ S is uncountable then their intersection B ∩ V is also uncountable

If f is a non-constant function, use the fact that ℝ w/ the usual topology is Hausdorff to get two disjoint open neighborhoods.

[prime elbow]

here is underlying set S?

i got the 1 point but don't get it 2

wait

[alpine nest]

Nice, I like this more than my argument

[prime elbow]

why B cap V is uncountable?

say it is countable

then S/ B cap V is open

so it is uncountable

[fickle tendon]

Oki thanks, i see the "over C" now

[prime elbow]

let f(x) \neq f(y), so there exists disjoint open set U containing f(x) and open set V containing f(y), if f is continuous then f^-1(U) and f^-1(V) is uncountable so by 2 their intersection is non-empty implies it is not disjoint.
correct?

[mighty hull]

If U and V are disjoint then their pre-images must also be disjoint, which means those preimages can't be open in the co-countable topology.

So f isn't continuous.

[prime elbow]

Yes but I don't get the 2nd point

I don't understand how B cap V is uncountable

[alpine nest]

It's because the complement of V is at most countable.

(assuming V is nonempty)

2. indeed doesn't hold if V is empty (as usual the empty set ruins everything)

Neither does 1., actually

(this isn't an obstacle in the actual argument, mind you; we're only interested in nonempty open sets)

[prime elbow]

What do you mean by at most countable?

[alpine nest]

Finite or countably-infinite (I'm fine just calling such sets "countable", but I wanted to be extra-precise)

[prime elbow]

Yeah its complement is countable

But how it helps me to prove 2nd

[alpine nest]

Consider the points of B which are not in V

[prime elbow]

X/B cap V is countable so B cap V needs to be uncountable because X is uncountable, correct?

[cedar jungle]

How can I prove that an isometry is an injection?

[prime elbow]

say f(x) = f(y)

Now use isometry property

[alpine nest]

You've got the sets switched, it's X\V that's countable, but the argument is along those lines.

[prime elbow]

Oh

The points which are in B but not in V are countable

[mighty hull]

In general, if you have two non-empty subsets A,B ⊆ X then A ⊆ B if and only if A ∩ (X\B) = ∅. That's just a basic fact about sets.

Well, if V is open and B ∩ V = ∅ then B ⊆ X\V, but X\V is countable, so B has to be countable, too.

(Here I'm using "countable" in the broader sense @alpine nest mentioned, meaning "finite or countably infinite".)

[prime elbow]

So we just need non-empty, right?

[cedar jungle]

If x = y then f(x) = f(y), and if x is not equal to y, then f(x) is not equal to f(y). So f is injective. But is that enough?

[prime elbow]

No this is the definition of well defined functions

a function is injective f(x) = f(y) implies x = y

[prime elbow]

But you used contrapositive

But how you showed if x is not equal to y, then f(x) is not equal to f(y)

[cedar jungle]

It is an isometry, so if f(x) = f(y), then the distance is 0 and so is the dictance between x and y, so x = y

[cedar jungle]

but is that enough?

[prime elbow]

Yes it is correct here you used isometry property d(f(x), f(y) ) = d(x,y)

[prime elbow]

Yes because it is contrapositive

[cedar jungle]

thank you

[prime elbow]

@mighty hull sorry to ping, so if X is uncountable and co-countable topology then every open set intersects we just need that to prove that f is constant mapping.

Thank you

[mighty hull]

Yes, this shows that if f is non-constant then it's not continuous.

And constant maps are always continuous, regardless of topology.

[prime elbow]

Yes thank you

[heady skiff]

say we have a mapping defined by coordinate functions into a product space which is indexed by an arbitrary set

if all of those coordinate functions are continuous, is the original function continuous

[quick crane]

yes. In fact, a net converges in the product topology if and only if each component converges.

[tender halo]

also this isnt really a product topology property per se, product topology is the weak topology generated by the projections, and a function into a space with the weak topology is continuous iff it is continuous in each of the generators

[prime elbow]

How can I show that every T_3 space is T_2.5 space?

T_3, when (X, T) is regular with the T_1 axiom.

T_2.5, if for any two distinct points x,y in X there are open sets U and V containing x and y respectively such that cl U cap cl V is empty.

Yes T_3 implies space is Hausdorff

[tender halo]

take the nbhds from hausdorff, in each of them take nbhd from regularity

[prime elbow]

If f is a continuous function from X to Y then cl f^-1(A) \subset f^-1(cl A), right?

[wise mist]

right

[prime elbow]

Thank you

[queen prism]

initial topology

[wise mist]

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

[gentle osprey]

nastasya

[wise mist]

check your statement from here

this is correct

[tender halo]

that is equivalent to being continuous

[prime elbow]

yes it is equivalent to continuous

[prime elbow]

okay

[prime elbow]

i am working on your hint

[wise mist]

well what is Tn ?

[prime elbow]

you mean T1?

[wise mist]

oh yeah u have used a lot of numbers other than 1 but yeah that one

countable ?

[prime elbow]

no it is equivalent to every point set is closed

[wise mist]

ok where can i get this concept in munkres

[tender halo]

hausdorff is t2, regular is t3, tychonoff is t3.5, normal is t4, hereditarily normal is t5, completely normal is t6

[prime elbow]

every single set is closed in that topology

[wise mist]

alr

its the finest one right ?

[prime elbow]

yes

[prime elbow]

what do you mean?

[wise mist]

im asking where, whats the name of the T

[wise mist]

T1 seems to be the finest

[prime elbow]

finest in which sense?

no it is not

[queen prism]

Tn is a classification scheme

[prime elbow]

what about T0?

[queen prism]

do you know what the Tn axioms are?

[wise mist]

i was thinking its the same as when every singleton set is open set

[tender halo]

T0 is for every pair of points there is an open set containing only one of them

[prime elbow]

yes

[tender halo]

that just means the space is discrete

[wise mist]

is that equivalnet to T1

[prime elbow]

you mean first i take open set U containing x and V containing y such that they are disjoint, then?

[prime elbow]

no

[queen prism]

https://en.wikipedia.org/wiki/T1_space

[prime elbow]

R with co-finite topology is T1 but not discrete

every metric space is T1 but not every metric space is discrete

[wise mist]

ok, i have never come across this Tn fuckery

[queen prism]

welcome to the club

[wise mist]

i am doing metric space topology so the book doesnt generalise much topological concept

[tender halo]

its all throughout non metric topology

[wise mist]

thats insane

#point-set-topology message

would u verify this for me ?

[tender halo]

idk i like it, except t3.5 i guess, that one i like to use "tychonoff"

[tender halo]

thats the first half

[prime elbow]

Yes but I don't understand what I do after that

[tender halo]

use regularity

two times

[alpine nest]

metric spaces are in fact perfectly normal

[alpine nest]

True, although I'd also say that none of the separation properties are as important as T2

The qualitative difference between Hausdorff and non-Hausdorff is huge, and the other ones are more about various degrees of "niceness"

[tender halo]

tychonoffness is pretty huge too

at least if you want any of the nice compactifications

[alpine nest]

What are the important spaces that are Hausdorff but not Tychonoff?

Not being snarky, genuinely curious since I tend to interact with reasonably nice spaces

[tender halo]

hmm

[alpine nest]

A compact Hausdorff space is immediately normal, isn't it?

Which is even stronger than Tychonoff

[tender halo]

yep

[alpine nest]

And locally compact Hausdorff are Tychonoff

[tender halo]

yep

[tender halo]

yea but if you want a nice compactification then you need to start with a tychonoff space, then compactification will be normal but the space itself doesnt need to be

[alpine nest]

I'd still say that intuitively the fact that you can separate points and that compact subsets are closed, is more impactful than what you get from the Tychonoff property on its own (which isn't to say that property isn't impactful, I just don't think it's as big of an impact)

As in I'd say there's more theorems that break without Hausdorffness than ones that break without Tychonoffness

[tender halo]

true

[alpine nest]

Fortunately I've almost never had to deal with non-metrizable spaces in practice

I enjoy general topology but more as a spectator

[tender halo]

i guess its just recency bias for me, im reading a book on rings of continuous functions into R and there everything needs to be tychonoff because you deal primarily with zero sets

and tychonoff gives you that zero sets are a closed base

[alpine nest]

Well yeah, in specific circumstances you will rely on the other Tn's hugely, which is one of the reasons why they're a thing

[tender halo]

¯_(ツ)_/¯

good book by the way, everyone should read it

probably the best math textbook ive ever read

[trim spade]

What’s the name of the book?

