1

I'm trying to show that the Hilbert cube is compact and want to know if I'm taking the right approach. My text defines the Hilbert cube as: $H=\{(x_1,x_2,...) \in [0,1]^{\infty} : for \ each \ n \in \mathbb{N}, |x_n|\leq \dfrac{1}{2^n}\}$

I need to show that it is compact with respect to the metric:

$d(x,y)=\underset{n}{\textrm {sup}}|x_n - y_n|$

In order for the sequence in $H$ to converge, we need each of it's components to converge.

Since, each component $x_n$ is bounded (by $0$ and $\dfrac{1}{2^n}$) for any point in $H$, a sequence (in $H$) of such points would form a sequence $(x_n)$ for each component $n$ and each of those sequences would be bounded.

Therefore, by Bolzano-Weierstrass, they have convergent subsequences.

We can show that for each sequence of points in $H$, we have a subsequence of the sequence of $x_1$s that converges, say ${x_1}_k$, we then have a sequence $({x_1}_k, {x_2}_k,...)$.

For the sequence of ${x_2}_k$s, there exists a convergent subsequence ${{{x_2}_k}_j}$. Since all subsequences of a convergent subsequence also converge, we have that both $({{{x_1}_k}_j})$ and $({{{x_2}_k}_j})$ converges in $({{{x_1}_k}_j}, {{{x_2}_k}_j},...)$.

Repeating this process ad infinitum, we get convergent subsequences for all $x_n$ so that the sequence in $H$ has a subsequence that converges as well and, therefore, $H$ is compact.

Is this correct? If it is, is that enough to show that $H$ is compact under the given supremum metric?

jackson5
  • 1,614
  • Since each point in $H$ has (infinitely) many components, in order for a sequence of points in $H$ to converge, we need to show that each of the component sequences converge. (x_n) is one such component sequence. Since each component sequence is bounded, it has a convergent subsequence. So we take such a subsequence for the first component, a convergent subsequence of that subsequence for the second component and so on. This way, each component converges and so does the sequence in $H$. – jackson5 Feb 01 '17 at 01:00
  • The sequence construction is similar to here: http://math.stackexchange.com/questions/567335/cartesian-product-of-compact-sets-is-compact – jackson5 Feb 01 '17 at 01:03

2 Answers2

3

No. This is not correct. As an aside, your paragraph

Since, each $x_n$ is bounded (by $0$ and $1\over 2^n$) for any point in $H$, a sequence of such points would form a sequence $(x_n)$ for each $n$ and each of those sequences would be bounded.

is so poorly constructed, I could not figure out what even the intent was until after I had read through the rest of the the post and had enough evidence to reconstruct it.

But the actual error in the argument is this: There is not necessarily a relation between the convergent subsequences in the different indices.

For example, the convergent subsequence chosen for $(x_{1n})$ may be $(x_{1(2k)})$, while the convergent subsequence chosen for $(x_{2n})$ may be $(x_{2(2k+1)})$. Thus the convergent subsequences for your first two indices would not ever come from the same points. There is no subsequence $(x_{n_k})$ of $(x_n)$ itself such that $\pi_1(x_{n_k})$ is the $k^{th}$ in your $(x_{1n})$ subsequence and $\pi_2(x_{n_k})$ is the $k^{th}$ in your $(x_{2n})$ subsequence.

You could try to work around this as follows:

Let $n^1_k$ define a subsequence of $(x_n)$ such that $\pi_1(x_{n^1_k})$ converges. Let $n^2_k$ define a subsequence of $(x_{n^1_k})$ such that $\pi_2(x_{n^2_k})$ converges. Etc.

But this still doesn't work. While $\pi_j(x_{n^m_k})$ is guaranteed to converge for each $j \le m$, there is no guarantee of a sequence that works for all indices. For example, it could be that $n^m_1 = m$ for all $m$. Thus any subsequence you tried for the points themselves would start out less than an infinite number of your index-wise convergent subsequences.

