272: Finite Product of Compact Topological Spaces Is Compact

<The previous article in this series | The table of contents of this series | The next article in this series>

A description/proof of that finite product of compact topological spaces is compact

---

Topics

About: topological space

---

The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

---

Starting Context

• The reader knows a definition of compact topological space.
• The reader knows a definition of product topology.
• The reader admits the proposition that the product of any finite number of topological spaces equals the sequential products of the topological spaces.

---

Target Context

• The reader will have a description and a proof of the proposition that the product of any finite number of compact topological spaces is compact.

---

Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

---

Main Body

---

1: Description

For any compact topological spaces, $T_1,T_2,...,T_n$, the product topological space, $T=T_1\times T_2\times ...\times T_n$, is compact.

---

2: Proof

1st, let us think of the case, $n=2$.

For any open cover, $\{U_{\alpha }\subseteq T|\alpha \in A\}$ where $A$ is a possibly uncountable indices set, of $T$, $U_{\alpha }={\cup }_{\beta \in B_{\alpha }}{U}_{1,\alpha ,\beta }\times U_{2,\alpha ,\beta }$ where $B_{\alpha }$ is a possibly uncountable indices set that depends on $\alpha$.

For each point, $p\in T_1$, there is the subset, $S_p:=\{U_{1,\alpha ,\beta }\times U_{2,\alpha ,\beta }|(\alpha ,\beta )\in C_p\}\}\subseteq \{U_{1,\alpha ,\beta }\times U_{2,\alpha ,\beta }|\alpha \in A,\beta \in B_{\alpha }\}$ where $C_p$ is the possibly uncountable indices set, such that $p\in U_{1,\alpha ,\beta }$ for each $(\alpha ,\beta )\in C_p$. $S_{2,p}:=\{U_{2,\alpha ,\beta }|(\alpha ,\beta )\in C_p\}$ covers $T_2$.

For $S_{2,p}$, there is a finite subcover, $\{U_{2,\alpha ,\beta }|(\alpha ,\beta )\in D_p\}$ where $D_p\subseteq C_p$ is a finite indices set, and there is corresponding $\{U_{1,\alpha ,\beta }\times U_{2,\alpha ,\beta }|(\alpha ,\beta )\in D_p\}\subseteq S_p$ for each $p$.

Let us take $U_{1,p}^{\prime }:={\cap }_{(\alpha ,\beta )\in D_p}{U}_{1,\alpha ,\beta }$, open on $T_1$ as a finite intersection of open sets. $\{U_{1,p}^{\prime }|p\in T_1\}$ covers $T_1$ and there is a finite subcover, $\{U_{1,p}^{\prime }|p\in E\}$ where $E\subseteq T_1$ is a finite subset.

Let us take $F:=\{\alpha \in A|\mathrm{\exists }\beta \in B_{\alpha }((\alpha ,\beta )\in D_p,p\in E)\}$, which is a finite indices set, because $E$ is finite and $D_p$ is finite for each $p$.

$\{U_{\alpha }|\alpha \in F\}$ covers $T_1\times T_2$, because for any $(p^1,p^2)\in T_1\times T_2$, $p^1\in U_{1,p}^{\prime }\subseteq U_{1,\alpha ,\beta }$ for a $p\in E$ and each $(\alpha ,\beta )\in D_p$ and $p_2\in U_{2,\alpha ,\beta }$ for a $(\alpha ,\beta )\in D_p$, because $\{U_{2,\alpha ,\beta }|(\alpha ,\beta )\in D_p\}$ covers $T_2$, so, $(p^1,p^2)\in U_{1,\alpha ,\beta }\times U_{2,\alpha ,\beta }$ for a $(\alpha ,\beta )\in D_p$, while $U_{1,\alpha ,\beta }\times U_{2,\alpha ,\beta }\subseteq U_{\alpha }$ where $\alpha \in F$.

So, $\{U_{\alpha }|\alpha \in F\}$ is a finite subcover of $\{U_{\alpha }|\alpha \in A\}$.

So, $T_1\times T_2$ is compact.

For any general $n$, $T_1\times T_2\times ...\times T_n=(...(T_1\times T_2)\times ...)\times T_n$, by the proposition that the product of any finite number of topological spaces equals the sequential products of the topological spaces, but $T_1\times T_2$ is compact by the above paragraph, then, $(T_1\times T_2)\times T_3$ is compact by the above paragraph, and so on (we can use the induction principle to be exact). So, $T_1\times T_2\times ...\times T_n$ is compact.

---

References

<The previous article in this series | The table of contents of this series | The next article in this series>