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I have to prove the following:

We have two metric spaces $(X_1,d_1)$ and $(X_2,d_2)$ and their product-space $X=X_1 \times X_2$ with metric $d=d_1 \times d_2$ (so $d(x)=d(x_1,x_2)=d_1(x_1)+d_2(x_2)$ ). We have a projection $\pi_i$ which projects $X$ on $X_i$ for $i=1,2$.
First question was proofing the projection is continuous, which was fairly simple. The next question was where I got stuck upon: $\\$

Let $K=K_1 \times K_2$ with $K_i \subset X_i$. Proof:

$K$ is compact in $X$ $\iff$ $K_i$ is compact in $X_i$ for $i=1,2$.

"$\Rightarrow$" gave no problems: Let $K$ compact in $X$, $\pi_i$ is continuous, which implies that $\pi_i(K)=K_i$ is then compact. (Weierstrass)

"$\Leftarrow$" This is the problem. I tried with collection of open sets, but this went wrong. I hope someone can give a nice proof for this.

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  • $\begingroup$ I had success with the problem by proving the contrapositive: $K$ is not compact in $X$ implies $K_i$ is not compact in $X_i$ for $i=1$ or $i=2$. $\endgroup$ – graydad Oct 2 '14 at 16:27
  • $\begingroup$ Did you use an infinite open cover of $K$ for that? $\endgroup$ – user3704690 Oct 2 '14 at 16:38
  • $\begingroup$ Well, I figured if $K$ is not compact then there must be an open cover $\mathcal{C}$ with no finite subcover. Since each $C \in \mathcal{C}$ is of the form $C=U \times V$ for open sets $U \subset X_1$ and $V \subset X_2$, then you should be able to get to your conclusion by looking at $\pi_1$ and or $\pi_2$... If that makes any sense :) Or while doing the proof by contrapositive, you could do a proof by contradiction and suppose that $K_1, K_2$ are compact, yet $K$ is not. $\endgroup$ – graydad Oct 2 '14 at 16:42
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Proof by contrapositive: Suppose $K$ is not compact in $X$. Then there is an open cover $\mathcal{C} = \{C_\alpha \}_{\alpha \in \lambda}$ with no finite subcover. Each cover element is of the form $C_\alpha = C_{1_{\alpha}} \times C_{2_{\alpha}}$ where $C_{1_{\alpha}} \subset X_1$ and $C_{2_{\alpha}} \subset X_2$ are open sets. Now for the sake of contradiction, suppose $K_1$ and $K_2$ are compact. We know $\{C_{1_\alpha} \}_{\alpha \in \lambda}$ and $\{C_{2_\alpha} \}_{\alpha \in \lambda}$ are open covers of $K_1$ and $K_2$ respectively, so each has a finite subcover, say $$\{C_{1_i} \}_{i=1}^N \quad \text{and } \quad \{C_{2_j} \}_{j=1}^M$$ Further, for each $C_{1_{i}}$ we know there exists $C_{\alpha_{i}} \in \mathcal{C}$ such that $C_{\alpha_{i}} = C_{1_{i}} \times U$ for some open set $U \subset X_2$ (U is not important here.) Now we can make two finite subsets of $\mathcal{C}$ which we'll call $$\mathcal{C}_1 =\left\{ C_{\alpha_{i}} \in \mathcal{C}: C_{\alpha_{i}}=C_{1_{i}} \times U \space \text{where} \space C_{1_i} \in \{C_{1_i} \}_{i=1}^N\right\}_{i=1}^N$$and $$\mathcal{C}_2 =\left\{ C_{\alpha_{j}} \in \mathcal{C}: C_{\alpha_{j}}=V\times C_{2_{j}} \space \text{where} \space C_{2_j} \in \{C_{2_j} \}_{j=1}^M\right\}_{j=1}^M$$ By construction $\mathcal{C}_1 \cup \mathcal{C}_2 \subset \mathcal{C}$ and it can easily be shown that $\mathcal{C}_1 \cup \mathcal{C}_2$ is a finite subcover of $K$. Once you show that, you will have reached your contradiction.

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  • $\begingroup$ U has to be open right? Otherwise you cannot have an open cover of K. (Just a detail) $\endgroup$ – user3704690 Oct 2 '14 at 22:48
  • $\begingroup$ Yup! That is correct. $\endgroup$ – graydad Oct 2 '14 at 22:56
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Just use sequences... if $(x_n,y_n)$ are points in $K_1 \times K_2$, then one can find a subsequence $(x_{n_k}, y_{n_k})$ where the $x_{n_k}$ converge to some $x \in K_1$. Then take a further subsubsequence $(x_{n_{k_l}}, y_{n_{k_l}})$ for which the $y_{n_{k_l}}$ converge to some $y \in K_2$. Hence $(x_{n_{k_l}}, y_{n_{k_l}})$ converges to $(x,y) \in K_1 \times K_2$.

So every sequence in $K_1 \times K_2$ has a convergent subsequence, which is equivalent to compactness for metric spaces.

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