Since convergence of sequences depends on the tails of the sequences, you might by careful argument find a way past those obstacles, but it would be tricky to do so. I suggest you try to prove this by other means than sequential compactness.

Instead, I suggest trying to prove that $H$ is complete and totally bounded.

Paul Sinclair
  • 43,643
  • In the proof for "Cartesian product of compact sets is compact" found here: http://math.stackexchange.com/questions/567335/cartesian-product-of-compact-sets-is-compact, sequences are constructed as follows: If we are given a sequence $(u_n)$ of $A$x$B$, then you can write $u_n=(a_n,b_n)$. Since $A$ is compact, you can find a sub-sequence $(a_{f(n)})$ with a limit in $A$. Then, since B is also compact, you can extract a sub-sequence $(b_{f(g(n))})$ of $(b_{f(n)})$ with a limit in B. Thus, the sub-sequence $(u_{f(g(n))})$ of $(u_n)$ has its limit in $A$x$B$. This proves that $A$x$B$ is compact – jackson5 Feb 01 '17 at 01:41
  • That's for just two components, but wouldn't repetition of that method work here as well? i.e. we find subsequences of subsequences That is what I was attempting here, sorry about my poor articulation of that – jackson5 Feb 01 '17 at 01:42
  • The difference is the argument in the other thread is for a finite number (2) of indices. It generalizes to any finite number of indices. But it fails - badly - when the number of indices becomes infinite. The cause is basically the same as what allows the intersection of an infinite number of open sets to not be open, even though the intersection of any finite number is. – Paul Sinclair Feb 01 '17 at 01:59
  • Oh ok I see, that makes sense. – jackson5 Feb 01 '17 at 02:00
  • By the way, I pointed out the problem in the one paragraph simply to indicate something you should work to improve on in your explanations. As you can tell, I struggled too to find a way to explain my intent. When dealing with sub-indices of indices, it can be really difficult to be clear. – Paul Sinclair Feb 01 '17 at 02:03
0

Just to add to Paul's beautiful discussion, let's say you find a subsequence (of indices) such that first coordinates converge, and out of that a further subsequence for which also the second components converge (by being a subsequence of the first one, we get first coordinate for free), now out of this one a further subsequence for which first, second, and third components will converge. Inductively, we can repeat this ad infinitum: having built level $n$, obviously, you can continue to $n+1$. So, it is not that the process will halt or anything.

So, where lies the problem and intricacy? The problem is that at the end you may be left with no indices at all!! You may not be able to give one common sequence of indices that is simultaneously a subsequence to all of the ones at stages 1, 2, 3,... Imagine, first sequence you find is all the even indices, and in the second you are forced to consider only those that are multiples of 3 as well, next, at 3rd run you also must get rid of all except for those that are multiples of 5 as well. Already at this point, you are considering only indices that are multiples of 30. So, it is easy to imagine a process that quickly discards many many indices, so that at the end you will not be able to extract one unified subsequence of all stages. (Draw an infinite matrix with each row representing one element in the cube and columns representing the the the same coordinate of sequence of points in the cube. Then what I described is precisely circling a special sequence in the first column that runs down and converges, then discarding all the rows that do not see a circled item. Then doing the same to the second column (of this revised matrix) and circling certain sequence...)

So, what is the way out? The quite famous Diagonal Trick. To build a sequence, you do not wait until eternity, you choose the index of the first item of the sequence for the first column to be your first index, then second index of the working sequence for the second column, third index of...

This will work, because the tail of this ultimate sequence will lie in the $j$'th column for any $j$ -- precisely after $j$'th component all will be inside $j$'the sequence. And this is as good as the unicorn sequence (an imagined common subsequence to all) would have been because convergence only concerns the tail.

Hope, this confuses and then clarifies, just a bit more! :))

Behnam Esmayli
  • 5,024