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Is it true: If all linear subspaces of a Banach space are closed, then the space is of finite dimensions?

My attempt to prove this:

For contradiction, suppose $X$ is an infinite-dimensional Banach space. Then one can construct an infinite, strictly decreasing sequence $\{Y_n\}$ ($Y_0=X$, $Y_1=$ Kernal$(f_1)$, where $f_1\in Y_0^*$, $Y_2=$ Kernal$(f_2)$, where $f_1\in Y_1^*$ , ... ) of infinite-dimensional closed linear subspaces of $X$ and an infinite sequence of points, $\{x_n\}$ in $X$, such that $\|x_n\|=1/2^n$, $x_n\in Y_{n-1}$ and $x_n\notin Y_n$. These sequences of linear functionals and points are constructed using Hahan-Banach Theorem. Let $W=$ span$\{x_n\,:\,n\in\mathbb{N}\}$, then $W$ is clearly a linear subspace of $X$. We want to prove that $W$ is not closed. For contradiction, suppose $W$ is closed. How can I get a contradiction and show that $W$ must be open?? I think one should prove that $W$ is a proper subspace of $X$, and $\overline{W}=X$ and hence $W\neq \overline{W}$.

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The result holds, and $X$ need not even be Banach. To see this, note that if $X$ is an infinite dimensional normed space, then there exists a discontinuous linear functional $\varphi$ defined on $X$. Since a linear functional is continuous if and only if its kernel is closed, it follows readily that the kernel of this functional $\varphi$ is a subspace of $X$ which is not closed.

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  • $\begingroup$ Thanks for the quick answer Yuri! However, am not very familiar with the Theorem you mentioned above, and therefore, I would like an answer that completes what I started which I think is doable. The sequence of points, $\{x_n\}$ mentioned above were constructed with an additional property I didn't mention $f_n(x_n)\neq 0$. $\endgroup$ – Ruzayqat Apr 23 '16 at 21:33
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    $\begingroup$ Got it. I'm convinced that $y := \sum_{n=1}^\infty x_n \in \overline{W}$ is such that $y \not\in W$. Indeed, let us consider the map $T : (\eta_n)_{n=1}^\infty \in \ell_\infty \mapsto \sum_{n=1}^\infty \eta_nx_n \in X$. Verify that $T$ is well defined, and that it is linear, continuous and injective. It is easy to see that $W \subseteq T(\ell_\infty)$. Now, $y= T(1,1,1,...)$, so by the injectivity of $T$ it follows that $y \not\in W$. I hope this helps you someway! $\endgroup$ – user179514 Apr 23 '16 at 22:23
  • $\begingroup$ I don't see how by the injectivity of $T$ it follows that $y\notin W$. Can you explain this point please? $\endgroup$ – Ruzayqat Apr 23 '16 at 22:37
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    $\begingroup$ If $w \in W$, then $w = c_1x_1+...+c_m x_m = T(c_1,...,c_m,0,0,...)$, for some interger $m\geqq 1$ and some scalars $c_1,...,c_m$. That is, $w \in W$ implies $w \in T(c_{00})$. Since $u:= (1,1,1,...) \not\in c_{00}$ is the unique sequence in $\ell_\infty$ for which $y=T(u)$, we deduce that $y \not\in T(c_{00})$, hence $y \not\in W$. Pardon me, I should have written "It is easy to see that $W \subseteq T(c_{00})$'' in the previous comment. $\endgroup$ – user179514 Apr 23 '16 at 22:45
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    $\begingroup$ No, the span of a set $S$ is usually defined as the intersection of all subspaces that contain this set. It turns out to be the set of all $\textit{finite}$ linear combinations of the elements of $S$. In this particular case you could form "infinite linear combinations" as a consequence of the way you chose the $x_n$'s, but as I argued, it can happen that they end up not in the span. In general, "infinite linear combinations" won't converge. $\endgroup$ – user179514 Apr 23 '16 at 23:00

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