Topological Product of Compact Spaces/Finite Product

Theorem
Let $T_1, T_2, \ldots, T_n$ be topological spaces.

Let $\ds \prod_{i \mathop = 1}^n T_i$ be the product space of $T_1, T_2, \ldots, T_n$.

Then $\ds \prod_{i \mathop = 1}^n T_i$ is compact all of $T_1, T_2, \ldots, T_n$ are compact.

Proof
Proof by induction:

For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
 * $\ds \prod_{i \mathop = 1}^n T_i$ is compact all of $T_1, T_2, \ldots, T_n$ are compact

Basis for the Induction
$\map P 1$ is the case:
 * $T_1$ is compact $T_1$ is compact

which is trivially true.

Thus $\map P 1$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is the induction hypothesis:
 * $\ds \prod_{i \mathop = 1}^k T_i$ is compact all of $T_1, T_2, \ldots, T_k$ are compact

from which it is to be shown that:
 * $\ds \prod_{i \mathop = 1}^{k + 1} T_i$ is compact all of $T_1, T_2, \ldots, T_{k + 1}$ are compact

Induction Step
This is the induction step:

We have:
 * $\ds \prod_{i \mathop = 1}^{k + 1} T_i = \paren {\prod_{i \mathop = 1}^k T_i} \times T_{k + 1}$

Hence:

So $\map P k \implies \map P {k + 1}$ and thus it follows by the Principle of Mathematical Induction that for any $n \ge 1$:


 * $\ds \prod_{i \mathop = 1}^n T_i$ is compact all of $T_1, T_2, \ldots, T_n$ are compact

Also see

 * Tychonoff's Theorem, where this result is extended to the topological product of any infinite number of topological spaces.