[tender halo]

rings of continuous functions by gillman

[prime elbow]

Yes

For disjoint non-empty closed sets A and B we can construct function x to d(x,A)/(d(x,A) + d(x,B), right? Then it gives the result

[wise mist]

looks like urysohn lemma

[paper wedge]

yes

[sharp aurora]

what is metric space topology

[mighty hull]

The topology generated by the open balls in the metric space.

See the Basic Notions section on Wikipedia: https://en.wikipedia.org/wiki/Metric_space#Basic_notions

[wise mist]

need a source to get bette grip on tietze extension theorem

[sharp aurora]

thanks

[prime elbow]

how can i show that if X is perfectly normal then for every closed set C there exists continuous function f:X-> [0,1] such that C = f^-1(0)?

X is topological space

[prime elbow]

i want to prove this, if X is perfectly normal then every closed set of X is G_delta set.

Since X is perfectly normal so for any closed set C, there exists continuous function f:X->[0,1] such that C = f^-1(0) = f^-1(cap [0, 1/n) over n in N) = cap f^-1[0, 1/n) over n in N, since f is continuous therefore f^-1([0, 1/n)) is open.

hence C is G_\delta set, is it correct?

[heady skiff]

when we talk about the "path components of an open subset U" where U is a subset of a topological space X, are we talking about the path component which intersects it, or the path component which contains it?

or are we talking about the path components contained in U viewed with the subspace topology

[wise mist]

difference of topology between $\Bbb{Q}$ and $\Bbb{R} \setminus \Bbb{Q}$

[gentle osprey]

nastasya

[wise mist]

seperablity?

connectedness

[prime elbow]

what do you mean?

[wise mist]

oh im not into these ideas yet, in anyways can i get hint to get into the differences

[prime elbow]

both are disconnected

[wise mist]

i mean those are two topological ideas idk yet

from whatever i know they're quite identical as a set

[prime elbow]

one is countable other is not so i think if we define T- co-countable topology on both sets they are different

[wise mist]

what about cardinality?

[wise mist]

what is even that

so R\Q is like R

if that makes any sense

[prime elbow]

a topology in which complement of open sets is countable with empty set

[wise mist]

i have never read that T.....

isn't it just countable compliment or co-countable

[prime elbow]

kind of but co-finite is coaser than co-countable

[wise mist]

made a mistake there

[wise mist]

yeah right

[queen prism]

cardinality is different so they are bound to be drastically different as topological spaces as well

[prime elbow]

how can i prove a => b and c => a ?

[wise mist]

which text is this ?

[prime elbow]

Topology course, University of Toronto

[low flame]

why is 4 and 5 false?

i would think if you have a connected space, and you take a subset of it, it would be connected. however as I type this I think of an extreme where you take a subset of singletons which wouldnt be connected

[paper wedge]

take {-1,1}

[queen prism]

you can take literally any subset and turn it into a subspace

[paper wedge]

for 5) Q is not path connected

[low flame]

its not even connected so yeah i saw that as i was typing it

[paper wedge]

yeah

[low flame]

im a bit slow, i dont get things very fast haha. thank you :=)

[sonic crane]

Same bro 😎

we in this together

[sonic crane]

why are those equivalent?

If an open set U intersects every other open set, that set is dense? And density means closure(U) = X right

[quick crane]

yes

[opaque scroll]

Notice that the complement of closure(U) is an open set that doesn't intersect U

[sonic crane]

oh, bruh.

Cool. thanks

[mighty hull]

There are many equivalent characterizations of "dense". Worth proving their equivalence if you haven't.

[sly geyser]

I never thought about it before about it before but R^2/{(0,0)} is an open set

[queen prism]

\

[sly geyser]

R^2/{(0,0}} = U(x∈R^2/{(0,0)})({|y-x|<=|x|/2)

Why is discord making the 's invisible

is it some sort of formatting thing

well I'll use / I guess then

[queen prism]

\ often signifies an escape character

meaning it combines with the next character to make something else

[sly geyser]

should I just use two then?

\

yeah that works

[queen prism]

yea so if you write \\ it means "I don't want this \ to cancel something pls put an actual backslash"

[sly geyser]

lmao

[queen prism]

easiest workaround is to just put a space

or put it in code tags like I did

[heady skiff]

is the weak topology on a space the basis formed by taking the preimages of all open sets in Y (say we only have one continuous map f: X \to Y)

or would that be the sub-base

[chrome trout]

Is there an uncountable meager set in the cantor space?

[kind sluice]

Hi, does anyone know if in a topological group the path-connected component of the identity is a closed normal subgroup? (additional assumptions such as compactness are welcome if needed)

[teal nova]

if X and Y are top. spaces, and we consider X x Y with the product topology. Why do we need Y to be compact for the projection morphism on X to be closed?

[kind sluice]

Isn't the cantor set uncountable and meager itself?

[unreal stratus]

it is unmeager in itself right

just meager within R

[opaque scroll]

It's definitely a normal subgroup, since you can shift and conjugate paths.

Path components are not in general closed, but topological groups are quite rigid, so maybe that holds there...

[kind sluice]

right, my bad

[waxen bough]

Hi, I'm with Maikel with the same question. What we want to know most is if the path-connected component of the identity is closed 🙂

Our group is compact, hausdorff and abelian if that helps

[unreal stratus]

Pretty sure as well that like the original thing is an iff

uhh

oh yeah i remember uh

i remember reading the nlab stuff around compactness and trying to get super constructive and frugal proofs of everything lmao

[opaque scroll]

I think maybe this group will be a counter example
https://search.app?link=https%3A%2F%2Fen.m.wikipedia.org%2Fwiki%2FSolenoid_(mathematics)&utm_campaign=aga&utm_source=agsadl1%2Csh%2Fx%2Fgs%2Fm2%2F4
But I'm not quite sure what the path component of the identity is

But if I understand it correctly, you take something like for example the 2-adic numbers, then connect x and x+1 by a path. So the path component will be the integers and the paths between them, and then the closure should be everything.

[waxen bough]

looks like it could indeed be a counterexample. We were trying to see if the Bohr compactification of R is path-connected... I think it totally should be

[chrome trout]

Does anyone know an example of an uncountable meager set in the cantor space 2^N? Ive been trying to figure this out for hours

Or a way to show such set exists

[heady skiff]

I'm trying to show that I with the endpoints identified is homeomorphic to the circle, and I know that if I can show $h: I \to S^1$ given by $x \mapsto e^{2 \pi i x}$ is open or closed then I'm done (since it's a continuous surjection). Any ideas?

[gentle osprey]

okeyokay

[queen prism]

can you picture in your head what that would look like

what do the open and closed sets in I look like, and what about their images in S^1

what does the mapping h look like

[heady skiff]

yeah i can picture what it would look like

but ig i'm trying to rigorize

i saw somewhere that i can use the open mapping theorem

ig that makes sense

https://math.stackexchange.com/questions/1245995/simplest-way-to-prove-that-eix-is-an-open-mapping-into-s1

altho i'm not too sure if this would work since they define the map on all of R and say that therefore it maps open sets of R to open sets of C

but i don't know if we can restrict and place subspace topology on [0, 1] and assert the same

ah smart man

thank you

[queen prism]

true

[heady skiff]

is this theorem used a lot in practice?

to me it seems like these are all lemmas leading up to this theorem:

[vernal gate]

hi , I havent done topology yet, but I want to check if I have the intuitive idea.
compactness if when we have like a closed shape , and if we can cover in pieces every subsection of the shape, then there is a theorem I think that shows we can cover the entire shape , it doesnt matter if the shape is infinite.
is this correct? (in the realm of oversimplification)

[tacit basin]

For a space $(X, \mathcal{T})$ being $T_1$, i.e., for all $x, y\in X$, there exists a nbhd $U$ of $x$ and $V$ of $y$ st $x\notin V$ and $y\notin U$, is it equivalent to replace $U$ and $V$ with closed sets?

[gentle osprey]

Sara

[mighty hull]

Finite sets are to functions in general as compact sets are to continuous functions.

