Topological Product of Compact Spaces/Finite Product

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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 if and only if 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 if and only if 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 if and only if $T_1$ 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 if and only if 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 if and only if 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:

\(\ds \prod_{i \mathop = 1}^{k + 1} T_i\) \(\text {is}\) \(\ds \text {compact}\)
\(\ds \leadstoandfrom \ \ \) \(\ds \prod_{i \mathop = 1}^k T_i, T_{k + 1}\) \(\text {are}\) \(\ds \text {compact}\) Topological Product of Compact Spaces
\(\ds \leadstoandfrom \ \ \) \(\ds T_1, T_2, \ldots, T_k, T_{k + 1}\) \(\text {are}\) \(\ds \text {compact}\) Induction Hypothesis


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 if and only if all of $T_1, T_2, \ldots, T_n$ are compact

$\blacksquare$


Also see


Sources