Uncountable Product of Sequentially Compact Spaces is not always Sequentially Compact

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Theorem

Let $I$ be an indexing set with uncountable 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 it is not necessarily the case that $\struct {S, \tau}$ is also sequentially compact.


Proof

Let $\mathbb I$ denote the closed unit interval.

Let $T = \struct {\mathbb I, \tau}$ be the topological space consisting of $\mathbb I$ under the usual (Euclidean) topology.

Let $T' = \mathbb I^{\mathbb I} = \struct {\ds \prod_{\alpha \mathop \in \mathbb I} \struct {\mathbb I, \tau}_\alpha, \tau'}$ be the uncountable Cartesian product of $\struct {\mathbb I, \tau}$ indexed by $\mathbb I$ with the product topology $\tau'$.


From Closed Real Interval is Sequentially Compact, $T$ is sequentially compact.


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But from Uncountable Cartesian Product of Closed Unit Interval is not Sequentially Compact, $T$ is not a sequentially compact space.

Hence the result.

$\blacksquare$


Sources

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