Proof of The Subsequence Criterion for Convergence in a Metric Space

Conventions. Order facts about $\mathbb{N}$ are those of Properties of the Order on the Natural Numbers, and $S$ denotes the successor map, so that $S(l)=l+1$ by the definition of addition on $\mathbb{N}$. Elementary order facts about $\mathbb{R}$ are those of Elementary Order Arithmetic in an Ordered Field; the order $\le$ of the ordered field $\mathbb{R}$ is a total order, so its reflexivity and totality are axioms of that definition. No choice principle is used: the indices constructed in the proof of claim 2 are least elements, supplied by The Natural Numbers Are Well Ordered.

Proof of claim 1. Let $(n_{k})_{k\in\mathbb{N}}$ be strictly increasing and let

The number $1$ lies in $B$ vacuously: every $k\in\mathbb{N}$ satisfies $1\le k$ by claim 4 of Properties of the Order on the Natural Numbers, that is $1<k$ or $1=k$, and in either case $k<1$ is excluded by the trichotomy of claim 3 of that lemma.

Let $l\in B$ and let $k\in\mathbb{N}$ satisfy $k<S(l)$. Then $k\le S(l)$ and $k\ne S(l)$ by claims 1 and 3 of Properties of the Order on the Natural Numbers, hence $k\le l$ by claim 5, that is $k<l$ or $k=l$. Since $(n_{k})$ is strictly increasing we have $n_{l}<n_{l+1}=n_{S(l)}$. If $k=l$ this already gives $n_{k}<n_{S(l)}$; if $k<l$ then $n_{k}<n_{l}$ because $l\in B$, and $n_{k}<n_{S(l)}$ follows by the transitivity of claim 1 of Properties of the Order on the Natural Numbers. Hence $S(l)\in B$, and $B=\mathbb{N}$ by Principle of Induction for the Natural Numbers. This is the first assertion of claim 1.

Now let $(k_{l})_{l\in\mathbb{N}}$ be strictly increasing. For every $l$ we have $k_{l}<k_{l+1}$, hence $n_{k_{l}}<n_{k_{l+1}}$ by the assertion just proved, so $(n_{k_{l}})_{l\in\mathbb{N}}$ is strictly increasing. Consequently, if $(x_{n_{k}})_{k\in\mathbb{N}}$ is a subsequence of $(x_{m})_{m\in\mathbb{N}}$ and $(x_{n_{k_{l}}})_{l\in\mathbb{N}}$ is a subsequence of it, then $(x_{n_{k_{l}}})_{l\in\mathbb{N}}$ is a subsequence of $(x_{m})_{m\in\mathbb{N}}$, by the definition of a subsequence.

Proof of claim 2. Suppose, seeking a contradiction, that $(x_{m})_{m\in\mathbb{N}}$ does not converge to $x$ in $(X,d)$. Then there is a positive $\varepsilon_{0}\in\mathbb{R}$ such that for every $N\in\mathbb{N}$ there is $m\in\mathbb{N}$ with $N\le m$ for which $d(x_{m},x)<\varepsilon_{0}$ fails. For such an $m$ we have $\varepsilon_{0}\le d(x_{m},x)$: the order of $\mathbb{R}$ is total, so $d(x_{m},x)\le\varepsilon_{0}$ or $\varepsilon_{0}\le d(x_{m},x)$, and in the first case $d(x_{m},x)\ne\varepsilon_{0}$ would give $d(x_{m},x)<\varepsilon_{0}$, so that $d(x_{m},x)=\varepsilon_{0}$ and the second alternative holds by reflexivity. Put

so that for every $N\in\mathbb{N}$ there is $m\in T$ with $N\le m$; in particular $T$ is nonempty.

Define a sequence $(n_{k})_{k\in\mathbb{N}}$ in $\mathbb{N}$ by recursion on $k$, at each step taking a least witness. Put $n_{1}=\min T$, which exists by The Natural Numbers Are Well Ordered. Given $n_{k}$, the set $\{m\in T:n_{k}<m\}$ is nonempty: there is $m\in T$ with $S(n_{k})\le m$, and $n_{k}<S(n_{k})$ by claim 5 of Properties of the Order on the Natural Numbers, so $n_{k}<m$ by the transitivity of claim 1 of that lemma. Put $n_{k+1}=\min\{m\in T:n_{k}<m\}$, again by The Natural Numbers Are Well Ordered. Then $n_{k}<n_{k+1}$ for every $k\in\mathbb{N}$, so $(n_{k})_{k\in\mathbb{N}}$ is strictly increasing, and $n_{k}\in T$ for every $k$.

By hypothesis, the subsequence $(x_{n_{k}})_{k\in\mathbb{N}}$ has in turn a subsequence converging to $x$: there is a strictly increasing $(k_{l})_{l\in\mathbb{N}}$ in $\mathbb{N}$ such that $(x_{n_{k_{l}}})_{l\in\mathbb{N}}$ converges to $x$ in $(X,d)$. Applying the definition of convergence with the positive real $\varepsilon_{0}$ produces $L\in\mathbb{N}$ with $d(x_{n_{k_{L}}},x)<\varepsilon_{0}$. But $n_{k_{L}}\in T$, so $\varepsilon_{0}\le d(x_{n_{k_{L}}},x)$, and the mixed transitivity of claim 2 of Elementary Order Arithmetic in an Ordered Field gives $\varepsilon_{0}<\varepsilon_{0}$, contradicting the irreflexivity of the strict order. Hence $(x_{m})_{m\in\mathbb{N}}$ converges to $x$ in $(X,d)$. $\blacksquare$