Basic Properties of Compact Sets in a LCTVS

Table of Contents

Proposition 1: Let

$E$

be a locally convex topological vector space. If

K \subseteq E

is compact then

$K$

is precompact and hence bounded.

Proof: Let $$K$$ be compact and let $$U$$ be an absolutely convex neighbourhood of the origin. Then there exists an open set $$V$$ such that $V \subseteq U$ . Then $\{ a + V : a \in K \}$ is clearly an open cover of $$K$$ . So by the compactness of $$K$$ there exists $a_1, a_2, ..., a_n \in K$ such that:

(1)

\begin{align} \quad K \subseteq \bigcup_{i=1}^{n} (a_i + V) \subseteq \bigcup_{i=1}^{n} (a_i + U) \end{align}

So by definition, $$K$$ is precompact. By the result on the Boundedness of Precompact Sets in a LCTVS page we have that $$K$$ is thus also bounded. $\blacksquare$

Proposition 2: Let

$E$

be a locally convex topological vector space. Then:
(1) If

$K$

is compact and

\lambda \in \mathbf{F}

then

\lambda K

is compact.
(2) If

$K_1, K_2, ..., K_n$

are compact then

\displaystyle{\sum_{i=1}^{n} K_i}

is compact.
(3) If

$K$

is compact and

$C$

is closed then

$K + C$

is closed.

Proof (1): Let $$K$$ be compact. If $\lambda = 0$ the result is trivial as $0K = \{ o \}$ , so suppose that $\lambda \neq 0$ Let $\{ U_{\alpha} : \alpha \}$ be an open cover of $\lambda K$ . Then observe that $\{ \lambda^{-1} U_{\alpha} : \alpha \}$ is an open cover of $$K$$ . By the compactness of $$K$$ there exists $\alpha_1, \alpha_2, ..., \alpha_n$ such that $\{ \lambda^{-1} U_{\alpha_1}, \lambda^{-1} U_{\alpha_2}, ..., \lambda^{-1} U_{\alpha_n} \}$ covers $$K$$ so that $\{ U_{\alpha_1}, U_{\alpha_2}, ..., U_{\alpha_n} \}$ covers $\lambda K$ . So $\lambda K$ is compact. $\blacksquare$

Proof of (3): Let $$K$$ be compact and let $$C$$ be closed. Suppose that $x \not \in K + C$ . For each $k \in K$ , since $$E$$ is a locally convex topological vector space we have that $$k + C$$ is closed.

Since $$k + C$$ is closed and $x \not \in k + C$ , there exists an open and absolutely convex neighbourhood $$U(k)$$ of the origin for which:

(2)

\begin{align} \quad (x + U(k)) \cap (k + C) = \emptyset \end{align}

Then $x \not \in k + U(k) + C$ , for if instead $$x = k + u_k + c$$ for some $u(k) \in U(k)$ and $c \in C$ then $$x - u(k) = k + c$$ with $-u(k) \in U_k$ from the absolute convexity of $$U(k)$$ , which shows that $x - u(k) = k + c \in (x + U(k)) \cap (k + C)$ , a contradiction.

Now the collection $\{ k + \frac{1}{2} U(k) : k \in K \}$ is an open cover of $$K$$ . So by the compactness of $$K$$ there exists $k_1, k_2, ..., k_n \in K$ such that $\{ k_1 + \frac{1}{2} U(k_1), k_2 + \frac{1}{2} U(k_2), ..., k_n + \frac{1}{2} U(k_n) \}$ covers $$K$$ . Let:

(3)

\begin{align} \quad V := \bigcap_{i=1}^{n} \frac{1}{2} U(k_i) \end{align}

Then observe that:

(4)

\begin{align} \quad K + V \subseteq \bigcup_{i=1}^{n} (k_i + \frac{1}{2} U(k_i)) + \bigcap_{i=1}^{n} \frac{1}{2} U(k_i) \subseteq \bigcup_{i=1}^{n} \left (k_i + \frac{1}{2} U(k_i) + \frac{1}{2} U(k_i) \right ) \subseteq \bigcup_{k \in K} (k + U(k)) \end{align}

Since $x \not \in k + U(k) + C$ , from above we have that $x \not \in K + V + C$ . Hence $(x + V) \cap (K + C) = \emptyset$ , for otherwise, if $y \in (x + V) \cap (K + C)$ then $$y = x + v = k + c$$ for some $v \in V$ , $k \in K$ , and $c \in C$ . So $$x = k - v + c$$ . But $$V$$ is absolutely convex, so $-v \in V$ and $x \in K + V + C$ , a contradiction.

Since $$x + V$$ is a neighbourhood of $$x$$ that does not intersect with $$K + C$$ we have that $x \not \in \overline{K + C}$ .

Thus if $\overline{K + C} = K + C$ , i.e., $$K + C$$ is closed. $\blacksquare$

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Basic Properties of Compact Sets in a LCTVS