Let $X$ be a separable infinite-dimensional Banach space, $U$ be non-empty and open in $X$ and $E$ be finite-dimensional linear subspace of $X$.

I would like to know if there is a non-empty open subset of $U$ which does not intersect $E$

What I've done so far:

Conjecture: Yes

Since $X$ is infinite dimensional and $E$ is finite-dimensional, then $E$ must have empty interior (since no ball can be contained in $E$). Since $E$ has an empty interior and every open subset $V\subseteq U$ is open, then $E$ cannot contain $U$ or any of its open subsets $V$. Hence, $$ V \not\subseteq E \qquad (\forall V\subseteq U,\, \mbox{V open}) $$

Let us show that there is a non-empty open subset $U^*\subseteq U$ which does not intersect $E$.

In particular, let $Ball(x,r)$ be an open ball in $U$ of radius $r>0$ centered at some $x \in U$. By translation, $Ball(0,r)$ and $E-x$ are both of the same for as above, expect now $Ball(0,r)$ contains a countable basis $\{x_n\}_{n \in \mathbb{N}}$ (by separability) of $X$.

In particular, there are $\{r_n>0\}_{n \in \mathbb{R}}$ such that the open balls $\{Ball(x_n,r_n)\}_{n \in \mathbb{N}}$ are disjoint, contained within $Ball(0,r)$, and the vectors in $Ball(x_n,r_n)$ are linearly independent from $span\{x_i:i=1,\dots,n\}$. (I assume this should be possibly by inductively choosing a small enough radius)

If $E-x$ intersected each of these then it would contain a basis for $X$, contradicting its finite-dimensionality. Therefore, there is some $N\in\mathbb{N}$ for which $E-x \cap Ball(x_n,r_n)=\emptyset$; whence $$ E \cap Ball(x_n+x,r_n)=\emptyset. $$

Comments: I feel that the existence of the open balls was a bit unconvincing and likewise for the fact that $E$ must contain a basis since is does intersect all of those balls.


2 Answers 2


An open set $U$ minus a closed set $C$, regardless of which topology, is always open. You are taking the intersection of an open set, and the complement of a closed set, which is open, and thus the resulting set $U \setminus C$ is open.

The only question is, can the resulting set be empty? The empty set is a perfectly valid open set. Note that $U \setminus C = \emptyset$ if and only if $U \subseteq C$, which is to say, either $U = \emptyset$ or $C \subseteq \operatorname{int} U \neq \emptyset$.

As you've pointed out, finite-dimensional subspaces of infinite-dimensional spaces are closed and have empty interior. Thus, we cannot have $U \setminus C = \emptyset$; thus $U \setminus C$ is the open subset of $U$, disjoint from $C$, that you're looking for.

  • $\begingroup$ Oh I feel so silly now. It's totally obvious actually $\endgroup$
    – ABIM
    Commented Jun 23, 2019 at 12:31

Since $E$ is finite-dimensional, it is a closed subset of $X$ and therefore you can just take $U=E^\complement$.


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