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Let $V$ be a real normed vector space.

Let $K$ be a compact set from $V$. Show that the set $2K =$ {2$x$: $x \in K$} is also compact.

A topological space is compact if every open covering has a finite sub-covering.

A norm on V is a function || · || : V → R that satisfies the following three conditions:
(i) ||v|| ≥ 0, ∀ v ∈ V, and ||v|| = 0 ⇔ v = 0;
(ii) ||αv|| = |α| ||v||, ∀ v ∈ V, α ∈ C;
(iii) ||v + w|| ≤ ||v|| + ||w||, ∀ v, w ∈ V

I attempted to prove this using proof by contradiction and assuming the set was not compact but this approach did not get me very far. Any pointers are greatly appreciated!

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Let $U_\alpha$ be an open cover of $2K$. Then ${1 \over 2} U_\alpha$ is an open cover of $K$ and there is a finite subcover ${1 \over 2} U_{\alpha_1},...,{1 \over 2} U_{\alpha_n}$. Hence $U_{\alpha_1},..., U_{\alpha_n}$ is a finite subcover of $2K$.

A notationally simpler approach is to note that continuous function map compact sets into compact sets, and since $x \mapsto 2x$ is continuous, we see that $2K$ is compact.

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  • $\begingroup$ Thank you, that's much cleaner than doing a proof by contradiction. I appreciate your insight! $\endgroup$ – Iff Dec 7 '15 at 21:38

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