Proof of The Diagonal Subsequence Lemma for Bounded Real Arrays

Throughout, a strictly increasing sequence is a sequence in $\mathbb{N}$ that is strictly increasing in the sense of Subsequence of a Sequence in a Set; for such $\sigma=(\sigma_{i})_{i\in\mathbb{N}}$ we write $\sigma(i)=\sigma_{i}$. By A Subsequence of a Subsequence is a Subsequence, the composition $\sigma\circ\rho$ of two strictly increasing sequences is strictly increasing (claim 2), $\sigma$ preserves strict order (claim 1), and $i\le\sigma(i)$ for all $i$ by Strictly Increasing Sequences of Natural Numbers Dominate Their Index. Convergence of real sequences is that of Limit of a Sequence of Real Numbers, which by claim 1 of Convergence, Limits and the Cauchy Condition for Real Sequences Agree with Those in the Real Line as a Metric Space, which is Complete agrees with convergence in the real line $(\mathbb{R},d_{\mathbb{R}})$ of The Absolute Value Metric on the Real Numbers; by A Subsequence of a Convergent Sequence Has the Same Limit a subsequence of a convergent real sequence converges. This proof uses Axiom of Dependent Choice.

Step 1: one column. Let $\sigma$ be a strictly increasing sequence and $k\in\mathbb{N}$. The real sequence $(a_{\sigma(i),k})_{i}$ takes values in the closed interval $[-|R_{k}|,|R_{k}|]$, because $|a_{\sigma(i),k}|\le R_{k}\le|R_{k}|$ (claim 3 of Properties of the Absolute Value in an Ordered Field) and claim 6 there. This interval is sequentially compact in $(\mathbb{R},d_{\mathbb{R}})$ by A Closed Interval is Sequentially Compact in the Real Line, so there is a strictly increasing $\rho$ such that $(a_{\sigma(\rho(i)),k})_{i}$ converges.

Step 2: dependent choice. Let $\Sigma$ be the set of all strictly increasing sequences, and let $S\subseteq\mathbb{N}\times\Sigma$ be the set of pairs $(k,\sigma)$ such that $(a_{\sigma(i),k'})_{i}$ converges for every $k'\in[k]$. Applying Step 1 to the identity sequence $\sigma(i)=i$ and $k=1$ gives $\sigma^{\ast}=\rho$ with $(1,\sigma^{\ast})\in S$, so $S\ne\varnothing$. Let $\mathcal{R}\subseteq S\times S$ be the relation consisting of the pairs $\bigl((k,\sigma),(k+1,\sigma\circ\rho)\bigr)$ of elements of $S$ with $\rho$ strictly increasing. For every $(k,\sigma)\in S$ there is $\tau$ with $\bigl((k,\sigma),(k+1,\tau)\bigr)\in\mathcal{R}$ and $(k+1,\tau)\in S$: take $\rho$ from Step 1 for $\sigma$ and the column $k+1$, and $\tau=\sigma\circ\rho$; then $(a_{\tau(i),k+1})_{i}$ converges by construction, and for $k'\in[k]$ the sequence $(a_{\tau(i),k'})_{i}$ is a subsequence of the convergent $(a_{\sigma(i),k'})_{i}$, hence converges. By Axiom of Dependent Choice there is a sequence $\bigl((k_{j},\sigma^{(j)})\bigr)_{j\in\mathbb{N}}$ in $S$ with $(k_{1},\sigma^{(1)})=(1,\sigma^{\ast})$ and each consecutive pair in $\mathcal{R}$. By Principle of Induction for the Natural Numbers, $k_{j}=j$ for every $j$, and for each $j$ there is a strictly increasing $\rho^{(j)}$ with $\sigma^{(j+1)}=\sigma^{(j)}\circ\rho^{(j)}$.

Step 3: the diagonal. Put $n_{j}=\sigma^{(j)}(j)$. Then $n_{j+1}=\sigma^{(j)}\bigl(\rho^{(j)}(j+1)\bigr)$ and $\rho^{(j)}(j+1)\ge j+1>j$, so $n_{j+1}>\sigma^{(j)}(j)=n_{j}$ by order preservation; thus $(n_{j})$ is strictly increasing. Fix $k\in\mathbb{N}$. By induction on $j\ge k$ (the set of $j$ with $j<k$ or with the following property contains $1$ and is closed under successor), for every $j\ge k$ there is a strictly increasing $\pi_{j}$ with $\sigma^{(j)}=\sigma^{(k)}\circ\pi_{j}$: take $\pi_{k}$ the identity and $\pi_{j+1}=\pi_{j}\circ\rho^{(j)}$. Put $i_{j}=\pi_{j}(j)$ for $j\ge k$, so $n_{j}=\sigma^{(k)}(i_{j})$, and $i_{j+1}=\pi_{j}(\rho^{(j)}(j+1))>\pi_{j}(j)=i_{j}$ as above. Since $(k,\sigma^{(k)})\in S$, the sequence $(a_{\sigma^{(k)}(i),k})_{i}$ converges to some $L\in\mathbb{R}$. Let $\varepsilon>0$ and choose $I\in\mathbb{N}$ with $|a_{\sigma^{(k)}(i),k}-L|<\varepsilon$ for all $i\ge I$. The sequence $(\tilde{\imath}_{l})_{l\in\mathbb{N}}$ given by $\tilde{\imath}_{l}=i_{l+k-1}$ is strictly increasing, so $l\le\tilde{\imath}_{l}$ for every $l$ by Strictly Increasing Sequences of Natural Numbers Dominate Their Index. For $j\ge k+I$, claim 7 of Properties of the Order on the Natural Numbers (applied to $k-1<j$ when $k>1$, and trivially $l=j$ when $k=1$) gives a unique $l\in\mathbb{N}$ with $j=l+k-1$, and $l\ge I+1$ since $j\ge k+I$; then $i_{j}=\tilde{\imath}_{l}\ge l\ge I$, so $|a_{n_{j},k}-L|=|a_{\sigma^{(k)}(i_{j}),k}-L|<\varepsilon$. Hence $(a_{n_{j},k})_{j}$ converges to $L$ (with the threshold $k+I$ in Limit of a Sequence of Real Numbers). As $k$ was arbitrary, the strictly increasing sequence $(n_{j})$ has the required property.