Set of Inverse Positive Integers with Zero is Compact

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