Is there a sequence of points in $[0, 1]^{\mathbb N}$ that has no convergent sub-sequence?

Question

What tools would I use to answer the following topology question?

Is there a sequence of points in $[0, 1]^{\mathbb N}$ that has no convergent sub-sequence?

I am not sure what tools to use to answer this question. I was also wondering if the space is compact. I do not believe it is. $[0,1]$ is compact, so take an open cover of $[0,1]^{\mathbb N}$. On each coordinate, it has a finite subcover, but the there are a non-finite amount of coordinates, so the product cannot be a subcover. Is my reasoning correct?

The topology on $[0,1]^{\Bbb N}$ is important. Most likely, you're using the product topology, in which case absalon's answer is sufficient. — Cameron L. Williams, Sep 11 '15 at 00:57
@CameronWilliams I had never heard of Bolzano Weierstrass property before and then I skimmed the linked document in absalon's answer but it seems it's only about finite dimensional products. Do you know where I can find the BW property stated for infinite products? — Rudy the Reindeer, Sep 11 '15 at 01:01
@RudytheReindeer Oops I didn't notice that he mentioned the BW property. I figured he meant what you wrote, but alas. — Cameron L. Williams, Sep 11 '15 at 01:20

score 4 · Answer 1 · edited Apr 13 '17 at 12:21

4

Note that the countable product of metric spaces is metrisable. (see e.g. here)

For metrisable spaces compactness and sequential compactness are equivalent.

Finally, the space $[0,1]^{\mathbb N}$ is compact by Tychonoff's theorem.

edited Apr 13 '17 at 12:21

Community

1

answered Sep 11 '15 at 00:51

Rudy the Reindeer

48,301

So: compact implies sequentially compact implies every sequence has a convergent subsequence. – Rudy the Reindeer Sep 11 '15 at 00:55
Thanks! Such a simple solution, yet I somehow never would have thought to put the ideas together as such. – ThinkConnect Sep 11 '15 at 01:49
The Tychonoff -product topology of any set of compact spaces is compact.This is called the Tychonoff Theorem.It is not obvious. For the special case $I^K$ where $I=[0,1] $and the cardinal of $K$ is less, or equal to, the cardinal of the reals, it can be done using the Bolzano-Weierstrass theorem. – DanielWainfleet Sep 11 '15 at 03:35
@user254665 Thank you for your comment. I looked at the document linked to in the other answer but could not find Bolzano Weierstrass for cardinalities greater than finite. Where can I find it for countably infinite? – Rudy the Reindeer Sep 11 '15 at 03:44
I gave the wrong name .I was thinking of the following ,due to Weierstrass: Every continuous$ f: I \to R$ can be uniformly approximated by a polynomial $p:I\to R$. Now $p$ can be uniformly approximated by a polynomial $p^$ with rational co-efficients, and there are countably many such $p^$. – DanielWainfleet Sep 15 '15 at 20:07
@user254665 I see. Thank you for the comment. – Rudy the Reindeer Sep 16 '15 at 00:45

absalon · Answer 2 · 2015-09-11T02:22:41.017

1

By Tykhonov's theorem it is a compact metrizable space and every compact metrizable space has the Bolzano-Weierstraß property. Done.

edited Sep 11 '15 at 02:22

answered Sep 11 '15 at 00:50

absalon

357

Can you state the Bolzano Weierstrass property? – Rudy the Reindeer Sep 11 '15 at 00:52
Thank you. And which part in the linked document is relevant to the question? It seems to me that the document only deals with finite dimensional products but humour me and point me to the relevant theorem in the document. – Rudy the Reindeer Sep 11 '15 at 00:57
In page 2 the author defines the Bolzano-Weierstraß property. – absalon Sep 11 '15 at 01:01

Is there a sequence of points in $[0, 1]^{\mathbb N}$ that has no convergent sub-sequence?

2 Answers2