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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\in 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}\leq ||\cdot ||_{2}\leq 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\in M\,, M\, is compact in (X,||\cdot ||_{2})\,.

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