the basis of the topology have all elements having cardinality $|\Bbb R|$ right

I wanted to use induction but that doesn't work

induction only gets me as far as elements of $\omega$

not $\omega$ itself

Can you prove without CH that card. of Cantor set is N_1? I guess not... I can show it's the same size as R.

cantor set can biject to [0,1[

by ternary -> binary argument

Every uncountable closed has cardinality $\mathfrak c$. This is (consequence of) Cantor-Bendixson theorem.

then [0,1[ easily bijects to R

But I think the same is true for Borel sets. (I might have misrememberd this.)

I dunno this stuff. Well, I'm off

You can have a closed set with cardinality $\aleph_0$, for example the convergent sequence: $\{0\} \cup \{1/n; n=1,2,\dots\}$.

@MartinSleziak oh, I see

@DHMO Consider the open cover [0,2) formed by a countable union of [0,k) with k in the sequence {2-1/n} from n=1/2,2/2,3/2,.... Then no matter what finite union of subsets [0,b) in {[0,k)} with b arbitrarily close to 0, [0,1] will not be a subset of this union, hence there are no finite subcover for this particular open cover, thus [0,1] is not compact in this topology.

Hi all... If $\lambda < \frac{a}{b}$ .. then can i say this $\lambda < \frac{a}{b} <\frac{a+c}{b+c} ;;$ a,b,c are positive real numbers ??

@BAYMAX no

@Secret {[0,1.5)} is a finite subcover

speaking of shurikens, I recall playing around with the graphs of $x^s+y^s=1$

there is an $0<s<\frac{1}{2}$ that has it very close to (but of course distinct from) a circle, around 0.34 or something

@DHMO All the variables are $x$. Not $u$..you made a copying error I guess. :) It is single variable. And yet I don't think it is solvable :P

@anonymous bad handwriting.

if $\frac{a}{b} = \frac{3}{4}$ , then any counterexample ??

@DHMO Hehe :D

@DHMO

@BAYMAX no

that is any example for which above is not working ??@DHMO

@BAYMAX 5/4

Hello, please how to prove thae that $D=\{x\in E, d(x,A)<d(x,B)\}$ is open, where $E$ is a metric space and $A,B\subset E$

@Vrouvrou its complement is closed

thanks @DHMO

so we can do that for $\frac{a}{b} < 1$

yes

@DHMO without using that it's complement is closed

@Vrouvrou why?

@Vrouvrou use the definition of open set under this particularly topology

and the triangle inequality

Let $d \in D$.

let it be closer to $A$ WLOG

let $a = d(d,A)$

let $s=\frac a 2$

consider $Y$, the open-ball centered at $d$ with radius $s$

consider $y \in Y$

$d(d,y) < d(d,s) < d(d,A)$

oh shit

@Vrouvrou sorry, I got the wrong intuition

@satyatech I'm pretty sure you made a copying error in this question (math.stackexchange.com/questions/2161846/…). Please upload the picture of the problem directly from your textbook! The integral isn't solvable (atleast at your level).

(May I know the name of the textbook btw?)

@Vrouvrou here I start again

Let $d \in D$

let $a = d(d,A)$

let $s = \frac a 2$

consider $Y$, the open-ball centered at $d$ with radius $s$

we need to prove that $Y \subseteq D$

consider $y \in Y$

$d(d,y) < s$ by definition of $Y$

$s < a$ by definition of $s$

$a = d(d,A)$

we need to prove that $d(y,A) < d(y,B)$

@DHMO ok how about: $[0,3/2) = [0,1/2) \bigcup_i [a_i,a_{i-1}) \cup [1,3/2)$ where $a_0=1$ and $a_i=1/2^i$. Now if any finite union of this collection is picked, part of the [0,1] will not be covered hence there is no finite subcover and [0,1] is not compact in the lower limit topology.

we have $d(d,A) + d(d,y) > d(y,A)$

NB I found it hard to came up with a cover involving unions as disjoint as possible

@Secret what is the second interval in your union?

@Vrouvrou sorry, I start again

ok

Let $d \in D$ and $f = d(d,B) - d(d,A)$. $f > 0$ by definition.

consider $Y$ an open-ball centered at $d$ with radius $\frac f 2$

we need to prove that $Y \subseteq D$

let $y \in Y$

we need to prove that $d(y,B) > d(y,A)$

I will have to leave now. I will just mention that I copied the discussion about cardinality of closed and Borel subsets of $\mathbb R$ to set theory chat room. And I added also a bit to that.

