Continuous Mappings preserve Compact Subsets

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Theorem

Let $X, Y$ be normed vector spaces.

Let $K \subseteq X$ be a compact subset.

Suppose $f : X \to Y$ is a continuous mapping at each $x \in K$.


Then $\map f K$ is a compact subset of $Y$.


Proof

Let $\sequence {y_n}_{n \mathop \in \N}$ be a sequence contained in $\map f K$.

Then:

$\forall n \in \N : \exists x_n \in K : y_n = \map f {x_n}$

$K$ is compact.

By definition, there is a convergent subsequence $\sequence {x_{n_k}}_{k \mathop \in \N}$ convergent with the limit $L \in K$.

$f$ is a continuous mapping.

By Continuous Mappings preserve Convergent Sequences, $\sequence {\map f {x_{n_k}}}_{k \mathop \in \N} = \sequence {y_{n_k}}_{k \mathop \in \N}$ is convergent with limit $\map f L \in \map f K$.

By definition, $\map f K$ is compact.

$\blacksquare$

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