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6. Homework 2, Problem 3

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The following problem is problem 3 on the second homework. Understanding this argument is very important, so if you were missing details or didn’t fully comprehend the argument, it’s good to review it in full detail and make sure you understand each step! Part of the struggle, naturally, is figuring out good notation for the problem.

Let’s unpack the problem statement by drawing a helpful diagram: we may arrange the sequences in rows of a table as follows. \[\begin{array}{c|cccccc} & n=1 & n=2 & n=3 & n=4 & n=5 & \cdots \\ \hline S_1 & a _{1,1} & a _{2,1} & a _{3,1} & a _{4,1} & a _{5, 1} & \cdots \\ S_2 & a _{1,2} & a _{2,2} & a _{3,2} & a _{4,2} & a _{5, 2} & \cdots \\ S_3 & a _{1,3} & a _{2,3} & a _{3,3} & a _{4,3} & a _{5, 3} & \cdots \\ S_4 & a _{1,4} & a _{2,4} & a _{3,4} & a _{4,4} & a _{5, 4} & \cdots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \\ S_N & a _{1,N} & a _{2,N} & a _{3,N} & a _{4,N} & a _{5, N} & \cdots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \end{array} \]

The \(N\)-th row of the table represents the \(N\)-th sequence in our sequence of sequences, \(\left\lbrace S_N \right\rbrace _{N=1}^{\infty}\). The problem then says that for any fixed \(n\) — i.e., given a column of the table — there exists some \(M_n\) such that \(\left\lvert a _{n,N} \right\rvert \leq M_n\) for all \(N\). In other words, the columns of the table are bounded.

The question now asks for a subsequence \(\left\lbrace S _{N_j} \right\rbrace _{j=1}^{\infty}\) of \(\left\lbrace S_N \right\rbrace _{N=1}^{\infty}\) such that \(\left\lbrace a _{n, N_j} \right\rbrace _{n=1}^{\infty}\) converges. In other words, you are keeping a subsequence of rows on the table so that the columns each converge.

I’m being very explicit with the indices here so we can keep track of “which direction” our sequences go. There’s multiple indices here that mean different things, and simply saying that “\(a _{n, N_j}\)” converges is quite ambiguous because there are multiple limits that can be taken.

We can very broadly split up the argument into two steps: first, we take iterated subsequences of rows that converge in more and more columns. Then, we use these iterated subsequences to construct our final subsequence that does converge in every coordinate.

Iterated Subsequences

I think almost everybody was able to get this step in more or less one piece. The notation can be quite difficult, so we’ll try to be very careful with our indexing and clarity. The crucial statement we’ll need is:

We start by observing that \(\left\lbrace a _{1,N} \right\rbrace _{N=1}^{\infty}\) is a bounded sequence of real numbers. Then, by this very lemma, there is a subsequence \(\left\lbrace a _{1, N _{1, j}} \right\rbrace _{j=1}^{\infty}\) that converges. We can then take the corresponding rows of the table to get a subsequence of rows \(\left\lbrace S _{N _{1, j}} \right\rbrace _{j=1}^{\infty}\) that converge in the first column.

The idea is that \(\left\lbrace a _{2, N _{1,j}} \right\rbrace _{j=1}^{\infty}\) is a subsequence of \(\left\lbrace a _{2, N} \right\rbrace _{N=1}^{\infty}\), so it’s also bounded. By the lemma again, there is a converging subsequence \(\left\lbrace a _{2, N _{2,j}} \right\rbrace _{j=1}^{\infty}\). There’s a bit of abuse of notation so we don’t run out of letters to index by. Now \(\left\lbrace S _{N _{2,j}} \right\rbrace\) is a subsequence of rows of our table such that both the first and the second columns converge: \(\left\lbrace a _{1, N _{2,j}} \right\rbrace\) is a subsequence of \(\left\lbrace a _{1, N _{1,j}} \right\rbrace\), and subsequences of converging sequences also converge.

