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I want to prove the following proposition

If $(X,||\cdot||)$ and $(X,||\cdot||')$ are homeomorphic, then $(X,||\cdot||)$ is complete if and only if $(X,||\cdot||')$ is complete.

So, I only know the definition of homeomorphic, and can't figure out how only with that I can prove that proposition. Can someone help me to prove this please?

Thanks a lot in advance :)

gymbvghjkgkjkhgfkl
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user162343
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3 Answers3

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As a sketch$\ldots$

Well if you've a homeomorphism $f$, then if $\{x_n\}_{n \in \Bbb N}$ is cauchy wrt $\|\cdot \|$, and converges to $x \in X$, then $\|x_n-x\|<\delta$ for $n$ sufficiently large, and $\|x_n-x_m\|<\delta$, for $n,m$ sufficiently large. Since $f$ is continuous we can make $\|f(x_n)-f(x_m)\|'<\epsilon$ for $\|x_n-x_m\|<\delta$, so $\{f(x_n)\}_{n \in \Bbb N}$ is cauchy. Now since $f$ is continuous at $x$ we also know that $\|f(x_n)-f(x)\|'<\epsilon$ for $\|x_n-x_m\|<\delta$. So $f(x_n)$ will converge to $f(x)$. Then to prove the other way, you do the same thing, except with $f^{-1}$, which is continuous since $f$ is a homeomorphism.

snulty
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  • Let me think a little bit in this, If I get stuck can I tell you? – user162343 Aug 19 '15 at 15:55
  • yeah of course you can! And I can tighten up the proof in a while if you want, make it less of a sketch – snulty Aug 19 '15 at 16:45
  • Yes I apprwciate that please :) – user162343 Aug 19 '15 at 16:49
  • Is this a corollary of this question http://math.stackexchange.com/questions/1402775/prove-the-following-theorem-involving-homeomorphic-metric-spaces – user162343 Aug 20 '15 at 01:55
  • @user162343 yes the norm on a vector space gives a metric where $d(x,y)=| x-y|$ – snulty Aug 20 '15 at 10:37
  • Then Can Iedit that question so I can post the answer for it (it is my question, I've post it yesterday) then I ask you a verification please :) – user162343 Aug 20 '15 at 13:20
  • I am back, the thing is that I have told the teacher about this, and he told me that it is not a corollary :(, then what can be done here? – user162343 Aug 21 '15 at 15:57
  • @user162343 I'm not sure what you meant by yesterday's comment. With regards to the corollary thing, if you prove the exact statement about completeness of metric spaces, then normed spaced give rise to a metric, and the proof for metric spaces hold, but I suppose you don't really need to refer to any specific as the proof should stand alone as is. If you give me a few minutes I'll give you a more rigorous proof of this statement. We are talking about vector spaces though? – snulty Aug 21 '15 at 16:34
  • Well maybe, maybe not, they are only normed spaces, and yes please I appreciate that you helpme with that please :). – user162343 Aug 21 '15 at 16:51
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Take a Cauchy sequence in $(X,||\cdot||)$ and show that is also a Cauchy sequence in $(X,||\cdot||')$.

This proves the assertion since convergence is a topological property which is preserved by homeomorphisms.

principal-ideal-domain
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Since $(X, ||\cdot || ) $ and $ (X, ||\cdot ||' )$ are homeomorphic the topologies generated by these norms are identical. Hence the identity map $\mbox{Id} :(X, ||\cdot || ) \to (X, ||\cdot ||' )$ is continiuous so the norms are equivalent. And from the equivalence of the norms it easily follows your assertion.