If you have a finite set S then functions from that set to the real numbers is bounded. They attain a max and min. All sequences in S have a convergent subsequence (which, for a finite set, means a constant subsequence). The image of f is finite.

Replace "function" with "continuous function" and replace "finite" with "compact".

If I have a collection of sets that cover a finite set then there's a finite sub-collection which also covers them.

If I have a collection of open sets that cover a compact set then there is a finite sub-collection which also covers them.

(You can think about functions to sets other than the reals, but you have to introduce slightly more general concepts.)

@vernal gate Compactness is a kind of "second-best alternative" to being finite.

A lot of reasoning that applies to finite sets also applies to compact sets.

[vernal gate]

I see

[mighty hull]

(Whereas most reasoning that applies to finite sets doesn't apply to infinite sets in general.)

[plush folio]

Yeah I think that's equivalent, since if U is an open set containing x but not y, then X \ U is a closed set containing y but not x

[tacit basin]

oh duh ty

my bad

[red yoke]

It's what characterizes a quotient so yea

"Continuous map out of X/~" = "Continuous map out of X such that f(x) = f(y) if x~y"

[paper wedge]

no

or nvm

[wise mist]

The algebraic numbers are dense in the reals

[red yoke]

Yes

[wise mist]

Q+ a^{1/n}

does it follows from the fact that Q is already dense

or am i missing something

oh yeah any open ball has a Q so it also has an algebraic

[wise mist]

what should be argument for algebraic numbers are countable?

[merry geode]

Do you have some ideas for that?

[wise mist]

Q is countable then on the top that we are creating n dimensional vector space so its a countable copies of the Q

n is the degree, and let's say more generally we can add k such root

maybe this is a handwavy argument

[merry geode]

Ye that needs to be more concrete

[wise mist]

does this cover all ov the algebraic stuff or am i still missing out something

k,n \in naturals

[merry geode]

There are algebraic numbers that cannot be expressed in terms of n-th roots, like many roots of degree 5 polynomials.

[wise mist]

haha, galois had me

any hint?
for the argument

[merry geode]

Maybe restrict the scope first - how many algebraic numbers are there from polynomial of degree 5?

[wise mist]

and for n degree poly the choices of coeff are from Q so its too countable
(so countable no of n degree poly)(n) = atmost possible root

[light herald]

i never saw it in practice, but i think that quotient map f is also characterized by Y having final topology induced by f
which seems a much nicer thing to check than all maps to all other spaces from Y

[light herald]

set of polynomials P with integer coeff is countable so there exists an injection f from P to N. For any finite polynomial the set of its roots is finite so we have injection f_2 from set of roots to N
now take set M of all pairs (p, r) ordered lexicographically by (f, f_2) where p \in P and r is a root of p and set g(p, r) = (f(p), f_2(r)). g is obviously an injection M -> N x N. Now take an injection h mapping every algebraic number to its first occurence in M. We then have chain of injections (algebraic numbers -> M -> N x N -> N) so we compose them and get an injection into N

sth like that

[abstract quail]

Is there any criterion to decide if a normed space is compact?

[hidden fiber]

how i prove that

Ac=C\A

[paper wedge]

what is your definition of hte boundary

is it the closure - the interior?

if so then unpack the definition

what does it mean to be a limit point?

what does it mean to NOT be an interior point?

[hidden fiber]

oh gotcha thx

[iron bolt]

this statement is much weaker than you might think btw

because X or whatever you call the space you're working in is itself a neighbourhood of a and contains both A^c and A, so you would just have to show that both A^c and A are nonempty

the more interesting statement is that when a is in the boundary of A ⊆ X, any neighbourhood U of a contains points from both A^c and A

not just one specific one

[forest pasture]

Hello

Can anyone give me an example of a topology on a set in which the intersection of an infinite number of the subsets isn't in it?

Writing this question I found the answer

The intersection of the neighborhoods of √2 within Q

Thank you very much to Sam for answering

(I'm Sam)

[cerulean oriole]

Most of them TBH.

[forest pasture]

Yeah lol

[opaque scroll]

The intersection of
(0, 1/n) for n natural numbers

[forest pasture]

That's 0

Wait ni

[opaque scroll]

0 is in none of them

[forest pasture]

That's ∅

That still belongs to every topology

[opaque scroll]

....?

[forest pasture]

A topology on X contains X and ∅

[opaque scroll]

Yeah, but the intersection of sets is always a set.

Are you asking for intersection of open sets to not be open?

[forest pasture]

No

Lol

[opaque scroll]

Like the intersection of (-1/n, 1/n) is just {0}

[forest pasture]

I asked for an infinite intersection of subsets of X which isn't in a topology of X

[opaque scroll]

That doesn't make sense

[forest pasture]

Wait

[opaque scroll]

If you intersect subsets of X you get a subset of X

[forest pasture]

Only finite intersections

Proof: R

[opaque scroll]

No

[forest pasture]

Ok

[opaque scroll]

Intersection just means the set of elements they have in common

[forest pasture]

Intersection of all neighborhoods of √2 on Q with radii 1/n for all n

[opaque scroll]

Is the empty set

[forest pasture]

n in N

[forest pasture]

Isn't it Q?

3

[forest pasture]

Why finiteness in 3

Munkers

[opaque scroll]

The definition of a topology concerns open sets.

The intersection of infinitely many open sets need not be open

[forest pasture]

But its supremum (assuming closed neighborhoods) is in Q

[opaque scroll]

For example the intersection of (-1/n, 1/n) is not open

[forest pasture]

Oh

Subcollection is open?

[forest pasture]

I know, I have finished Ch 2 of baby rudin

[opaque scroll]

T is the collection of open sets.

So a subcollection of T is some collection of open sets

[forest pasture]

I am the warcriminal behind this

[forest pasture]

Oh ok thank you very much

[snow gull]

Can a first-countable set (and not second-countable) be separable?

[wispy veldt]

yes

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

[snow gull]

Thx!

[heady skiff]

is the best way to show that S^n for n >= 1 is path connected to use induction

oh wait nvm i could just normalize in the denominator i think

never mind the denominator would be zero if we used straigh tline

for S^1

damn

would it just be the two hemispheres or something?

project the coordinates onto S^n and then use the fact that S^n is path connected maybe?

ah, two points on S^n + 1 always lie on a homeomorphic copy of S^n

so we can use the path connectedness there

idk how to prove this tho lol

[plush folio]

Can you use the result that S^n is a quotient of [0, 1]^n ?

[heady skiff]

but then i have to show that [0, 1]^n / ~ is path connecte dright

which might be easier, assuming that continuous images of path connected spaces are path connected?

[plush folio]

Yep, it's fairly straightforward to prove

[heady skiff]

oh okay that's actually a lifesaver

but then I guess i would have to show that S^n is the quotient of [0, 1]^n for all n lol

which is hopefully an easier problem

[plush folio]

Not sure actually, it might be tricky 🙃 S^1 is relatively easy with the right lemmas, since you can parametrize the circle by its angle, but it might be more difficult for S^n, n > 1

[wise mist]

A (metric) space X is said to be connected if the only
sets which are both open and closed in X are \emptyset and the full space X, when X is a metric space

im working with just this definition now

next i was asked to prove the statement of not connected disconnectedness ig)

to which i assume A to be a nonempty subset (open and closed)

[gentle osprey]

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

[wise mist]

such we can construct disconnectedness ?

then,

[gentle osprey]

nastasya

[wise mist]

next question is When is a finite subset of a metric space connected?

\emptyset, {x}

is this all correct ?

(i dont have any grasp on connectedness yet)

how do i get of the notion of metric to talk about connectedness ?

i think i have to solve more problems, but im asking anyways

[queen prism]

I don't really understand what your questions are supposed to be

but a topological space is connected if it cannot be written as the disjoint union of two open subsets

you do not need a metric to give a meaning to this

[wise mist]

im trying to get rid of metric, but i love this book too much, its all exercise

i need to suffer through munkres ig

[worldly magnet]

It is called an exclusion. Often, these are by definition with regard to sweeping nets.

[wise mist]

why is there a notion of pseudometric ?

what do we gain by inferring 0 distance

to create equivalance class for quotienting out ?

[wispy veldt]

there is certain structures where the natural topology on them is induced from a psuedometric rather than a metric, the equivalence class turns them to actual metric spaces (or even better, normed spaces) which are much nicer to work with topologically

[quick crane]

sometimes you don't care that points with 0 distance aren't the same, and you are just using the pseudometric to quantify distances in some sense.

