Sum of Compact Subsets of Topological Vector Space is Compact
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Theorem
Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$.
Let $A$ and $B$ be compact (topological) subspaces of $X$.
Then $A + B$ is compact.
Proof
Since $X$ is a topological vector space, the map $+ : X \times X \to X$ defined by:
- $\map + {x, y} = x + y$
for each $x, y \in X$ is continuous.
We have:
- $A + B = + \sqbrk {A \times B}$
From Tychonoff's Theorem, $A \times B$ is compact.
From Continuous Image of Compact Space is Compact, $+ \sqbrk {A \times B} = A + B$ is therefore compact.
$\blacksquare$