@DHMO Something like this: $... \cup [1/2^n,1/2^{n-1}) \cup [1/2^{n-1},1/2^{n-2}) \cup ... \cup [1/4,1/2) \cup [1/2,1)$

we know that $d(y,d) < f$ by definition

It seems that several parallel discussions are going on in in this chat room at the moment, so it seemed like a good idea to take it elsewhere and not add another one.

@Secret in that case $[0,1/2)$ should be a typo

Guys any reference for the proof of convergence of Gauss-Seidel method ??

@DHMO now it's not. The sequence 1/2^n will converge (in the usual sense) to 0 thus it can make (0. It, however cannot reach [0 as no n will give 1/2^n=0

@Secret in that case, unfortunately {[0,1/2), [1/2,1), [1,3/2)} is a finite subcover

@Vrouvrou so $d(y,A) < d(d,A) + d(d,y)$ by definition of $Y$

and $d(y,A) < d(d,A) + \frac f 2$

also, $d(y,B) + d(d,y) > d(d,B)$

so $d(y,B) + \frac f2 > d(d,B)$

so $d(y,B) > d(d,B) - \frac f2$

@Vrouvrou are you still here?

@DHMO I cannot seemed to find any interval that will cover [0,n) for arbitrarily small n so to allow me to glue a countable sequence of intervals end to end (cause that will guarantee I cannot get a finite subcover)

@Secret maybe you're covering using the wrong way

@Vrouvrou Just notice that $f(x)=d(x,B)-d(x,A)$ is continuous. (It is a difference of two continuous function.) So you are asking about $f^{-1}(0,\infty)$, since $d(x,B)>d(x,A)$ is equivalent to $f(x)>0$. Since $f$ is continuous, preimage of any open set is open.

@MartinSleziak hey, I just proved it using a more elementary method lol

namely, by definition

It seemed long, so I was too lazy to read what you answered to Vrouvrou.

I am not sure whether they are needed, but I will also add links to posts about continuity of $x\mapsto d(x,A)$. For example, Prove that $f_A (x) = d({\{x}\}, A)$, is continuous. and Prove that $f(x)$ is continuous on $M$.

@MartinSleziak can you give me an example of something which is not a Borel set?

@DHMO Didn't we talk about something similar already? Or was it with secret? Existence of such sets can be shown with cardinality argument. But since it cannot be shown in ZF, we cannot expect "explicit" example.

@MartinSleziak I see

Probably for any reasonable interpretation of the word explicit. Any proof of existence of non-Borel set should contain a step which is not effective (cannot be proved in ZF).

definable

@DHMO How about $[0,1/n) \cup [1/n,1/(n-1)) \cup ... \cup [1/4,1/3) \cup [1/3,1/2) \cup [1/2,1) \cup [1,3/2)$. This is a disjoint union thus it should guarantee failure to find finite subcover?

@Secret what is n?

MO: Non-Borel sets without axiom of choice and math.SE: Constructing a subset not in $\mathcal{B}(\mathbb{R})$ explicitly

@MartinSleziak the most humbling thing I've read (paraphrase): there's only countably many definable objects, because there's only countably many finite strings to define objects

thus, many Borel sets are not even definable

@DHMO ok this is not working. I have all the sets, but I am always missing a [0,n) so I can cover [0

Even if I pick (a,b) where a < 0 will still not work. The issue is the b) end. There seemed to be no b that can prevent a countable cover to become finite

Does that mean I need to try uncountable covers instead?

no

it's still countable

But for the above, once I set n to be some fixed number 0<n<1 , then there will always exists a finite (however large) M such that n<1/M thus allowing a finite subcover of size n+1

so it's time to change your strategy

One thing I find strange on math.SE is that sometimes I see users who ask questions about some relatively advanced topics (homology, Morse theory) and at the same time in other posts you see the same user struggle with some basike question, like a simple limit or with proof that some relatively simple subset of $\mathbb R^2$ is open.

I am one such user, a lot of my question concern things I briefly read and does not have solid enough background on (things like proof theory for example)

I do not want to name names - in fact I would be able to name only one such user (I remembered the name because I interacted with this specific user a few times). But I'd say I've seem something similar with a few other users, whose username I do not remember.

