Basic Properties of Compact Sets in a LCTVS

# Basic Properties of Compact Sets in a LCTVS

 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}
 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. $\blacksquare$