# infinite-dimensional Banach spaces has linear subspaces of finite-codimension that are not closed

I want to show that

In every infinite-dimensional Banach spaces there are linear subspaces of finite-codimension that are not closed .

There is a hint for this question that says use Zorn lemma.

Can someone help.

• You could at the kernel of a linear functional that is not continuous (you can define such a functional using a Hamel basis). – David Mitra Feb 15 '15 at 17:05

Given a Hamel basis $\{e_\alpha\}$ you can make any vector space $X$ a normed space (norm properties can be checked) by setting $$x = \max{\{|c_1|,\ldots,|c_k|\}}$$ where $x = c_1e_1 + \cdots + c_ke_k$ is the unique expression of $x$ as a finite linear combination of basis elements.
Fix some basis element $e_\alpha$. Then define a linear functional $\phi$ on $X$ by $$\phi(e_\alpha) = 1$$ $$\phi(e_\beta) = 0$$ for all $\beta \neq \alpha$. The codimension of the kernel of $\phi$ is 1. However the kernel is not closed if $X$ is infinite-dimensional, as can be seen by taking any sequence of basis elements $e_1,e_2,\ldots$ all distinct, and distinct from $e_\alpha$, noting that they cannot converge to any element in $X$.
• Using that, I'll bet you can find a way to define the functional so that $|\phi(e_{\alpha_{n}})| \ge n\|e_{\alpha_{n}}\|$, which guarantees the function is unbounded and, hence, also discontinuous. – Disintegrating By Parts Feb 16 '15 at 19:16
• Every non-zero linear function is non-zero on a one dimensional subspace. If $\mathcal{N}(\phi)$ denotes the null space of $\phi$, and if $\phi(x) \ne 0$, then you can scale so that $\phi(x)=1$. Then, $\phi(y-\phi(y)x)=0$ for all $y$, which gives $y-\phi(y)x \in \mathcal{N}(\phi)$, or $X=[\{x\}]\oplus\mathcal{N}(\phi)$. So $\mathcal{N}(\phi)$ is of co-dimension $1$ in $X$. $\phi$ is continuous iff $\mathcal{N}(\phi)$ is closed. – Disintegrating By Parts Feb 16 '15 at 21:51
• If $\phi$ is continuous, then $\phi^{-1}\{0\}=\mathcal{N}(\phi)$ must be closed. Conversely, if $\mathcal{N}(\phi)$ is closed, then $\phi$ factors into continuous maps $X \rightarrow X/\mathcal{N}(\phi)\rightarrow \mathbb{C}$ (the last is continuous because all linear maps on finite-dimensional normed spaces are continuous.) – Disintegrating By Parts Feb 17 '15 at 2:32