@Secret I think that is slightly different. In the cases I saw it seemed that those users were studying a course in an advanced topic. (The nature of the question seemed to be that it might homework or an exercise from a textbook.) This is different from somebody who is learning basic stuff from some subject and asks about some more advanced question that comes to mind out of curiosity. And often such questions are very natural.

Ah I see

@MartinSleziak I remember asking you this:

do we need AoC to prove results containing beth numbers?

or

what results of beth numbers do we need AoC to prove?

I remembered that too. Some brief summary of that discussion is here.

@MartinSleziak but is there any?

Didn't we leave only comparabilty as unresolved?

why is that unresolved?

I thought Cantor's argument gives us $\beth_n < \beth_{n+1}$

In that case it is unresolved because I missed an easy argument at the time.

But notice that you are not finished yet. You should also do something about limit ordinals.

But that is probably again from Cantor's theorem.

in that case

let's agree to work with beth numbers and ditch AoC

lol

can $\Bbb R$ be well-ordered?

If we change axioms, why not take ZFC+GCH or ZFC+V=L or something that simplifies thing even more.

Is there a known well ordering of the reals?

@DHMO Yes if you work in ZFC. Not in you work in ZF. So again, you cannot expect an explicit example.

@MartinSleziak so we can't ditch AoC

It depends on what you want to do.

If you decide to study what results hold in ZF, you have to work without AC.

how can we prove {{a},{a,b}} = {{c},{c,d}} $\implies$ a=b and c=d?

Of course, for well-ordering of reals you do not need full strength of AC, you could probably restrict it a bit. I guess that this can be found in literature and very likely there are some posts on math.SE and MO. My guess would be that choice on $\mathcal P(\mathcal P(\mathbb R))$ could be enough, but that is really only a guess.

@DHMO Some proof is given in the Wikipedia article Ordered pair. In the part about Kuratowski definition.

@MartinSleziak a chain of inclusion of R can be as long as 2^R?

So you are asking about maximal cardinality of a chain in $(\mathcal P(\mathbb R),\subseteq)?

yes

[0,1/2) {s} [1,3/2). I need {s} to converge on the left to e.g 1/4 and converge to the right to e.g. 5/4

Something very similar was discussed not that long ago. This answer to a more general question says that the answer is yes if you assume continuum hypothesis.

he he @MartinSleziak is that user me in case you want to disclose it??

@MartinSleziak I see

@BAYMAX No I was not talking about you.

@DHMO I am not sure whether we can show this without CH (or some other additional assumption about $\mathfrak c$) at least in the case of reals. (The questions I linked was about powerset of arbitrary sets $X$.

@MartinSleziak i was looking for a proof on Convergence of Gauss Seidel and could understand except the last line where the modulus of eigenvalue was < 1...

Can this be simplified somehow or broken into summation of two arcsines ? $$\sin^{-1}\left(\sqrt{\frac{2-\sqrt{3}}{4}}\right)$$

its about an inequality in the last which seems obvious but still looks complex to me

@anonymous $\sqrt{4-2\sqrt 3} = 1-\sqrt3$

so your thing = $\arcsin\left(\dfrac{1-\sqrt3}{2\sqrt 2}\right)$

$=\arcsin\left(\sin 45^\circ \cos 60^\circ - \sin 60^\circ \cos 45^\circ\right)$

@DHMO Ah, so 45 degrees and 60 degrees

@anonymous yes

Thanks :)

So now there are two users with the username anonymous even present in the room. (Which means that they both receive the pings.)

@MartinSleziak yes, you've said it 2 hours ago

@MartinSleziak I have a small "a" at the beginning of my name...

The other one has a capital "A"

@anonymous but both are still pinged

@DHMO Eh, how ?

Yes, I mentioned this now mainly for the benefit for the other user who has just joind the room.

I got 14 notification today

@DHMO LOL XD...that must be a bug

i think its the software of MSE

It is better to use the reply button directly I guess instead of pinging :P

that is even if someone types baymax i get pinned

I know that using @string pings all users who match the string. Just out of curiosity, how is it with direct replies.

@Anonymous Yep, the same one as in your picture!

@anonymous I posted this message as a reply. (I did not type @username.) Does this ping both of you?

@MartinSleziak It pinged me.

Yes, and it was supposed to - since I replied to your message.

@MartinSleziak can you do once again?

I am curious whether it pinged the other anonymous too.

Ok so he/she didn't get a ping ^ (Anon..)

I got

So reply button works

Let's try this other way round.

