Tychonoff's theorem and convergent subsequence

[Monocles]

Let $$X$$ be the countably infinite product of closed unit intervals under the product topology. By Tychonoff's theorem, this space is compact. Consider the sequence $$\{ x_n \}$$, where $$x_k$$ is the vector that is zero for all components except for the kth component, which is 1. Since this space is compact, this sequence must have a convergent subsequence. I cannot figure out what that could be though?

I asked my analysis professor earlier today and he could not answer either. I assume it must have something to do with the fact that we're using the product topology instead of the box topology, but my general topology is rusty and so I cannot tell what is going on here.

 

[radou]

I'm not really sure about this, but considering a specific metric could help.

For Rω (the infinitely countable product of R with itself, i.e. the set of all infinite sequences of real numbers), the metric that induces the product topology is given with:

$$D(x, y)=\sup \left\{\frac{d'(x_{i}, y_{i})}{i} \right\}$$,

where d'(xi, yi) = min{|xi, yi|, 1}. Now you have a metric space.

Since every metric space is Haussdorf, and since the subspace of interest is compact, then it must be closed. Another consequence of compactness is that every sequence has a convergent subsequence. The limit is uniqe (this holds for metric spaces), and must lie in your subspace, since it is closed.

Actually, let x0 be the zero sequence, and let ε > 0 be given. In the metric D, the distance between x0 and xn, where xn is the sequence whose n-th element is 1, and any other element is 0, equals 1/n. Clearly there exists a positive integer N such that 1/N < ε, so any sequence xn with n >= N belongs to the open ball of radius ε around x0. Hence your sequence converges to x0.

According to the definition of a subsequence, the original sequence itself is its own subsequence.

Again, I'm not really sure about this, so I hope someone else will comment on any potential mistakes.

 

[micromass]

Take a product space $$\prod X_i$$ we have the following theorem about converging sequences:

Take a sequence $$(x_n)_n \in \prod X_i$$, then we have

$$x_n\rightarrow x~\Leftrightarrow~\forall i\in I: pr_i(x_n)\rightarrow pr_i(x)$$

So a sequence converges in a product space iff all the projections converge.

Now,in $$[0,1]^\mathbb{N}$$ we have the sequence (1,0,0,...), (0,1,0,0,...), (0,0,1,0,...),... It is clear that every projection converges to 0 (for example, the second projection yields $$0,1,0,0 ... \rightarrow 0$$). So the overall sequence converges to (0,0,0,0,...)

In the box topology however, the sequence does not converge (as it does not enter the open set $$\prod ]-\frac{1}{2},\frac{1}{2}[$$ ) and it does not have a convergent subsequence. This is not a contradiction however, since this box topology is not compact. In fact, there is no Tychonoff theorem for box topologies, which is why the product topology is so interesting.

 

[mathwonk]

what is the definition of the product topology? i.e. make a guess as to the limit, and then ask whether every basic open nbhd of that limit eventually contains the sequence.

and i suggest your prof does know the answer to this.

 

[radou]

Take a product space $$\prod X_i$$ we have the following theorem about converging sequences:

Take a sequence $$(x_n)_n \in \prod X_i$$, then we have

$$x_n\rightarrow x~\Leftrightarrow~\forall i\in I: pr_i(x_n)\rightarrow pr_i(x)$$

So a sequence converges in a product space iff all the projections converge.

Now,in $$[0,1]^\mathbb{N}$$ we have the sequence (1,0,0,...), (0,1,0,0,...), (0,0,1,0,...),... It is clear that every projection converges to 0 (for example, the second projection yields $$0,1,0,0 ... \rightarrow 0$$). So the overall sequence converges to (0,0,0,0,...)

That's a useful theorem, I was asking myself what's the relation between the limits of the sequences and the limit of the overall sequence.

 

[mathwonk]

well that why I asked you to think about the definition of the product topology. I did not notice your question had already been completely answered however. But you might try using the definition of the topology to prove that useful theorem. It seems immediate.

 

[Landau]

It is immediate, once you recall that a basis of the product topology is given by finite intersections of things of the form

$$p_i^{-1}(U_i)$$

where p_i is the i-th projection and U_i is open in X_i (i.e. these things form a subbasis).

 

[mathwonk]

That is the reasoning I hoped to inspire in the OP. For everyone of us, remember it is always useful when giving a proof, to recall the definitions.

 

[Monocles]

Thanks for the help, the professor actually emailed me that he realized the solution was that it converged to zero and so I forgot about this thread.

 

[mathwonk]

is your professor a graduate student?

 

[Monocles]

No he is actually a full professor, I just think my question misled him. We were talking about space filling curves from the closed unit interval to I x I x ... x I. I asked if this were true for an infinite product of unit intervals, and he said no because an infinite product wasn't compact, and gave the sequence I mentioned in the original post as to why it wasn't compact. Then I mentioned the Tychonoff theorem says it will be compact, but neither of us remembered the detail at the time that the Tychonoff theorem requires the product topology as opposed to the box topology.

 

[mathwonk]

well anyone makes mistakes. i just remember this from first year grad school, so it seems pretty basic to me. Of course there are many basic things i don't know either.