Dilation of Compact Set in Topological Vector Space is Compact

Theorem

Let $k$ be a topological field.

Let $X$ be a topological vector space over $X$.

Let $K$ be a compact subset of $X$.

Let $t \in k \setminus \set {0_k}$.

Then $t K$ is compact.

Proof 1

Let $\family {U_\alpha : \alpha \in I}$ be open sets such that:

$\ds t K \subseteq \bigcup_{\alpha \mathop \in I} U_\alpha$

From Dilation of Union of Subsets of Vector Space, we have:

$\ds K \subseteq \bigcup_{\alpha \mathop \in I} \paren {t^{-1} U_\alpha}$

From Dilation of Open Set in Topological Vector Space is Open, we have that $t^{-1} U_\alpha$ is open.

Since $K$ is compact, there exists $t^{-1} U_1, t^{-1} U_2, \ldots, t^{-1} U_n$ such that:

$\ds K \subseteq \bigcup_{i \mathop = 1}^n \paren {t^{-1} U_i}$

Then, using Dilation of Union of Subsets of Vector Space again we have:

$\ds t K \subseteq \bigcup_{i \mathop = 1}^n U_i$

So every open cover of $t K$ has a finite subcover.

So $t K$ is compact.

$\blacksquare$

Proof 2

From Dilation Mapping on Topological Vector Space is Continuous, the mapping $c_t : X \to X$ defined by:

$\map {c_t} x = t x$

for each $x \in X$ is continuous.

From Continuous Image of Compact Space is Compact:

$\map {c_t} K = t K$ is compact.

$\blacksquare$