@Anonymous This is reply to the message posted by Anonymous (with the capital A), so they should receive the notification. Does anonymous (with the lowercase a) also get a notification?

@MartinSleziak I get a highlighted name but not a ping

what the hell

I can see the orange highlight on the name

@Anonymous

@anonymous yes boy?

But did not get the (1) ping symbol on my profile pic at the bottom left

Hilarious :"D

①

@MartinSleziak did I ask you about the lower limit topology?

will be posting now the proof of Gauss Seidel

Ok, so the conclusion of the experiment is: 1) Typing @xyz pings all users where username begins with xyz. 2) If a user with username xyz posts a message, then direct reply to that message pings only this particular user. (Even if there is another pingable user with the same username.)

@DHMO I don't recall. But I should probably go leave now anyway. (And I am not sure whether I would be able to give you an answer.)

@DHMO I am not entirely sure what you want to test, but this is reply to the message which you requested.

@MartinSleziak bye

bye @DHMO

cya!

@anonymous no, he's leaving

oh sorry

@DHMO How about the countable union of {[1-1/n,1-1/(n+1))} with n=1,2,3,4,... and the union of [1,3/2) ?

@Secret congratulations

Ok, have a nice day!

The edible root, the beetroot, $\sqrt[b]x$, pun on the b-th root.

@MatsGranvik god dag

Talar du svenska?

No, that's the only phrase I know :p

Ok

is Swedish close to German?

There are similar words, German and Swedish are closer to each other than Swedish and English.

ich kann kleinen Deutsch sprechen... aber ich habe viele Fehlern

@DHMO i am also guessing that in general, any disjoint union of intervals generated by a sequence with the property that $a_0=0$ and $\limit_{n\to\infty} a_n > 1$ will be an irreducible countable cover. This choice (of which one explicit example I provided previously) also have an interesting property that it is an open cover expressed with only the basis elements of the topology

@Secret indeed

now we can move on to q3 and q4

@MatsGranvik did you understand that?

@DHMO "Jag kan klen Tyska språka, men jag har/gör flera fel." If you translate with direct words.

I know maybe less than 10 words in Swedish

one of which is "jag"

so I can guess some of the words

@DHMO $(-\infty,a)\cup [a,\infty) =\Bbb{R}$ is clopen. $(a,b)\cup [b,-\infty)=(a,\infty)$ is open. $[a,b] \cup (b,\infty)=[a,\infty)$ is closed.

@Secret wonderful.

@anonymous Should be P(A) + P(B) = P(A cup B) + P(A cap B)

@BalarkaSen I wonder what happens in the order topology on $\Bbb Q$

Oh, right

Got it

I made a sign mistake there

I feel so silly sometimes :P

@DHMO but how can I prove they are the only ones?

@Secret of course they are not the only ones?

@DHMO It should be the standard topology.

@BalarkaSen is (-pi,pi) an open set?

Well, yes, since it is an open set in the standard topology on Q.

can we define the basis to be only the intervals with starting point and endpoint in Q?

I guess no, since that doesn't preserve arbitrary union

I wonder if that has any perfect set

What has any perfect sets?

the standard topology of Q

I know what a perfect set is :P

What about [0, 1] cap Q.

not closed

Of course it is!

By definition of subspace topology.

indeed

You should really get some textbook in topology than reading wikipedia and throwing out arbitrary questions :P

I don't have time

Then don't do the latter

is there any neither set in Q?

oh, [0,1) cap Q

what if I extract the rational numbers with odd denominator?

@BalarkaSen it is dense but it is neither open nor closed

I am not paying attention to your questions anymore

lol

@DHMO cannot think of any way to decompose [a,b] and (a,b) into a disjoint pair of open and closed sets Attempt to make a cut )[ or ]( will give you at least one neither, also () U () U () union with [] U [] U [] is cheating because these sets have a notion of union in them thus you technically unioning more than a pair of open and closed disjoint sets?

@Secret we have confused open sets with open intervals again

But in the standard topology, all open intervals are open sets

but not all open sets are open intervals

and that isn't cheating; that is exactly I want

I didn't say connected

@Secret give me an open dense disconnected set in R

Does the integral $$ \int_0^{2\pi} ((\sin(x))^3 + 2 \ln |\sin(x)| + (\tan (x))^5) dx$$ coverge/have a definite value? I'm pretty sure it doesn't due to last term. Yet, can someone verify it?