[light herald]

every metric space with induced topology is hausdorff, which means we can find disjoint neighbourhoods of any two given points, so if you have more than one point you can just pick two of them and a small enough neighbourhoods so that they don't contain any other points

[worldly magnet]

I think you raise a decent point, namely, that the so called, "Real," numbers aren't actually a field - they are a projective system of quotient spaces under modular reduction. Inferring zero distance only serves to embed paradox in the notational language. In reality, there are at least five different "zero-like," symbols differentiated based upon whether or not existential quantification is observed, acting, but unobserved, unobservable, and each of these acts within a field of numeric energy - topological energy numbers hold that is a priori to their mapping to the real or complex plane.

[snow gull]

I need to prove cos(2Pit) is continuous , but I'm kinda lost on how to start. Can someone give me a starting tip ?

I know that if f is continuous then the preimage of an open set must be open. But chosin an open set ]a,b[ in R, how do I show it's pre-image is open?

[worldly magnet]

you should be able to do it with this book: https://s2pnd-matematika.fkip.unpatti.ac.id/wp-content/uploads/2019/03/Elementary-Diffrential-Aquation-and-Boundary-Value-Problem-Boyce-DiPrima.pdf

[tender halo]

do you need to actually show that it is continuous by definition?

anyway if you need to show directly, just use the formula for subtracting cosine from cosine

otherwise its just a composition of continuous functions

[snow gull]

No I don't think I need to show by definition, just that it's continuous

[prime elbow]

is a topology generated by dictionary order on R^2 and usual topology on R^2 same? in usual topology on R^2 which is product topology therefore its basis elements are product of two open intervals in R.

but now i am thinking about basis element in topology generated by dictionary order on R^2

[mighty hull]

No, they are not the same.

[prime elbow]

how can i show that? does i need to consider an open set and show that it is not in other

[mighty hull]

Yes, that would suffice.

It's easy. Pick almost any two points and think about what the interval between them looks like in the order topology.

[prime elbow]

( (1,2), (1,3) ) is open in order topology on R^2

[mighty hull]

That's true. Now expand the definition.

What does that look like in the plane?

[prime elbow]

i have to show it is not open in usual plane

[prime elbow]

just straigth line

[mighty hull]

Yes, a little line segment with its endpoints missing.

[prime elbow]

straight spelling mistake i don't know

[prime elbow]

endpoints means width?

as rectangles are basis elements in usual toplogy

but there is no width

[mighty hull]

The endpoints of an interval

(a,b) doesn't include its endpoints, a and b

[prime elbow]

sorry but can you explain more?

[mighty hull]

Do you know what the endpoints of an interval are? Like on the real line?

[prime elbow]

yes

[mighty hull]

Ok...that's all I'm saying.

A is (1,2) and B is (1,3)

The interval between those two points looks like a line segment between those two points, but excluding the those two points.

[prime elbow]

yes

but how it helps me to show it is not open in usual topology

[mighty hull]

Are you unsure about whether it's open or not? If you understand what it means to be open in a metric space then you should have an immediate idea of why it can't be open.

It might be a little annoying to prove the details, if you're required to, but the path should be clear.

[prime elbow]

yeah i have an idea of metric space of it

thank you

[mighty hull]

??

[prime elbow]

i got it

yeah usual topology is on R^2 is equivalent to max metric space on R^2, so let (1, 2.5) so if i take any open ball around it must contain some points, take open ball of radius r>0, then it must contain ((r+2)/2, 2.5) which is not in given set

[mighty hull]

I dunno what "max metric space" means, but the metric topology just is the usual topology on R^2.

In any case, yes, that's the idea. Pick any point on the little line segment. Any ball centered at that point contains points in R^2 not on the segment, so it's not open.

[prime elbow]

i mean d( (x1,y1), (x2, y2) ) = max { | x1-x2|, |y1-y2| }

it was my mistake

[mighty hull]

That's...well, alright. That feels like pulling out a bazooka to kill a fly.

It's true that all norms on R^2 generate the same topology, but if you can take that for granted then surely you can take for granted the fact that the standard Euclidean metric ball contains contains points not on the line segment.

[prime elbow]

yes thank you

[plush folio]

I think you mean all norms? For example the discrete metric doesn't induce the same topology

[mighty hull]

Sorry, yes.

[worldly magnet]

$$
\left{\left\langle\partial \theta \times \vec{r}{\infty}\right\rangle \cap\left\langle\partial \vec{x} \times \theta{\infty}\right\rangle\right} \rightarrow$$ $1$

[gentle osprey]

\blue {kevlat}

[queen prism]

r u ok

[paper wedge]

$$H_n(X) = H^{n-k}(X)$$

[gentle osprey]

Moamen

[worldly magnet]

$I(X, Y) = \lim_{n \to \infty} \sum_{p \in X_n \cap Y_n} i(p, X_n, Y_n) \rightarrow 1$

[gentle osprey]

\blue {kevlat}

[snow shoal]

Hi, I had the following problem: Prove that every n-manifold with boundary can be embeded into an n-manifold (without boundary). I have no idea about how I should approach this question. In the book's deffinition, n-manifolds are T2 and M2. So we can probably use some metrizibality theorem (but I don't know how it can help). So if someone could give me a slight hint or direction, I would appreciate it.

Also, idk if it is a commonly used deffinition, but manifold with boundary refers to topological spaces with manifold properties, but insted of them being locally R^n, we can also let some points be locally n-dimensional half spaces.

[paper wedge]

i think this is just whitneys embedding theorem

the weak version

[light herald]

whitney's embedding theorem is for 2n though isnt it

[paper wedge]

oh he wants to embed into an n manifold

sorry

[opaque scroll]

Would just gluing the boundary times [0, 1) to it work?

[snow gull]

In this example, can someone please explain why what I highlighted is true pls?

[mighty hull]

A set S is open in R if and only if every point in S is contained in an open ball entirely inside S

[tender halo]

zero is in B because its in da box

since the preimage is open, it contains some nbhd of 0

[snow gull]

That makes sense. Thx!

[queen prism]

can you post something more meaningful

[light herald]

uh, what's the context of you sending these

[snow shoal]

thanks for the answers, gonna try it.

[plush folio]

I think this guy is just a spammer

[plush folio]

<@&268886789983436800>

[gentle girder]

don't post AI stuff in the math channels

seems like metal timed you out, but this is for when you get back ^^

[snow gull]

Following up on this example, how can I know if the functions is or not continuous in the uniform topology?
And why can't I use the same argument of the coordinate functions being continuous for the box topology? Is the identity not continuous in the box topo?

[paper wedge]

f(x) =(x,x,x,x,x,...) from R to R^omega with the box topology is not continuous

[snow gull]

Yes I understand that. I just wanted to know where I'm thinking wrong. For a function to be continuous all it's component functions must be continuous, in this case the component functions are all f_n(t) = t . Aren't they all continuous in the box topology?

[paper wedge]

"For a function to be continuous all it's component functions must be continuous," this is a characteristic of the product topology

the example i gave you is a example that shows that this statement fails in the box topology

[snow gull]

Oh ok so it's only for the product topo

[paper wedge]

yes exactly

thats why its more "prefered"

cuz of this strong theorem

f is continuous iff each component is continuous

infact, it is exactly how the product topology is defined

that is, it has subbasis of inverse image of projections of open sets

ie, it is the weakest topology such that the projections are continuous

so that f is continuosu iff pi_i o f is continuous for all i

[snow gull]

Ok I understand now, for some reason I thought that theorem would apply to all topologies on products

[paper wedge]

it is fine to think so

i think it is exactly why munkres gives u these examples

it is natural to think what u thought

if it wasn't then it wouldn't be a discussion point

to even define the box topology and show ui

etc

it just turns out that it is just wrong

alot of things go wrong with different topologies

f(x) = x is not always continuous if you choose the right topologies

it is cool

[snow gull]

So now to show that the func is not continuous in the uniform topology, the only way is to show that indeed the pre-image of an open set in uni topo is open?
I kinda have no clue about how I would do this

[paper wedge]

you can show that the preimage of an open set need not be open

[snow gull]

is that easier in some way than proving it's open?

that is, If i want to know if the function is continuous

[paper wedge]

i am not following

[snow gull]

So the basic question is : is f continuous in the uniform topology?
I'm struggling to understande what the pre-image of V looks like when V is an open set in the uniform topo

Tbh I'm kinda struggling to understand what the pre-image looks like in all the topologies in product spaces

[paper wedge]

look at a subbasis element

or a basis element

to understand a topology you should see what the basis or the ssubbasis look like

in general

by defintion the product topology has subbasis pi_i^-1(U_i) where U_i is an open set in you ith factor

to go from a subbasis to a basis you consider finite intersections

you also know that pi_i^-1(U_i) is X_1 x X_2 x ... U_i x X_i+1 x ....

once you look at finite intersections of those you get that a basic open set would be a product of open basic open sets finitely many times

so for example in R omega

you would have products of open intervals and then R x R x R x ..

