Intersection of Compact and Closed Subsets of Normed Finite-Dimensional Real Vector Space with Euclidean Norm is Compact

Theorem

Let $\struct {\R^d, \norm {\, \cdot \,}_2}$ be the normed finite-dimensional real vector space with Euclidean norm.

Let $K$ be a compact subset of $\struct {\R^d, \norm {\, \cdot \,}_2}$.

Let $F$ be a closed subset of $\struct {\R^d, \norm {\, \cdot \,}_2}$.

Then $F \cap K$ is compact in $\struct {\R^d, \norm {\, \cdot \,}_2}$.

Proof

By assumption, $K$ is compact.

We have that a compact subset of normed vector space is closed and bounded.

Hence, $K$ is closed and bounded.

Since $K$ is bounded:

$\exists C \in \R_{> 0} : \forall \mathbf x \in K : \norm {\mathbf x}_2 \le C$.

Then:

$\forall \mathbf x \in K \cap F : \norm {\mathbf x}_2 \le C$

Hence, $K \cap F$ is bounded.

By assumption, $K$ and $F$ are closed.

We have that intersection of closed sets is closed in normed vector space.

Therefore, $K \cap F$ is closed.

By Heine-Borel theorem, $K \cap F$ is compact.

$\blacksquare$

Sources

• 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): Chapter $\S 1.5$: Normed and Banach spaces. Compact sets