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.

  • $\begingroup$ The textbook I am using assumes this part. $\endgroup$ – Jungleshrimp Apr 19 at 2:50
  • $\begingroup$ I am sorry. I misread. $\endgroup$ – amsmath Apr 19 at 2:54
  • 2
    $\begingroup$ 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 $X\times Y\subset$ ??? $\endgroup$ – amsmath Apr 19 at 3:06
  • $\begingroup$ But how do we know that $U_{\alpha_i} \times V_{\alpha_j}=U_{\alpha} \times V_{\alpha}$ for any $\alpha$? $\endgroup$ – Jungleshrimp Apr 19 at 3:37
  • 1
    $\begingroup$ 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 $\bigcup_{\alpha}(U_\alpha \times V_{\alpha})$? $\endgroup$ – 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

$$\{U^{\alpha_1}_1 \times \ldots \times U_n^{\alpha_n}: \forall i: \alpha_i \in F_i\}$$

is a finite cover of the product, it is not a subcover of the original one.

You can use the tube lemma (or variations thereof) to show that the product of two compact (not necessarily metric, general spaces will do..) spaces is compact and the finite product case then follows by standard finite induction. The infinite product case does require new techniques.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.