[snow gull]

yeah I know what the open sets look like, I don't really know what f^-1 (v) looks like.

So following ur example. An open set in product topo would be V = (U_1,U_2, ... , U_n , R , R , R , ...) . But what does f^-1 (V) look like?

I need to understand what the pre-images look like so I can decide if the functions are continuous or not

[paper wedge]

well thats up to you to decide

f^-1(something) is the set of x such that f(x) is in something

[queen prism]

initial and final topologies my plural beloved

[snow gull]

So in this case the pre images would be all the sets in R such that f(t) = (t,t,t,...) are in V = (U_1,U_2, ... , U_n , R , R , R , ...)

does that mean that a set A belongs to the pre-image of f(V) if for i=1,...,n A is in U_i ?

[paper wedge]

preimages ofw hat

of what

yes

all t such that f(t) are in V

this would be the preiamge of V

[paper wedge]

yes, f(A) to be contained in V means that every element of f(A) is in U_1 x U_2 x ... U_n x R x R x R ...

[snow gull]

So since here we are in R omega each U_i is an open set in R with usual topo. So now A is such that ]a,b[ is in ]a_i , b_i[ . It looks to me then that the pre-image of open sets are open sets, so it's continuous.

[paper wedge]

if its the box topology then those would be infinite ones

idk how u got that A

A is all elements x such that (x,x,x,x) is in U_1 x U_2 x U_3 ...

so u would want x is in U_1 and U_2 and U_3...

which u would want x is in the intersection of all U_is

now if this were the box topology then this would be an infinite intersection which need not be open

[snow gull]

yes I should have written in that way

[snow gull]

makes sense

[paper wedge]

and hence f(x) = (x,x,x,..) is not continuous

thats what munkres is trying to say

thats it

i want u to

as an exercise for yourself

unfold the defintiions and how would a basic open set in the product topology look like

good luck

[snow gull]

Thx for the help!

[paper wedge]

np man

[plain gale]

Let $U$ be a Subset of Y, $f: X \to Y$ a continuous function and $U_i$ a open cover of U. Is $f^{-1}[U_i]$ then a open cover of $f^{-1}[U])$?

[gentle osprey]

neuron

[plain gale]

I would proof it as follows: if we have $x \in f^{-1}[U])$ then $f(x) \in U$ so for some i $f(x) \in U_i$ and thereby $x \in f^{-1}[U_i])$

[gentle osprey]

neuron

[plain gale]

is that proof correct?

[queen prism]

that tells you it’s a cover

is it an open cover?

[plain gale]

well because f is continous all the $f^{-1}[U_i]$ are open so its an open cover

[gentle osprey]

neuron

[queen prism]

yea that’s all

ur good

[plain gale]

no u are

[plain gale]

thanks

[prime elbow]

Let X = (0,1/2) and define topology on X as subspace topology of R.

Let f:X -> R by x -> x^2, then for all x≠y in X d(f(x), f(y) ) < d(x,y) and it is not contraction mapping.

Because if it is then for all x≠y, d(f(x), f(y)) ≤ cd(x,y) for some c in (0,1)

We have, |x+y| ≤ c, for all x,y in X but it is not true.

Is it correct?

[light herald]

yea

[prime elbow]

Thank you

[prime elbow]

Is there any other way to prove Uyrsohn lemma than 3 long page proof?

[paper wedge]

no

joking

[red yoke]

Is that T4 implies disjoint closed subsets are separated by a real function

[prime elbow]

what defination are you using for T4, yes normal space implies disjoint non empty sets are separated by continuous function

should i go through the long proof?

[red yoke]

The idea should be simple though

[prime elbow]

yes i got the idea little bit

[red yoke]

I don't know if it deserves to take up 3 pages

[prime elbow]

If X is not compact then there is a sequence with no convergent subsequence and {x_n | n in N} is infinite, but how can i say it has no limit point?

[tender halo]

are you talking about metric spaces?

because what you are saying is not true, there are non compact spaces that are countably compact

for example omega 1

[prime elbow]

Yes metric space

[tender halo]

having a converging subsequence is the same as having a limit point (in first countable spaces, which metic spaces are)

[sonic crane]

Why is the circle not homeomorphic to the annulus?

Im trying to understand what the crucial issue/difference is

[queen prism]

you mean... a circle in R^2 versus a solid ring in R^2?

[prime elbow]

Yes but what if there is a convergent sequence by rearranging the elements of {x_n} ?

[tender halo]

any rearrangement for a sequence has the same limit points

[sonic crane]

The size of the set of homotopy equivalence classes between some {a} and a top space Y would be like the number of path-connected components of Y, right?

[paper wedge]

what

number of path connected components of Y probably would be the rank of the 0th sinuglar homology group

[sonic crane]

Idk, im just learning this stuff rn

We had an example of “a path is a homotopy” where he did some h: {a} x [0,1] -> Y

where h(a,0) = f(a) = c1
h(a,1) = g(a) = c2

Idk just a weird way of writing a path in Y

So i mean i was thinking then shouldnt f is homotopic to g be when there is a path in Y from f(a) = c1 to g(a) = c2?

So then shouldnt the equivalence classes be the different path connected components of Y?

[paper wedge]

im not following im sorry

a path is a continuousu function from the interval to the space

with such and such endpoiunts

h is not that

you can have a homotopy of paths

a path is a "homotopy"

homotopy betwen what

im also sleepy so i will let someone else help u out

i gtg

gl

[paper wedge]

yeah a homotopy between two maps is more like a

family of paths

[sonic crane]

A homotopy between maps in that example is a path in Y

One path

[snow gull]

I need some help understanding what the elements of a product look like.
If we interpret the set of functions f : X -> Y as the following product.

Then each x in the product is something of the form x = (f(a_1),f(a_2),...) (so kinda the images for all elements in the domain X if I'm not mistaken)
Now when we take a sequence of functions does x_n = (f_n(a_1),f(a_2),...) ?(again the image for all elements in the domain X but this time the functions is f_n)

To try to simplify I think I could choose f: N -> R and now the elements in the product would look like this:
x = (f(1),f(2),...)
x_n = (f_n(1),f(2),...)

Is this really what the elements look like or am I completely misunderstanding everything?

[queen prism]

the elements of a cartesian product are functions, more or less
an element of R x R x R looks like (x, y, z)
but you can interpret this as a function on {1, 2, 3} sending 1 to x, 2 to y, and 3 to z
similarly, an element of R^N is a sequence of real numbers
and we actually define sequences to be functions N -> R assigning a real number to each integer index

[snow gull]

Yes and how would an element from a sequence of functions look like, that the thing I can't get my head around.

I think the x examples I gave make sense with what you wrote. What about the x_n's

[queen prism]

so if your product is P = 🥧_{x in X} Y
that means an element of P is a particular function f : X -> (union of a family of copies of Y, one for each x in X)
but the codomain in this case is actually just equal to Y
hence every element of P is an ordinary function f : X -> Y, or in other words an element of Y^X

now not every element of a cartesian product looks like a sequence
you have to give up a lot of ideas you take for granted, such as your elements (functions) having an order to their values or having discrete, clearly separated values

[snow gull]

So the only thing I really know is that the f_n in a sequence of functions is an element in Pi_{x in X} Y

[queen prism]

you know the difference between a function and a function evaluated on an input right?

oh hey I'm blue again and it actually matches my avatar this time

[snow gull]

The actual problem I was trying to solve is a bit more concrete but I wanted to make sure I understand the basics first.
So given the sequence of functions in the image I want to calculate the limit in the product topology. So as n -> infty the function goes to 0 so I guess it sounds like a good ideia to try and prove it converges to 0.
So I guess now I need to show that for any neighbourhood U of 0 there exists and N such that n > N => f_n is in U. Is this correcT?

