Tychonoff's Theorem/General Case/Proof 1

Theorem

Let $I$ be an indexing set.

Let $\family {X_i}_{i \mathop \in I}$ be an indexed family of non-empty topological spaces.

Let $\ds X = \prod_{i \mathop \in I} X_i$ be the corresponding product space.

Then $X$ is compact if and only if each $X_i$ is compact.

Proof

First assume that $X$ is compact.

From Projection from Product Topology is Continuous, the projections $\pr_i : X \to X_i$ are continuous.

It follows from Continuous Image of Compact Space is Compact that the $X_i$ are compact.

Assume now that each $X_i$ is compact.

By Equivalence of Definitions of Compact Topological Space it is enough to show that every ultrafilter on $X$ converges.

Thus let $\FF$ be an ultrafilter on $X$.

From Image of Ultrafilter is Ultrafilter, for each $i \in I$, the image filter $\map {\pr_i} \FF$ is an ultrafilter on $X_i$.

Each $X_i$ is compact by assumption.

So by Equivalence of Definitions of Compact Topological Space, each $\map {\pr_i} \FF$ converges.

By Filter on Product Space Converges iff Projections Converge, $\FF$ converges.

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$\blacksquare$

Axiom of Choice

This theorem depends on the Axiom of Choice, by way of Filter on Product Space Converges iff Projections Converge.

Because of some of its bewilderingly paradoxical implications, the Axiom of Choice is considered in some mathematical circles to be controversial.

Most mathematicians are convinced of its truth and insist that it should nowadays be generally accepted.

However, others consider its implications so counter-intuitive and nonsensical that they adopt the philosophical position that it cannot be true.