Set of Inverse Positive Integers with Zero is Compact
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Theorem
Let $K$ be the set of inverse positive integers with zero:
- $\ds K := \set {1, \frac 1 2, \frac 1 3, \dots} \cup \set 0$
Let $\struct {\R, \size {\, \cdot \,}}$ be the normed vector space of real numbers.
Then $K$ is compact in real numbers.
Proof
We have that $K \subset \closedint 0 1$.
Hence, $K$ is bounded.
Furthermore:
- $\ds \R \setminus K = \openint {-\infty} 0 \cup \paren {\bigcup_{n \mathop = 1}^\infty \openint {\frac 1 {n + 1}} {\frac 1 n}} \cup \openint 1 \infty$
By Union of Open Sets of Normed Vector Space is Open, $\R \setminus K$ is open.
By definition, $K$ is closed.
By Heine-Borel theorem, $K$ 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