Product of compact metric spaces is again compact.

Proof Let $$(X_j,d_j)$$ be a compact metric space for $$j=1,...,n$$. Denote $$X=X_1 \times X_2 \times...\times X_n$$ to be the product of compact metric spaces.

Assume the property that the open sets in $$(X,d)$$ are unions of product sets of the form $$U_1 \times ... \times U_n$$, where $$U_j$$ is an open subset of $$X_j$$.

Suppose $$X\subset \bigcup_{\alpha\in A} U^{\alpha}$$, where $$U^{\alpha}=U_1^{\alpha}\times ... \times U_n^{\alpha}$$, and $$U_j^{\alpha}\subset X_j$$ are open.

Now, we have that $$X_j\subset \bigcup_{\alpha\in A} U_j^{\alpha}$$, for if $$x_j\in X_j$$, $$\exists y\in X$$ such that $$y_j=x_j$$, but $$y\in U^{\alpha}=U_1^{\alpha}\times ... \times U_n^{\alpha}$$ for some $$\alpha$$.

$$\implies y_j=x_j\in U_j^{\alpha}$$.

So by compactness of $$X_j$$, $$\exists$$ a finite subcover $$X_j\subset \bigcup^{m(j)}_{i=1} U_j^i$$, where $$X_j$$ is covered by $$m(j)$$ open sets for each $$j=1,...,n$$.

Finally, $$X\subset \bigcup_{1\leq i_j\leq m(j)} (U_1^{i_1} \times U_2^{i_2} \times ...\times U_n^{i_n})$$ is a union of $$\Pi^n_{i=1}m(i)<\infty$$ open subsets of $$X$$, hence $$X$$ is compact. end Proof

Is this a valid way of arguing? My main concern is that the open cover at the end of the proof is not actually a subcover... or that it contains a subcover plus some surplus open sets.

• The textbook I am using assumes this part. – Jungleshrimp Apr 19 at 2:50
• I am sorry. I misread. – amsmath Apr 19 at 2:54
• You can work with two spaces $X$ and $Y$. Then the rest is induction. Now, $X\times Y\subset \bigcup_\alpha (U_\alpha\times V_\alpha)$ and you infer correctly that $X\subset \bigcup_{i=1}^nU_{\alpha_i}$ and $Y\subset\bigcup_{j=1}^mU_{\alpha_j}$. Then $X\times Y\subset$ ??? – amsmath Apr 19 at 3:06
• But how do we know that $U_{\alpha_i} \times V_{\alpha_j}=U_{\alpha} \times V_{\alpha}$ for any $\alpha$? – Jungleshrimp Apr 19 at 3:37
• I guess what I was afraid of originally was that the sets $U_{\alpha_1},...,U_{\alpha_n}$ which cover $X$ do not always coincide with the $V_{\alpha_1},...,V_{\alpha_m}$ which cover $Y$. Is it possible that none of the $mn$ different combinations are actually within $\bigcup_{\alpha}(U_\alpha \times V_{\alpha})$? – Jungleshrimp Apr 19 at 3:55

We are starting out with a cover of the form $$U^\alpha_1 \times \ldots U^\alpha_n$$, where the $$\alpha \in A$$ vary. These are specific combinations of open sets in the factor spaces that are combined to form basic open sets of the product.
If you then independently (as you do) take a subcover $$U_i^\alpha$$ for $$X_i$$, $$\alpha$$ from some finite subset $$F_i\subseteq A$$ of indices, there is no guarantee at all that the sets you got from the finite subcover combine to the original products that you started with. It is true that
$$\{U^{\alpha_1}_1 \times \ldots \times U_n^{\alpha_n}: \forall i: \alpha_i \in F_i\}$$