[snow gull]

Since I don't really know how to answer your question, I'd probably say no xD

[queen prism]

alright well
a function is a thing that takes inputs in a domain and sends them to outputs in a codomain
an example of this would be f : R -> R defined by f(x) = x^2, sending x in R to x^2 in R
now we think of f as the function, not f(x)
because f(x) implies we've chosen a particular input x, say, x = 5, and we're calculating the value f(x)
working with just f removes this bias

so let's say C(R) is a function space and you consider the product pi_{n in N} C(R)
each element in this product is a mapping (i.e. a function) assigning to each integer n a continuous function f_n : R -> R

[snow gull]

That makes sense

[rare saffron]

Taking a point-set topology course rn, and it seems like a lot of topologies you learn about are really only useful for counter-examples (lower limit topology, finite complement topology, K-topology, etc...). Are most topologies like this, or do students only learn about these niche topologies first b4 moving on to more useful topologies?

[opaque scroll]

Well, if you want to illustrate a counter example you usually want to reach for an example that is easy to construct / demonstrate. So then you might end up with some somewhat arbitrary things.

But I'm not sure I agree that most topologys one usually learns about in such a course are just for counter examples.

Like I'm hoping you learned about, the standard topology on R / metric spaces, the product topology, subspace topology, quotient spaces, discrete and indiscrete topology.

It also wouldn't be unusual to learn about the weak topology or the order topology.

[plush folio]

I think most of the topologies that are useful/interesting to an undergrad are actually metric spaces, while the useful non-metrizable topologies like the Zariski topology requires more background knowledge. So some parts of topology may feel kind of useless until you encounter it later on (and of course some mathematicians never need general topology beyond metric spaces)

[quick bough]

if i have a pushout square, where the left arrow is the inclusion of a strong neighborhood deformation retract, is the right one also one?

[prime elbow]

I have to prove that every open subspace and closed subspace of a sequential space is sequential space.

Now if U is closed set in X, where X is sequential space.

Let V\subset U which is sequentially closed in U, then it is also sequentially closed in X, because there is no sequence in V which convergent in X\U, because X\U is open so it does not intersect with V.

Now V = U cap V which is closed in U.

But I have a problem when U is open.

[prime elbow]

What does it mean by the discrete subgroup of R in the topological sense? Does it mean if we treat that subgroup as subspace of R then it is discrete space

[red yoke]

Yes

[quick bough]

does anyone know🥲

[wanton heart]

I was hoping someone could check my proof, especially for the second part because I am not entirely sure what it means by "or they are not comparable"..

[dire dove]

"not comparable" means neither is a subset of the other

(which you correctly showed cannot be the case)

[wanton heart]

I see, thank you

[prime elbow]

I know an open continuous mapping of a Baire space is a Baire space.

Can the above result help me to show an open subspace of the Baire space is Baire space?

[sonic crane]

For a set of subsets to be a basis for a topology, do we require that the set is closed under intersection?

I thought that was a requirement to be a basis for a topology. But my topology prof docked points off my hw and he said that “such a property makes it a basis, but not all bases satisfy this”

I think his definition of basis is that: a set of subsets B of X is a basis if, for any open set U in X, U = union of elements in B

[plush folio]

I think closure under intersection is slightly too strong, you just need that for any two sets B1 and B2 in the basis there exists a basis set B3 such that B3 \subseteq B1 \cap B2. So you need a set in the intersection, not necessarily the intersection itself

[fringe thorn]

it might also be useful to know that there is a difference between a basis for a topology on X and a basis for the topology on X

[plush folio]

For example, {(a, b) | a, b in Q} U {(0, pi)} is a basis for the topology on R, but the intersection (1, 4) \cap (0, pi) = (1, pi) is not in the basis

[fringe thorn]

the former is defined for sets not yet endowed with a topology, while the latter is for sets with topologies

the former is the one you’re talking about in your first message, and the latter is the one referred to in the second message (the prof’s def)

[plush folio]

hmm, is there really a difference? Both follow the same rules, namely to generate the topology under unions, right?

[queen prism]

one generates a new topology and one generates a specific, given topology

[plush folio]

yep, that is my understanding too

[sonic crane]

Yeah, right, i actually remember that. That was probably the property i was actually thinking of

Does that follow from the definition i gave for a basis for a topology?

For The question on my homework, we had the topology given to us, by defining it as the topology generated by a given subbasis. The question then asked why isnt the subbasis a basis?

I thought it was because the subbasis elements arent closed under intersection, which was not the correct reason i guess

[junior granite]

Hello! I am looking for assistance with the structure of a proof I need to write. The problem is as follows:

Let X be a Hausdorff space in which every subspace is compact. Prove that X is finite.

I am considering a proof by contradiction, in which I assume X is infinite, and use some property of the closure of a set containing infinitely many points from X to show that it, in fact, cannot be infinite. But I'm not sure if the logic is right, or what properties I can assume of the set to make such a conclusion.

Thank you in advance for any help!

The definition of compact is what is troubling me, so perhaps I should've started there.

[plush folio]

I think so, you want the basis to generate a topology under unions, and for the generated collection of sets to be a topology you need it to be closed under finite intersections. So if every point in the intersection A \cap B is contained in a basis set (which itself is contained in A \cap B), then the union of those basis sets ensures that A \cap B is also open

[plush folio]

Yeah, I guess it's not quite correct, since a basis also doesn't have to be closed under intersection

[steep kite]

Do you need to prove it like that?, if not I think is easier to prove it directly. In a Hausdorff space every compact subspace is closed, use this to prove that X is discrete, then it follows that X is finite.

[junior granite]

i don't need to, i just like making things difficult for myself i guess lol

I will think about it more using your hint

[steep kite]

Everyone needs that time to time to feel alive, don’t worry haha

Good luck

[gentle osprey]

okeyokay

[cerulean oriole]

Given a locally compact Hausdorff topological group which is a directed union (as a set) of a family of closed subgroups, does it necessarily have the final topology with respect to the inclusions of those subgroups?

[junior granite]

Let $X$ be a Hausdorff space in which every subspace is compact. Prove that $X$ is finite.
\begin{proof}
Let $X$ be a Hausdorff space in which every subspace is compact. We want to show that $X$ is finite. Since every subspace $Y$ of $X$ is compact, $Y$ is closed in $X$. The complement of any given $Y$ must be open, and will be contained in $X$. This implies $X$ has the discrete topology, thus making $X$ finite.
\end{proof}

[gentle osprey]

Vic

[junior granite]

Would this proof suffice for the problem?

i am missing something but what

[cerulean oriole]

Yes (and I believe it was mentioned above).

[cerulean oriole]

I don't think so.

[plush folio]

There are some details missing, but what matters is whether you feel those details are trivial or not. Like, do you know why X having the discrete topology implies that X must be finite?

[junior granite]

is it because the discrete topology is the collection of subsets of X, of which there are finitely many?

[plush folio]

Not quite. The important assumption here is that X is compact

But yeah, the discrete topology consists of all subsets of X, but that is not enough to show that X is finite

I recommend trying to prove that if X is compact with the discrete topology then X is finite

[junior granite]

OH the finite subcover

[prime elbow]

let $A$ and $B$ be disjoint compact sets, now let fix $a\in A$ since $X$ is Hausdroff space. Now let $b\in B$ then there exist open set $b\in V_{a,b}$ and $a\in U_{a,b}$ and $V_{a,b} \cap U_{a,b} =\emptyset$.

Now $V\subset \bigcup_{b\in B} V_{a,b}$, since $B$ is compact so $V\subset \bigcup_{i=1}^{n_a} V_{a,b_i}$

take $U_{a}= \bigcap_{i=1}^{n_a} U_{a,b_i}$ containing $a$ and $U$ is open set because it is finite intersection of open sets.

Now take $A\subset\bigcup_{a\in A} U_{a}$.
Since $A$ is compact so $A\subset\bigcup_{j=1}^{m} U_{a_j}$

now again take $\bigcap_{k=1}^{m} (\bigcup_{i=1}^{n_{a_{k}}} V_{a_k,b_i})$, it is open set because finite intersection of open sets and each $\bigcup_{i=1}^{n_a_k} V_{a_k,b_i}$ contains $B$.

