How to prove that $A \cap B$ is compact using the definition of compact set

Question

Suppose that $A$ and $B$ are compact set in $\mathbb R^n$. Then $A \cap B$ is compact.

Using Heine-Borel Theorem, since $A$ and $B$ are both closed and bounded, $A \cap B$ is also closed and bounded, so it is compact.

But how can it be proved by 'definition' of compact set?

Answer 1

PART 1.Let $X$ be a compact space and let $Y$ be a closed subspace of $X$. Then $Y$ is compact.PROOF. Let $F$ be a family of open subsets of $Y$ with $\cup F=Y$. Let $F^*=\{f^* : f\in F\}$ where, for each $f \in F$ we have $f=f^*\cap Y$ with $f^*$ being an open subset of $X$. Now $ G=F^*\cup \{X\backslash Y\}$ is an open cover of $X$ so there exists a finite $H\subset G$ with $\cup H=X$. We must have $H\subset (\{f^*:f\in J\}\cup \{X\backslash Y\}$ where $J$ is a finite subset of $F$. Since $\cup H=X$ we must have $\cup \{f^* :f\in J\}\supset Y$. Therefore $Y=Y\cap (\cup f^* :f\in J\})=\cup \{Y\cap f^* :f\in J\} =\cup J. $ So the open cover $F$, of $Y$, has a finite sub-cover $J. $ PART 2. Let $X$ be a Hausdorff space and let $Y$ be a compact subspace of $X$. Then $Y$ is closed in $X$.PROOF.For $Y=\phi$ this is trivial.For $Y\ne \phi$ we show that any $p\in X\backslash Y$ has a nbhd in $X$ which is disjoint from $Y$ : For each $q\in Y$ let $U_q$ and $V_q$ be disjoint open sets of $X$ with $q\in U_q$ and $p\in V_q$.Then $\{U_q :q\in Y\}$ is an open cover of $Y$, so there exists a non-empty finite $J\subset Y$ such that $\cup_{j\in J}U_j\supset Y. $ Now $\cap_{j\in J}V_j$ is a nbhd of $P$ which is disjoint from $Y. $ PART 3. If $A,B$ are compact subsets of $R^n$ then $A,B$ are closed in $R^n$ by Part 2. So $A\cap B$ is a closed subspace of $A$. So by Part 1, $A\cap B$ is compact.

Answer 2

You can prove this more generally for any Hausdorff space - If $A$ and $B$ are compact subsets of a Hausdorff space $X$ then $A\cap B$ is compact.

This follows from the facts that -

1. A compact subset of a Hausdorff space is closed. (hence $A$ and $B$ are closed)

2. A closed subset of a compact space is compact. (hence $A\cap B$ which is closed in $A$ as well as $B$ is compact)

The proofs of 1 and 2 are by "definition"