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Suppose I have a Hilbert space $H$ and a linear subspace $V\subset H$ with its inner product inherited from $H$. The spaces are in particular sets, so I may consider the inclusion map $i:V\hookrightarrow H$. Of course for any element $v\in V\subset H$, $||i(v)||=||v||$ since the norm is inherited. But this means that $i$ is an isometry, whence $V$ is a complete subspace and also therefore closed.

But we know there are subspaces of Hilbert spaces which are not closed. These are not open either, but there must be some error in my arguments above. Where is my mistake?

Taking $H,V$ to be Banach spaces doesn't change anything I think.

plebmatician
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    Very catchy title! – Ben Grossmann Dec 08 '21 at 14:55
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    I'm missing something. Why should the fact that $i$ is an isometry imply that $V$ is a complete subspace of $H$? – Ben Grossmann Dec 08 '21 at 14:56
  • Can you please elaborate how you concluded V to be complete? – Jean Dec 08 '21 at 14:57
  • If $i$ were a surjective isometry it would follow that $V$ was complete... – David C. Ullrich Dec 08 '21 at 15:00
  • Thank you! I'm thinking like this. $H$ is complete, so its Cauchy sequences converge. If we take a Cauchy sequence in $V$ and then consider the norm of the difference of adjacent elements far along, we get $\epsilon>||v_{j+1}-v_j||=||i(v_{j+1}-v_j)||=||i(v_{j+1})-i(v_j)||$ by linearity. So ${ i(v_j) }$ is Cauchy in $H$, where it converges to some $h\in H$. Ah. Do I now need that $h$ is in the image of $i$.. – plebmatician Dec 08 '21 at 15:05
  • I see, I think I confused the whole space $H$ with the image of $i$, meaning $i(V)=V$. – plebmatician Dec 08 '21 at 15:10

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