If $A \subset \Bbb R^n$ is compact and $x_o \in \Bbb R^m$, then $A \times \{x_o\}$ is compact.

Question

I am having difficulty understand a proof in my Real Analysis textbook.

Statement: If $A \subset \Bbb R^n$ is compact and $x_o \in \Bbb R^m$, then $A \times \{x_o\}$ is compact.

Proof: Let $U^*$ be an open cover of $A \times \{x_o\}$, and $V^* = \{V | V=\{ y|(y, x_o) \in U \}, U \in U^* \}$ (I realize the notation for this set is awkward but this is what is written in my textbook). It then states without proof that $V^*$ is an open cover of $A$. I am not sure why this is true. Any hints as to why appreciated.

Answer 1

Consider the canonical projection $p:\mathbb{R}^n\times\mathbb{R}^m\rightarrow \mathbb{R}^n$ defined by $p(x,y)=x$. Let $U$ be an open subset of $A$, there exists an open subset $U'$ of $\mathbb{R}^n$ such that $U=U'\cap A$. Since $p$ is continuous, $p^{-1}(U')=V'$ is an open subset of $\mathbb{R}^n\times\mathbb{R}^m$, $V=V'\cap A\times\{x_0\}$. We deducce that $V$ is an open subset of $A\times\{x_0\}$.

Answer 2

If you would like to use a theorem, its just one-liner. We will use Tychonoff's theorem which states that the product of any collection of Compact topological spaces is compact w.r.t the product topology.

Here since $\{x_0\} \in \mathbb{R}^{n}$ is compact and $A$ is compact, by the Tychonoff's theorem their product is also compact.

Answer 3

At present, only alternative proofs have been shared, without answers to your original question: Why is this $V^*$ an open cover of $A$?

To prove this, we need to prove two things:

1. Every $V \in V^*$ is an open subset of $A$
2. Every element $a \in A$ is in some $V \in V^*$.

I think you can prove those things, but they are not immediate, and it takes some work to unpack that definition. Use the fact that $U \subset A$ is open $\iff$ $U \times \{x_0\} \subset A \times \{x_0\}$ is open (this is not the definition of the product topology but is true for this case).

To clarify, your confusion may be a typo in the textbook. It should say $V^*$ is an open cover for $A$ (emphasis on the $*$).