So let $U = \bigcup_{j=1}^{m} U_{a_j}$ and $V= \bigcap_{k=1}^{m} (\bigcup_{i=1}^{n_{a_{k}}} V_{a_k,b_i})$

we will show that $U \cap V=\emptyset$.

I think the idea is correct?

[gentle osprey]

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

[tender halo]

are you trying stuff in as much \cap and \cup's as you can or what

[prime elbow]

sorry i don't get it

[knotty vine]

Can anything be said about the number of connected components of the stone--cech compactification of a discrete (infinite) space?

[grave solstice]

Is that $\Spec(\prod_\kappa\mathbb F_2)$?

[gentle osprey]

croqueta3385

[grave solstice]

I think it is the case, but don't quote me here. Maybe you can use this then

maybe this is unnecessarily complicated tho, maybe there is a straightforward way to do it. But now at least it sounds reasonable.

[knotty vine]

Sorry, I'm quite oblivious when it comes to commutative algebra. I remember the spectrum usually not being Hausdorff, while the Stone--Cech compactification is always Hausdorff

[cerulean oriole]

It's totally disconnected - for any two distinct ultrafilters U, V on a set X, there is an A ⊆ X which is one but not the other, and then {ultrafilters containing A}, {ultrafilters not containing A} are disjoint open sets with union bX one of which contains U and the other V.

(More generally, any topological space with a basis of clopen sets is totally disconnected.)

[cerulean oriole]

I'm not sure if this can easily be shown from the universal property though.

[grave solstice]

That Spec is Hausdorff

[cerulean oriole]

Yep.

[knotty vine]

Cool!

[grave solstice]

so, every prime in that ring is maximal

and thus your answer is the size of Spec prod_kappa F_2. Which yeah is what Raghuram said

[knotty vine]

Nice, thanks!

[tender halo]

well its extremally disconnected

each point is its own connected component

the cardinality of \beta D(\kappa) is 2^2^kappa

[knotty vine]

"extremally" very strange word

[tender halo]

it really is

i was reading it as extremely for like a year

\beta D(\kappa) is just a weird space in general

[knotty vine]

What is \D here?

oh discrete

[tender halo]

discrete space of cardinality \kappa

for example every infinite closed subspace of \beta N contains a copy of \beta N

[grave solstice]

btw, do we know the size of beta(kappa)?

[low flame]

I defined f: A -> R : x -> p(x, X-U)

then since this is continous it gets its minimum and distances are positive.

[knotty vine]

Yes, it's 2^2^kappa if kappa is infinite, and kappa if finite

[tender halo]

its pretty easy to prove too

well. i guess easy-ish

by Hewitt-Marczewski-Pondiczery, the space [0; 1]^2^\kappa has a dense subset A of cardinality \kappa. take an arbitrary surjective function f from D(\kappa) to A. then the function f: D(\kappa) -> [0; 1]^2^\kappa has a continuous extension to g: \beta D(\kappa) -> [0; 1]^2^\kappa (g is necessarily surjective), so cardinality of \beta D(\kappa) is at least | [0; 1]^2^\kappa | = 2^2^\kappa

and it cant be more than 2^2^\kappa because the density = \kappa is too small to allow more points

its just corollary of Hausdorffness or regularity or something

[sonic crane]

If f is continuous and h is continuous, is a composition like h(f(x),t) continuous?

Assuming all the maps make sense

[sonic crane]

Ty!

[sonic crane]

How do you guys think of continuity? What is continuity?

Could you think of it simply as, the right condition to impose on functions so that topological properties are preserved ?

Smth like that?

[rare saffron]

I think that at some point u gotta abandon any traditional notions of continuity and just get down to the abstract definitions.

Especially as you look at rlly abstract topologies that aren't equivalent to the standard topology

[tardy carbon]

I think of it as you can get arbitrarily close to the functions value at $x_0$ by evaluating it at points near $x_0$

[gentle osprey]

L

[tardy carbon]

so it's a quantitative thing at the core

[rare saffron]

The notion of closeness becomes pretty weird in other topologies though. Take the finite complement topology for example.

In fact, if your space isn't metrizable, you don't even really have a notion of closeness

[tardy carbon]

The only spaces I've had to deal with (other than the space of distributions) were ones whose topology was defined by seminorms or by a metric.

[rare saffron]

Mmmm I getcha

[tardy carbon]

Even the product topology, although not always metrizable, is determined by seminorms (assuming the factors are all normed vector spaces).

[rare saffron]

I thought the product topology was metrizable? You can just take the uniform metric right? Unless you're including the box topology

Or I guess if you have a product of non-metrizable spaces

[cerulean oriole]

The uniform metric gives uniform convergence, not pointwise convergence (which is what the product topology gives) unless it's a finite product.

It is true that a countable product of metric spaces is metrisable (one valid metric looks like ∑_n 2^{-n} d_n/(1+d_n)), but an uncountable product (assuming each factor has more than one point) is not (it's not even first-countable).

[tender halo]

definition-wise, i think the most illustrative one is f(cl(A)) \subseteq cl(f(A))

[gritty widget]

here i have to encounter the problem, Suppose that (X, d) and (Y, p) are metric spaces, that f_n: X-> Y is continuous for each n, and that ( f_n ) converges pointwise to f on X. If there exists a sequence (x_n) in X such that x_n -> x in X but f_n(x_n) does not converges to f(x), show that ( f_n ) does not converge uniformly to f on X.

here, my doubt is they used f_n(x_n) does not converge to f(x), is that mean f_1(x1), f_2(x2),... this sequence does not converge to f(x)?

and i think we can take a help of contradiction, if it is uniformly continuous then f is continuous, but i have a doubt in f_n(x_n) notation

[tender halo]

well if they did converge uniformly, then the sequence would converge to f(x)

[gritty widget]

yes that i have to prove

[tender halo]

just take n where d(f, f_n) < \epsilon and look at f_n

all the functions after the nth are different from f_n by no more than 2\epsilon

[gritty widget]

yes triangle inequality

[tender halo]

that should be enough for what you claiming, you just need to move some epsilons around

[gritty widget]

how?

[tender halo]

im busy atm, if you still want some help in a couple of hours and ill write it out explicitly

[gritty widget]

okay fine

actually i searched on mse but it don't use d( f_n, f_m ) < 2\epsilon, now i am interested in your way

[gentle osprey]

RealTek

[low flame]

so im not really sure how to show this is surjective.

with B) theres gotta be some connection to compactness and every cover having a finite subcover.

I dont know what a covering map is honestly, but its prob related to the 3 big names of the question.. I can look more into it myself.

whats an approach to A?

[low flame]

Y-f(X) is empty since every point of f(x) maps to Y, thus f(X) = Y?

since Y is connected and the only seperation of Y is the entire space and the empty set?

because if this wasnt so then its a contradiction since Y is compact, it means its closed and its compliment would be open, not clopen. which would violate the connectedness

anyways ping me if you can help

[trim spade]

f is continuous so f(X) is closed since Y is Hausdorff and X is compact. Try and also show it is open. It then follows since Y is connected

[low flame]

is what what i said above?

[trim spade]

I’m not entirely certain what you were getting at

[low flame]

what im saying is this. This is compact so Y-F(X) is gonna be open if its non empty. If that occurs then theres a seperation of Y. Which contradicts connectivity. Thus Y -F(X) has to be empty, so F(X) = Y

[trim spade]

yeah as long as you show f(X) is clopen that’s how it would go

[low flame]

well if a set is connected, which X is, then the only clopen set is X and the empty set. its a requirment to even be connected.

[gritty widget]

Let A and B closed subsets of R then I have to find an example such that A + B is not closed.

Idea: Both must have to be unbounded because if one of them is bounded say A then A is compact then A + B is closed.

I have been trying for the last two hours, I tried different examples such that N and {n+ 1/k | n,k in N}

But I don't get a solution, I am not interested in the solution.

I am interested in the thinking process.

And also I am thinking, are there any closed sets A and B of R such that R\( A + B ) is non-empty finite because R\( A + B ) is open and if it is non-empty finite then it cannot be closed because R is connected

[tender halo]

what does A + B mean

symmetric difference?

