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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.

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    $\begingroup$ You could at the kernel of a linear functional that is not continuous (you can define such a functional using a Hamel basis). $\endgroup$ – David Mitra Feb 15 '15 at 17:05
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I don't know if this is exactly what David had in mind, but here is one construction:

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 $X$.

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    $\begingroup$ Yes, you define it on a basis and extend by linearity. $\endgroup$ – Matthew C Feb 16 '15 at 18:19
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    $\begingroup$ 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. $\endgroup$ – Disintegrating By Parts Feb 16 '15 at 19:16
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    $\begingroup$ 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. $\endgroup$ – Disintegrating By Parts Feb 16 '15 at 21:51
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    $\begingroup$ Oh I understand now. The statement at the end is new to me. Thanks! $\endgroup$ – Matthew C Feb 17 '15 at 0:47
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    $\begingroup$ 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.) $\endgroup$ – Disintegrating By Parts Feb 17 '15 at 2:32

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