@anonymous you can use desmos

and I'm sure that the last term is 0

@DHMO Desmos to find integral ?

@anonymous yes

@DHMO Huh, () is open [] is closed [) is closed if the ) end is infinity, otherwise it is neither (] is closed if ( is infinity otherwise neither. Arbitrary Union of open sets are open, singletons are closed. That should be everything?

@DHMO Isn't 0 to pi/2 of the last term undefined as it doesn't converge?

Then how is 0 to 2pi defined?

@anonymous because pi/2 to pi will cancel out 0 to pi/2

@Secret of course not, but that's all we need

@DHMO Wolfram alpha gives no value (wolframalpha.com/input/?i=integrate+0+to+2pi+(tan+(x))%5E5))!!!

Ah, is that a valid logic?

@anonymous that depends on the definition of integral

Two undefined terms cancelling out...

@DHMO complement of the rational?

@DHMO What type of definition

@Secret not open

@anonymous whether we can let the two terms cancel out

I don't remember the technical details

This came in my test and I left it assuming it a wrong question :P

As I thought two undefined terms can't cancel out

Hmm, I need to study about this

BTW what is the conventional definition?

@BalarkaSen ^

@BalarkaSen Need your expert advice :P ^

@DHMO actually are the rationals open? Naively it seemed to be a countable union of the singletons {p/q}. But since this is an arbitrary union of closed sets, it is possible to make open intervals when we perform this union. Then, we have open intervals centred at every rational. Since arbitrary union of open sets are open, it means the rationals will be open?

@Secret the singletons are not open

closed sets can make open intervals doesn't mean every union of closed sets is open

But we can always get a sequence of rationals such that they converge at a rational or irrational, thus producing (a,b) for irrational a,b? for any rational?

as the rationals are dense so they can get arbitrarily close to each other and the irrationals

you cannot get the irrationals by having the union of the singletons of the rationals

you don't get to automatically include the limit points

@DHMO Umm, using Queen's rule (qph.ec.quoracdn.net/…) of definite integration it is certainly cancelling out to zero. However, I don't know if it okay to blindly apply Queen's rule. Should I ask on the main site?

@DHMO In the above, a,b are irrationals and they are not included, or do you mean because of the dense property of rationals I always end up with some irrational c such that a<c<b?

@Secret I don't understand what we are trying to prove/disprove

Consider the singletons {1/n}, for n=1,2,3,.... This represent a sequence of rationals that will converge to 0. Thus this sequence will lie in the interval (0,1] so there will be countable many rationals lying arbitrarily close to 0 hence the open end at the 0. A different sequence {2-1/n} can be picked which converges to 2, thus lying within the interval [1,2). Combining these two unions, we have the resulting sequences of rationals to be within (0,2) thus it converges to 0 on the left and 2

yes

@anonymous On what? The integral? No idea.

@BalarkaSen no, on whether $\displaystyle \int_0^\pi \tan x \ \mathrm dx$ converges

@BalarkaSen I asked it here math.stackexchange.com/questions/2161990/…

on the right, making the sets near the open ends not closed?

Lets see if we get an answer..

@Secret what?

which set is not closed?

$\displaystyle A = \left(\bigcup_{n=1}^\infty\left\{\frac1n\right\}\right) \cup \left(\bigcup_{n=2}^\infty\left\{2-\frac1n\right\}\right)$ is not closed.

@DHMO This has a small singularity around $x = \pi/2$, but you can always break it up to $\int_0^{\pi/2 - \epsilon} + \int_{\pi/2 + \epsilon}^\pi$ and let $\epsilon \to 0$.

This is called Cauchy principal value of the integral.

@BalarkaSen and this is what @anonymous is interested in

I think then this would converge fine

The reason being that those two integrals will cancel up

@BalarkaSen I see. Thanks. However, I guess Wolfram Alpha uses a different definition (wolframalpha.com/input/?i=integrate+0+to+pi+tan(x))

I don't trust W|A on these stuff. They probably use some kind of numerical integration technique which makes it fail.

I mean, $\int_0^{\pi/2} \tan(x) dx$ diverges after all

@BalarkaSen I guess that'll do for my purpose. I have to use that principal/assumption in my exam questions probably. Otherwise I have to leave them unanswered :P

@anonymous actually

your site says that the Cauchy principal value is 0

@BalarkaSen Yes 0 to pi/2 diverges

scroll to the bottom of your site

@DHMO Yes I saw that. Didn't know the meaning of that earlier :)

@anonymous Right.

