For regular spaces (such as Banach spaces) we know that non-empty, open sets contain closed subsets. The reverse shouldn't always be true (e.g. if we take singletons). Now, I have a non-trivial, closed, linear subspace $M$ of a Banach space $X$ and would like to find an open subset (not even subspace) of $M$. Is this possible in general?
Alternatively, I would like to show that given a dense subset $S$ of $X$, $M\cap S$ is never empty.
Many thanks in advance.
On
I think the answer to both your questions is no if $M$ has positive codimension.
Simply take $X=\mathbb{R}^2$ and $M=\mathbb{R}\times \{0\}$. This is closed but does not contain an open set of $X$. Likewise, you can take $S=\mathbb{Q}^2\setminus (\mathbb{Q}\times \{0\})= \mathbb{Q}^2\setminus M$ is dense but does not intersect $M$.
You can't even do it with the simplest examples. Try $M=\Bbb R$ and $X=\Bbb R^2$. In fact you cannot do it at all, unless $M$ is equal to the whole space $X$. Why?
Well, if $M$ is a proper subspace then there is at least one basis vector $v$ not contained in $M$. If you pick some starting $x\in M$ and try to fit an open ball inside, say of radius $\delta$, then $x+v\cdot\delta/2$ cannot be in $M$ (otherwise, subtracting $x$, you'd find $v\in M$ which is false). But $x+v\delta/2$ is in this open ball - so you cannot achieve what you want.
This observation is conceptually related (for me) to the open mapping theorem. It's often stated as (under suitable conditions) a surjective operator is open, but in fact any open operator is necessarily also surjective for this very reason. In fact we could restrict ourselves to saying, if the image of the operator has nonempty interior, then the operator is surjective.