Iterate this subsequence construction. If \(\left\lbrace S _{N _{m,j}} \right\rbrace _{j=1}^{\infty}\) is the \(m\)-th iterated subsequence, which we may assume to converge in the first \(m\) columns, then we know that \(\left\lbrace a _{m+1, N _{m,j}} \right\rbrace _{j=1}^{\infty}\) is a subsequence of the \(m+1\)-th column. In particular, the \(m+1\)-th column of the \(m\)-th iterated subsequence is bounded, so it hase a convergent subsequence \(\left\lbrace a _{m+1, N _{m+1,j}} \right\rbrace _{j=1}^{\infty}\). Keeping the corresponding rows of the table yields the \(m+1\)-th iterated subsequence of sequences, \(\left\lbrace S _{N _{m+1,j}} \right\rbrace _{j=1}^{\infty}\).

To recap the abhorrent notation: \(N _{m, j}\) is the \(j\)-th index of the \(m\)-th iterated subsequence. Then \(S _{N _{m, j}}\) is the \(j\)-th row of the \(m\)-th iterated subsequence, and \(a _{n, N _{m,j}}\) is the \(n\)-th column of the \(j\)-th term of the \(m\)-th iterated subsequence. Awful!

There are two key features of this iterated construction. First, when \(n\leq m\), we have that \(\left\lbrace a _{n, N _{m,j}} \right\rbrace _{j=1}^{\infty}\) converges. That is, if you go down the \(n\)-th column of the \(m\)-th iterated subsequence, those numbers will converge as long as \(n\leq m\). Second, whenever \(m’\geq m\), \(\left\lbrace S _{N _{m’,j}}\right\rbrace _{j=1}^{\infty}\) is a subsequence of \(\left\lbrace S _{N _{m, j}}\right\rbrace _{j=1}^{\infty}\). That is to say, these iterated subsequences are all “nested” in the subsequences that come before it, like an infinite Matryoshka doll.

The “Diagonal” Subsequence

It’s tempting to just “take the limit” of these iterated subsequences, but it’s not always possible to make sense of this. If one continues taking iterated subsequences infinitely, then it’s possible to “run out” of rows in the table. For instance, it’s possible that the \(m\)-th iterated subsequence discards every \(m\)-th row of the table (for \(m>1\)). Then, every row will eventually be discarded by some iterated subsequence!

To address this problem, we consider for \(k=1,2,\ldots\) the subsequence of rows given by \(S _{N_k}= S _{N _{k, k}}\), i.e. the \(k\)-th term of the \(k\)-th iterated subsequence. This really is a subsequence of the rows! We claim that for all \(n\), the sequence \(\left\lbrace a _{n, N_k} \right\rbrace _{k=1}^{\infty}\) converges, which would conclude the proof.

Fix any \(n\). It’s not true that \(\left\lbrace a _{n, N_k} \right\rbrace _{k=1}^{\infty}\) is a subsequence of the \(n\)-th row of the \(k\)-th iterated subsequence! This is because the \(a _{n, N_1}\) comes from the first iterated subsequence, which need not be a part of the \(k\)-th subsequence.

This is not a completely bad thing. \(N_k\) is the index of the \(k\)-th row of the \(k\)-th subsequence; in particular, when \(k\geq n\), since the iterated subsequences are all nested in each other, \(S_{N_k}\) is a part of the \(n\)-th iterated subsequence! Therefore, \(\left\lbrace S _{N_k} \right\rbrace _{k=n}^{\infty}\) is a subsequence of the \(n\)-th iterated subsequence — notice that we started at \(k=n\) and chopped off the first \(n-1\) terms. So, since the \(n\)-th column of the \(n\)-th iterated subsequence converges, so too does \(\left\lbrace a _{n,N_k} \right\rbrace _{k=n}^{\infty}\). But the first \(n-1\) terms do not affect the convergence of the sequence, so \(\left\lbrace a _{n, N_k} \right\rbrace _{k=1}^{\infty}\) converges as well.

This is true for all \(n\in \mathbb{N}\), and we conclude that \(\left\lbrace a _{n, N_k} \right\rbrace _{k=1}^{\infty}\) converges for all \(n\). \(\square\)