[queen prism]

looks like minkowski sum

[gritty widget]

Sorry, A + B = {a + b, a in A and b in B }

[gritty widget]

Let (A_k) be a sequential of invertible matrices in M(n, R) converging to an A in M(n, R). Is it necessary that A is invertible?

I think no

Take n = 2, A_k = [ 1/k 0, 0 1/k ] then det(A_k) = 1/k^2 and A_k converges to [0 0, 0 0 ], right?

[paper wedge]

yeah

or note that the set of invertible matrices is the preimage of the determinant map of <0 and >0

nvm

[gritty widget]

Yes but that's why I took the matrices which determinant goes to 0

[paper wedge]

🤛

[gritty widget]

I forget to mention a metric space on M(n, R) but it is equivalent to Euclidean metric on R^(n^2)

[heady skiff]

would S^2 be the quotient of I x I / {(1, 1), (0, 1), (1, 0), (0, 0)} then

and in general would we just identify all points which have each coordinate either 0 or 1

[swift sluice]

Not that it matters but I just tried this and you end up having to glue the sides in the process

[heady skiff]

lmao i just tried this before ur message and convinced myself it was 😂

oops

THERE'S GOTTA BE AN EASIER WAY TO SHOW THAT S^N IS PATH CONNECTED

yeah but i'm having trouble seeing S^n as the union of two path-connected spaces with non-empty intersection

but isn't that like the same problem, bc then I have to show hemispheres are path connected

oh

wait holup

the top hemisphere would be just the points in S^n whose coordinates are all nonnegative right

i think they would be homeomorphic each to I^n+1, since we could just take a point and normalize it

nvm I^n

intuitively we can bend the top hemisphere of S^1 into a straight line

and the top hemisphere of S^n into a square

just flatten them both

nvm lol

ah u right

i'll try that, thank you

didn't even think of that 😂

yeah i'll try it if i can show this

this problem has been kicking my ass

[wise mist]

$$f: [0,1] \to \Bbb{R}^2$$
[0,1] is connected, so $\forall$ continuous $f, f([0,1])$ is connected.

\ \

if we are trying to connect two separated branches($x<0, x>0$) of any hyperbola of the form $x^2 -y^2 = r$

\ \

$f([0,1])$ will have to intersect $y$ aixs (IVT)
but the hyperbola doesn't contain $x=0$ so no such $f$ can exists. so....

is this enough to show the thingy is not path connected ?

[gentle osprey]

nastasya

[real gazelle]

I went to the compact connected 1-manifold store and all they sold was circles!🤬

[gritty widget]

Consider B[0, 1], the closed unit ball in C[0, 1] under the sup norm. Show that it is not compact.

Because there is a sequence which has no convergent subsequence, f_n(x) = nx, works right?

[tough hamlet]

that isn't in the unit ball

[gritty widget]

Oh right

I misread the question

I thought, the question was about to show C[0,1] is not a compact space

[gritty widget]

f_n(x) = nx/ (1 + n^2 x^2 ) , works? It is pointwise convergent but not uniformly therefore it is not convergent in C[0, 1]. But yes we still need to show it has no convergent subsequence

[tough hamlet]

yes

[gritty widget]

I think for the same reason we pick any subsequence it is still pointwise convergent to 0 but not uniformly convergent

[tough hamlet]

yes

[gritty widget]

Thank you

[unreal stratus]

Intuition

Any two points in R^n \ 0 can be joined by a path avoiding 0

Now divide that path by its norm

[rare tapir]

Hey is anyone familiar with Perron Frobenius theorem and eigen values?

[river granite]

I know some about it, post your question in #linear-algebra though

assuming it's about applying it, if it's about a proof of it using a fixed point theorem it might be fine here

[heady skiff]

bro uses \mathbb{N} for N

[queen prism]

\mathbb{N} supremacy

[unreal stratus]

My brain hurts from it lol

[heady skiff]

cursed notation fr

[queen prism]

hmm?

seems like overreacting to me

[sonic crane]

Bro u cant be fr

[cerulean oriole]

Another way: any two points lie in a 2-dimensional linear subspace (it's just their span unless they're antipodal), homeomorphic to ℝ^2 such that the intersection of S^n with it is the unit circle (under the homeomorphism, if you like). So the problem is reduced to the case of S^1.

[cerulean oriole]

That answer has been given...

[keen seal]

ah didn't see

mb

[vague hinge]

Hey, could anyone help me with proving that $d(x,y) = \sum_{n=1}^{\infty} \frac{|x_n-y_n|}{2^n}$ is a metric on Hilbert’s cube? Ive been stuck on this, I have some kind of solution but I dont understand it at all.

[gentle osprey]

imobump

[paper wedge]

the hilberts cube is like product [0,1] right?

[vague hinge]

yes

[paper wedge]

so your dealing with sequences that have norm<=1

sequences that have components*

[vague hinge]

correct

[paper wedge]

so im assuming your having trouble with the triangle inequality

but i want u to observe that |x_n-y_n| <= 2

literally

[vague hinge]

ahh damn it ive done it again

[paper wedge]

done what

[vague hinge]

i asked the wrong question

[paper wedge]

u wanna show its complete?

[vague hinge]

i actually meant to ask how to prove that topology generated by this metric is equal to product topology on Hilberts cube

[paper wedge]

well to do this its kinda systematic but just has details

you know if you have T and T' topologies on a space

T' is finer than T iff for every basis element of T, call it B, and for every point in that B you can find B', basis in T'' such that B' is a subset of B

so you do it bothwise to show that T = T'

this is lemma 13.2 in munkres iirc

and just can be easy to prove

the details are going to be mainly which epislon to choose

so u should know what a basis element looks like in both cases ( the metric top and the product top )

work it out

[vague hinge]

i think ive tried doing something similar

I got a set U from the product space basis and tried to show that for each x in U you can fit in an open ball

But its extremely confusing because I cant imagine the balls

[paper wedge]

yes you can do that

the only thing you need to do is just

workout the epislon

you need

(the radius of that ball)

so that it fits inside

you know that this set U in the product space is equal to only finitely many open sets in [0,1] and then [0,1] x [0,1] x ...

these are open sets and hence you can fit an interval intersec [0,1] in there

ie you can fit an epislon interval

[queen prism]

hilbert cube is [0, 1] x [0, 1/2] x [0, 1/3] x ... tho?

[paper wedge]

take the minmimum of these epislons

[paper wedge]

yeah that

[queen prism]

ok ig it doesn't matter

[paper wedge]

lmao

[vague hinge]

oh teacher lied to me then

[paper wedge]

no ur rihgt

[vague hinge]

but I meant [0,1]^inf

[paper wedge]

its whatever

[paper wedge]

now which epislon should you take?

intuition just says you take the minimum of these epislons to that your guranteed ur in every open set

[vague hinge]

its so confusing

I dont usually have problems with understanding those 'complex' topics

But Im lost this time. At least now I have some direction to go

So thanks a lot

So if $\mathcal{U} = [0,1] \times [0,1] \times \mathcal{U}_1 \times ... \times \mathcal{U}_k \times [0, 1] \times ...$ can I assume that any 'part' of the ball will fit inside the $[0, 1]$'s? Since $y \in [0, 1]^\mathbb{N}$

[gentle osprey]

imobump

[vague hinge]

I dont really know how to phrase what I mean, but basically I think that the ball consists of intervals/points/whatevers in each of the sets of which the product is made of

So I wanted to only focus on the U_1, ... U_k

[paper wedge]

yes

[rapid olive]

Why is a topological space defined as it is?

[queen prism]

because it simplifies all the proofs while still generalizing the properties of a metric space well enough

can you make your question more precise?

[rapid olive]

as in, why this generalisation as opposed to any other

why require e.g. arbitrary unions instead of just countable unions

[paper wedge]

thats a hard questiona

it took a while to get down to how to define a topology

there was topology b4 the definition of a topology

like for example there was a time where a topology meant a hausdorff space

[cloud kindle]

The sort of most correct but not very satisfying answer is that it turns out those are the properties that work

[paper wedge]

and then kuratwoski did something too aswell

yeah basically think of it as you are doing a problem that involvles finding a counterexample

u start with something intuitive, it doesn't work

you keep cooking untill you get the finished product

and by finished product they meant it in the sense Nope said

you wanted to generalize ideas from analysis

they had a whole classification of surfaces b4 knowing what a top space is

analysis situs was 1898