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 $W\neq \overline{W}$.
Best Answer
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.