Countable Product of Sequentially Compact Spaces is Sequentially Compact

Theorem
Let $I$ be an indexing set with countable cardinality.

Let $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be a family of topological spaces indexed by $I$.

Let $\ds \struct {S, \tau} = \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the product space of $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$.

Let each of $\struct {S_\alpha, \tau_\alpha}$ be sequentially compact.

Then $\struct {S, \tau}$ is also sequentially compact.

Proof
Let $\sequence { {\mathbf x}_n}$ be a sequence in $S$.

That is, for each $n \in \N$, ${\mathbf x}_n$ is a tuple $\family {x_n^i}_{i \mathop \in I}$ indexed by $I$.

We claim there is a strictly increasing sequence $\sequence {n_r}$ in $\N$ such that the subsequence $\sequence { {\mathbf x}_{n_r} }$ converges in $\struct {S, \tau}$.

By Sequence on Product Space Converges to Point iff Projections Converge to Projections of Point, it suffices to construct a strictly increasing sequence $\sequence {n_r}$ in $\N$ such that the subsequence $\sequence { \map {\pr_i} { {\mathbf x}_{n_r} } }$ converges in $\struct {S_i, \tau_i}$ for every $i \in I$.

Let $\set {i_0, i_1, \ldots}$ be an enumeration of $I$.

By induction, for each $m \in \N$ we construct a strictly increasing sequence $\sequence {n^m_r}$ in $\N$ and an element $x^{i_m} \in S_{i_m}$ as follows.

By sequential compactness of $\struct {S_{i_0}, \tau_{i_0} }$, there is a strictly increasing sequence $\sequence {n^0_r}$ in $\N$ such that the subsequence $\sequence { x^{i_0}_{n^0_r} }$ converges in $\struct {S_{i_0}, \tau_{i_0} }$ to some point $x^{i_0} \in S_{i_0}$.

Suppose inductively that the strictly increasing sequence $\sequence {n^m_r}$ has been defined for some $m \in \N$.

Applying sequential compactness of $\struct {S_{i_{m+1} }, \tau_{i_{m+1} } }$ to the sequence $\sequence { x^{i_{m+1} }_{n^m_r} }$, there is a subsequence $\sequence {n^{m+1}_r}$ of $\sequence {n^m_r}$ such that the subsequence $\sequence { x^{i_{m+1} }_{n^{m+1}_r} }$ converges in $\struct {S_{i_{m+1} }, \tau_{i_{m+1} } }$ to some point $x^{i_{m+1} } \in S_{i_{m+1} }$.

Moreover, by each $\sequence { x^{i_m}_{n^{m'}_r} }$ is a subsequence of $\sequence { x^{i_m}_{n^m_r} }$ whenever $m,m' \in \N$ such that $m' \ge m$.

Since $I$ is countable, consider separately the cases where $I$ is finite and $I$ is countably infinite.

Case I
Suppose $I = \set {i_0, \ldots, i_M}$ is finite.

Then the above inductive process must terminate after $M + 1$ steps.

Define $\sequence {n_r} := \sequence {n^M_r}$.

Define $\mathbf x := \family {x^i}_{i \mathop \in I} \in S$.

Let $m \le M$.

We claim that the subsequence $\sequence { \map {\pr_{i_m} } { {\mathbf x}_{n_r} } }$ converges to $\map {\pr_{i_m} } { \mathbf x }$.

By construction, the subsequence $\sequence { x^{i_m}_{n^m_r} }$ converges to $x^{i_m}$.

By Subsequence of Subsequence, $\sequence { x^{i_m}_{n^M_r} }$ is a subsequence of $\sequence { x^{i_m}_{n^m_r} }$.

Then:

The result follows.

Case II
Suppose $I = \set {i_0, i_1, \ldots }$ is countably infinite.

Then by axiom of dependent choice, the inductive process gives a strictly increasing sequence $\sequence {n^m_r}$ in $\N$ and an element $x^{i_m} \in S_{i_m}$ for every $m \in \N$.

Define $\sequence {n_r} := \sequence {n^r_r}$.

Define $\mathbf x := \family {x^i}_{i \mathop \in I} \in S$.

Let $m \in \N$.

We claim that the subsequence $\sequence { \map {\pr_{i_m} } { {\mathbf x}_{n_r} } }$ converges to $\map {\pr_{i_m} } {\mathbf x}$.

By Subsequence of Subsequence, $\sequence { x^{i_m}_{n^{r}_r} }$ is a subsequence of $\sequence { x^{i_m}_{n^m_r} }$.

Then:

The result follows.