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Let $(E, \lVert \cdot \rVert )$ a normed vectorial space. If every subspace of $E$ is closed, then $E$ has finite dimension.

I have seen this as an excercise of a Functional analysis book and I am wondering how to prove it. My first thought was to do it by contraposition, supposing that $E$ is infinite dimensional, which implies that the unit closed ball $\overline{B} (0 ; 1) \subseteq E$ is not compact. I had the intuition that this could be used to construct a subspace which is not closed, but I am not getting it.

Any possible help or suggestion on the way to prove this would be appreciated.

Superdivinidad
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    My first thought was that since a linear functional is continuous if and only if its kernel is closed, it suffices to prove that on an infinite-dimensional normed vector space, there always exists a discontinuous linear functional. – P. P. Tuong Mar 01 '24 at 17:09
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    And to that end, see https://math.stackexchange.com/questions/99206/discontinuous-linear-functional (discussing the existence of discontinuous linear functionals on infinite dimensional normed vector spaces, and linking to other answers/discussion), in particular the accepted answer – leslie townes Mar 01 '24 at 21:52
  • @P.P.Tuong I see, that is very interesting, thank you so much! – Superdivinidad Mar 01 '24 at 23:01
  • @leslietownes this is really helpful, thanks!! – Superdivinidad Mar 01 '24 at 23:01

1 Answers1

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$\def\ed{\stackrel{\text{def}}{=}}$ Thank you to Chad K for pointing out an embarrassing blunder in an earlier version of this answer. Using his observation I've produced the following simple proof that any infinite dimensional normed vector space must contain a subspace that isn't closed.

Suppose $\ (E,\|\cdot\|)\ $ is an infinite dimensional space over the field $\ \mathbb{F}\ ,$ where $\ \mathbb{F}=\mathbb{C}\ $ or $\ \mathbb{F}=\mathbb{R}\ ,$ as required, and $\ \big\{u_i\big\}_{i=1}^\infty\ $ a countably infinite sequence of independent unit vectors in $\ E\ .$ Let \begin{align} W\ed\left\{\left.\sum_\limits{i=1}^nx_iu_i\right|\,n\in\mathbb{N}, \big(x_1,x_2,\dots,x_n\big)\in\mathbb{F}^n, \sum_\limits{i=1}^nix_i=0\right\} \end{align} and $$ w_n\ed u_1-2^{-n}u_{2^n} $$ for $\ n=1,2,\dots\ .$ Then $\ w_n\in W\ $ for all $\ n\ $ and $\ \lim_\limits{n\rightarrow\infty}w_n=$$\, u_1\in$$\,\text{cl}(W)\ .$ But $\ u_1\not\in W\ $, so $\ W\ $ is not closed.

lonza leggiera
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    ${\sum_{i=1}^nx_i u_i\colon n\in\Bbb{N},(x_1,\ldots,x_n)\in\Bbb{F}^n,\sum_{i=1}^n i x_i = 0}$ is a proper subspace of $V$ that is not closed in $V$. It's the kernel of the linear functional that maps $u_n$ to $n$. – Chad K Mar 02 '24 at 18:47