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