It is just a semantics problem then :P

$\{1/n\}$ is not closed

hence it is open <-- typical garbage

This is actually a 1-dimensional version of a 2-dimension phenomenon, where you "perturb contours" and integrate complex functions on easier, homologous contours.

@DHMO Hehe :'D

@BalarkaSen I see :) I have never done integrals in 2D (contour integrals)

Those are cool, and a gateway to a lot of topological ideas. Try it sometime when you're free.

@BalarkaSen I'm seriously waiting for my entrance exams to be over by June. I will spend my 2 months learning these cool stuff!

Thanks a lot to both of you btw

@DHMO yeah, the principal value is 0 by symmetry. (tan^5 is a antisymmetric around x=pi).

Entrance exams are nightmares. I guess mine are next year though.

@BalarkaSen You said you wont give entrance exams a few weeks back on hbar! Or will you? I think you should join one of the maths institutes in India or abroad rather than engineering as you are so much interested in maths. You know so many topics more than me!

( I am more inclined towards applied physics so I guess engineering is for me)

Oh, I'm not going to take JEE/IIT but I have to take the institutional exams for admission.

@BalarkaSen Institutional exams like for CMI/ISI ? Oh very well then :) All the best!

Yeah, like those.

I feel you will be a very good mathematician :)

Good luck

Thanks. But I need to get off my ass and work - I am a lazy man.

@anonymous the topology on $\omega^2$ seems interesting

@DHMO I know nothing about topology. You pinged the wrong person it seems!

@anonymous oh, I thought Balarka mentioned topology to you

@DHMO ok just to quick check. In the usual topology, [a,b] are closed. Now we know that $\bigcup_{i \in [a,y)} [a,i]=[a,y)$. $\bigcup_{i\in \{a,a_1,a_2 ...\}(\text{converge to y})}[a,i]=[a,y)$. $\bigcup_{i\in [a,y)} [i,i]=[a,y)$ But $\bigcup_{i\in \{a,a_1,a_2 ...\}(\text{converge to y})}[i,i]$ is a union of closed sets that is closed?

@Secret neither set is closed

assuming that $y \notin (a_n)$

@DHMO Actually I don't know what I will get for the tailing end when I take the complement of the above sequence. For the non tailing end I know I will get a union of disjoint intervals of the form $(a_i,a_{i+1})$ thus end up with a set where each of these intervals are disconnected from one another, but I am not sure what happened to the tailing end. You said the sequence is not a closed set, so what is the complement near the y) end of the sequence became?

@Secret it still looks like disjoint open sets

except that in the end you get $[y,\infty)$ which is closed

O wow, so any bounded countable sequence is a neither set because its complement is a countable union of open sets and a closed set $[y,\infty)$?

that is not a good "because"

as I said, a union of open set and closed set can be open

and it can also be closed

and neither

and both

But this set () U () U () U ... U [,...) is not any of the cases we covered previously?

alright

@Astyx bonjour

Hi

@Secret just saying that it is not a good "because"

but it does help your intuition

A lot of stars are given these time ..

I am not sure how one can find neither sets rigorously and systematically. That set that is the bounded sequence is already not intuitive enough

@Secret a set is closed iff it contains all its limit points

I thought it was closed intuitively because there are singletons no matter how far one go, but I guess it being dense near the y end screwed it

it is far from "dense" but I get what you mean

What is the correct term to describe what I just said about its behaviour near y?

y is a limit point...

Hmm, so I am guessing the rationals is actually a neither set since it is not closed because the limit points the irrationals are missing, and not open because its complement are the irrationals (which I don't know how to describe it since rationals are not well ordered in the usual ordering and is dense thus there is no (p,q) where q is the successor of p, thus I cannot say it is a countable union of open intervals pl) which is also not closed because the limit points the rationals are missing.

yes

@Secret that raises an interesting point

the rationals are a countable union of closed intervals (namely, the singletons)

but the irrationals are not a countable union of open intervals

the above is fact 1

fact 2: by induction, the complement of union of closed intervals is a union of open intervals

are fact 1 and 2 contradictory? @Secret

@DHMO Oh. Sorry I don't know what "topology on $\omega^2$" means. Hope to learn it in the near future.

@anonymous sure

Someone would like to help with number theory question here or here.

?