# Is a subspace of a vector space always closed?

Let $E$ a vector space (let say over $\mathbb R$). If $F$ is a subspace of $E$ will it be closed ? If $E$ has finite dimension, then I proved that it's correct.

• If $E$ has infinite dimension and $F$ has finite dimension I proved it as follow : Let $(e_1,...,e_n)$ a basis of $F$. Then $$\Phi: F\to \mathbb R^n$$ defined as $$\Phi\left(\sum_{i=1}^n x_i e_i\right)=\begin{pmatrix}x_1\\ \vdots\\ x_n\end{pmatrix},$$ is an isomorphism, and thus an homeomorphism.

Q1) I know that completeness depend on the metric, i.e. there are metric that make a space complete, and other metric that make space not complete. But if two topological spaces are homeomorphic, if one is complete, will the second one be complete ?

If yes, we have that $F$ is complete, and thus closed in $E$ (since complete spaces in metric spaces are closed).

• Now if $F$ has infinite dimension, I guess it's not true but I can't find counter example. I imagined a space $E=\{(x_1,x_2,...)\mid \sup|x_i|<\infty \}$ and $F$ as $$F:=\{(x_1,x_2,...)\mid x_i\in E, x_i\neq 0\text{ for finite numbers of i}\}.$$

Now $x_1=(1,0,...), \quad x_2=(1,1,0,...),\quad x_3=(1,1,1,0,...)$...

and thus "$x_n\to (1,1,...,1,...):=x$" but unfortunately it doesn't converge in $E$ since $\sup\{|x_n-x|\}=1$. But may be there is a way to arrange that ? (like a similar construction in an other space).

• You actually do know some counterexamples. Surely you know examples of dense subspaces of Banach spaces - a (proper) dense subspace cannot be closed. – David C. Ullrich Aug 31 '18 at 13:10
• Hint: space of polynomials in... – peter a g Aug 31 '18 at 13:27
• E.g., take the space of all finite sequences in $\ell^2(\Bbb N)$. – amsmath Aug 31 '18 at 13:30

In fact, any infinite dimensional normed space $E$ has a non-closed subspace.
Proof: Pick a discontinuous linear functional (see Martin Sleziak's answer) $f:E\to \mathbb{R}$. Then $\ker f$ is a non-closed linear subspace of $E$.
Moreover, any subspace $F$ of countably infinite dimension is necessarily non-complete as a consequence of the Baire Cathegory theorem, and hence if $E$ is complete, $F$ is non-closed.