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Assume that c||\cdot ||_1\, and ||\cdot ||_2\, are two equivalent norms on X\,, and M \subset X\,. Prove that M\, is compact in (X, ||\cdot ||_1)\, if and only if M\, is compact in (X, ||\cdot ||_2)\,.


Since M\, is compact in (X,||\cdot ||_1)\,, every sequence \{x_n\}\, in M\, has a convergent subsequence \{x_{n_k}\}\, that converges to x\isin M\,. This means \forall \epsilon > 0, \exists K\, s.t. ||x-x_{n_k}||_1<\epsilon \forall k>K\,.


||\cdot ||_1\, and ||\cdot ||_2\, are equivalent norms iff \exists a,b\, s.t. a ||\cdot ||_1 \le ||\cdot ||_2 \le b ||\cdot ||_1\,.


So b||\cdot ||_1 < b\epsilon\,. Let \delta_\epsilon = b\epsilon\,.


Now, for any \delta_\epsilon\,, and for all n>N\,, ||x-x_n ||_2<\delta_\epsilon\, which means x_n\to x\,, and since x\isin M\,, M\, is compact in (X,||\cdot ||_2)\,.


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