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As I came across that completeness is equivalent to closeness of a subspace if the whole space is complete, I was wondering if I can prove or disprove that the completness of a subspace also can give completness of the whole space.

One problem was, as I tried different approaches, that the norm was not a valid one for the bigger space. I also tried artificial examples as $C([0,1])$ with the $L_2$-norm that is not complete to intersect with some $L_p$ space to get something. My intuition is telling me that it shouldn't be true.

Can someone help me or give me a hint? Thanks in advance!

EDIT: Sorry, I forgot to mention that the subspace should be NOT finite. Because, as Lord Shark the Unknown mentioned, every finite-dimensional subspace of a normed space is complete.

CostaZach
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    Any finite-dimensional subspace of a normed space is complete. – Angina Seng Dec 29 '17 at 12:48
  • As you see I tried for infinite-dimensional subspaces. I know that finite-dimensional subspaces have very nice properties. – CostaZach Dec 29 '17 at 12:52
  • Or, making Lord Shark's argument even more compelling: the trivial subspace ${0}$ is always complete, and of course that doesn't tell us anything about the whole space. –  Dec 29 '17 at 12:54
  • What if the subspace is not finite? – CostaZach Dec 29 '17 at 13:00
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    In general you can't expect to deduce anything about the completeness of $X$ from the completeness of $U$. But if $U$ has finite codimension and is complete, then I'm pretty sure that $X$ will be complete too. – Angina Seng Dec 29 '17 at 13:13
  • @LordSharktheUnknown Do you know a counterexpample? I am trying hours to find one but nothing works. – CostaZach Dec 29 '17 at 13:29
  • Let $U$ be a Banach space, and let $Y$ be an incomplete normed space. Let $X=U\times V$, with $||(u,v)|| = ||u||_U+||v||_V$. – David C. Ullrich Dec 29 '17 at 13:37

3 Answers3

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There is indeed a counterexample.

Take the field of $p$-adic numbers $\mathbb Q_p$, which is complete with the usual $p$-adic norm $\vert\vert \cdot\vert\vert_p$ (by construction if you construct it right).

Then the algebraic closure $\mathbb Q_p^a$ of $\mathbb Q_p$ is not complete.

Its completion is $\widehat {\mathbb Q_p^a}$ and it's called Tate's field.

E. Joseph
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  • Thank you for the answer. I have to admit that I am not so familiar with the space $\mathbb Q_p$. But it sounds interesting and I will have a look for sure. – CostaZach Dec 31 '17 at 19:28
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The example is from this question.

Consider the Banach space $\ell^2(\mathbb{Z})$ and define its subspace:

$$X = \{(x_n)_{n\in\mathbb{Z}} \in \ell^2(\mathbb{Z}) : x_n \ne 0 \text{ for at most finitely many } n < 0\}$$

$(X, \|\cdot\|_2)$ is not a Banach space as it is not a closed subspace of $\ell^2(\mathbb{Z})$.

However $$U = \{(x_n)_{n\in\mathbb{Z}} \in \ell^2(\mathbb{Z}): x_n = 0 \text{ for all } n < 0\} \le X$$

is a complete infinite-dimensional subspace of $X$, because obviously $U \cong \ell^2(\mathbb{N})$.

mechanodroid
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  • Thank you for that explicit example. The funny thing is that it is basically an explicit example of the answer above. I haven't seen it immediately but if you take $U=l^2(\mathbb N)$ and $V=c_e(\mathbb N)$ (sequences that are only not equal to zero for finite $n\in\mathbb N$) equipped with the $l^2$-norm you have actually the same problem as yours :) – CostaZach Dec 31 '17 at 19:35
  • @CostaZach I'm not sure I follow, the space $c_{\varepsilon}(\mathbb{N}) = c_{00}$ of finitely-supported sequences is a subspace of $\ell^2(\mathbb{N})$, and it is not complete. You could take something like $$X = {(x_n){n\in\mathbb{N}} \in \ell^2(\mathbb{N}): x_n \ne 0 \text{ for at most finitely many even numbers } n}$$ and $$U = {(x_n){n\in\mathbb{N}} \in \ell^2(\mathbb{N}): x_{2n} = 0 \text{ for all } n \in \mathbb{N}}$$ to obtain the same example. – mechanodroid Dec 31 '17 at 21:12
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Let $U$ be any Banach space with norm $\|\cdot\|_U.$ Let $V$ be a normed linear space that is not complete, with norm $\|\cdot \|_V$ and origin $0_V.$

Let $X=U\times V$ with $\|(u,v)\|_X=\|u\|_U+\|v\|_V.$ Then $U\times \{0_V\}$ is a closed linear subspace of $X,$ and is complete because it is isometrically isomorphic to $U.$ But $X$ is not complete.

DanielWainfleet
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