If $B \subset \mathbb R^m$ is compact and $x\in \mathbb R^n$ then $\{x\} \times B \subset \mathbb R^{n+m}$ is compact.

Question

If $B \subset \mathbb R^m$ is compact and $x\in \mathbb R^n$ then $\{x\} \times B \subset \mathbb R^{n+m}$ is compact.

If $ \mathcal O$ is open cover of $\{x\} \times B \subset \mathbb R^{n+m}$ then as $B$ is compact there is finite cover $ \mathcal O'$ of $B$ and hence finite open sets $U$ s.t. $ \{x\} \times U' \subset U \in \mathcal O $ where $U' \in \mathcal O'$.

Is this argument correct?

Answer 1

No. Of what cover is $\mathcal O'$ a subcover ?? Let $p: \mathbb R^m \to \mathbb R^{n+m}$ be defined by $p(y):=(x,y)$. Then $p$ is continuous. Hence $p^{-1}(G)$ is open for each open set $G \subseteq R^{n+m}$.

Now let $\{O_i: i \in I\}$ be an open cover of $\{x\} \times B$. Then $\{p^{-1}(O_i): i \in I\}$ be an open cover of $B$.

Can you proceed ?