Compact Sets.

**maxgunn555** · December 10th 2011, 07:34 PM

Let $(X,d_{x})$ be a continuous function. Suppose that $K\subset X$ is compact and that $A\subset\ \mathbb{R}^k$ is compact.
Prove that $f(K)\bigcap A\ =\ \emptyset$ implies dist $(f(K),A)>0$

It's analysis and there isn't an extra class for that unlike the other modules (even though it's considered the hardest).
Thnx in advance

**Drexel28** · December 10th 2011, 08:21 PM

Originally Posted by maxgunn555

Let $(X,d_{x})$ be a continuous function. Suppose that $K\subset X$ is compact and that $A\subset\ \mathbb{R}^k$ is compact.
Prove that $f(K)\bigcap A\ =\ \emptyset$ implies dist $(f(K),A)>0$

It's analysis and there isn't an extra class for that unlike the other modules (even though it's considered the hardest).
Thnx in advance

You know that $f(K)$ is compact, and so $f(K)\times A$ is compact, so the mapping $d:f(K)\times A\to\mathbb{R}$ obtains a minimum value $d(f(k_0),a_0)$ for some $(f(k_0),a_0)\in f(K)\times A$ . Clearly though $d(f(k_0),a_0)>0$ otherwise $f(k_0)=a_0$ which contradicts $f(K)\cap A=\varnothing$ .

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