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There is a conjecture: "The only subspace of a normed vector space $V$ that has a non-empty interior, is $V$ itself." (here, the topology is the obvious set of all open sets generated by the metric $||\cdot||$).

I have a proof for the case $V$ is finite dimensional. Because, let $V$ have dimension $n$ and a subspace $S$ of $V$ have dimension $m < n$. Let $\{v_1,v_2,\ldots,v_m\}$ be a basis for $S$, extended to the basis $\{v_1,\ldots,v_m,\ldots,v_n\}$ of $V$. Now, suppose that $p = b_1v_1+\cdots+b_mv_m$ is an interior point of $S$. Now, consider the norm $N(a_1v_1 +\cdots+a_nv_n)=\max(|a_1|,\ldots,|a_n|)$. Then, there is an $r>0$, such that $||x-p||<r$ and $x$ is in $V$ $\implies$ $x$ is in $S$, since on a finite dim. space, all norms are equivalent. Let $m<k\leq n$,and chose $v$ in $V$ as:

$$v=\left(b_1+\frac{r}{2}\right)v_1+\left(b_2+\frac{r}{2}\right)v_2+\cdots+\left(b_m+\frac{r}{m}\right)v_m+\frac{r}{2}v_k$$ Then, $N(v-p)\leq\frac{r}{2}<r$, so $v$ is in $S$ and by the subspace property of $S$, $v_k$ is in $S$ too, a contradiction to $m<n$.

I have primarily 2 questions:

(1) Is there a simpler method to proof the conjecture for the finite dimensional case?

(2) Is the conjecture true for the infinite dimensional case?

Sorry, if the question admits a very trivial answer. The motivation behind my question , is the fact that an open interval in $\mathbb{R}$ is not open in $\mathbb{R}\times\mathbb{R}$, etc.

Martin Sleziak
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Somabha Mukherjee
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3 Answers3

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Your conjecture is true in any normed vector space. They key is that you don't need to switch to an equivalent norm, as your proof does.

Suppose $S$ has a nonempty interior. Then it contains some ball $B(x,r) = \{y : \|y-x\| < r\}$. Now the idea is that every point of $V$ can be translated and rescaled to put it inside the ball $B(x,r)$. Namely, if $z \in V$, then set $y = x + \frac{r}{2 \|z\|} z$, so that $y \in B(x,r) \subset S$. Since $S$ is a subspace, we have $z = \frac{2 \|z\|}{r} (y-x) \in S$. So $S=V$.

A nice consequence of this is that any closed proper subspace is necessarily nowhere dense. So if $V$ is a Banach space, the Baire category theorem implies that $V$ cannot be a countable union of closed proper subspaces. In particular, an infinite dimensional Banach space cannot be a countable union of finite dimensional subspaces. This means, for example, that a vector space of countable dimension (e.g. the space of polynomials) cannot be equipped with a complete norm.

Nate Eldredge
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It's also true in any topological vector space. If $S$ contains a neighbourhood $U$ of $s$, then it also contains $U-s$ which is a neighbourhood of $0$. By continuity of scalar multiplication, for any $v \in V$ there is $\delta > 0$ such that $t v \in U-s$ for all scalars $t$ with $|t| < \delta$. And then $v = t^{-1}(tv) \in S$.

Robert Israel
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Suppose $W$ is a subspace of a normed vector space $V$ with nonempty interior. Let $w\in W$ be a point such that there is an open neighborhood $U$ of $w$ contained within $W$. Then the open set $U' = U - w = \{v\in V : v + w\in U\}$ is an open neighborhood of $0$ which is contained within $W$, since $W$ is closed under addition. In particular, some ball $B(0,r)$ is contained within $W$. If $v\in V$ is any point, then $$\frac{r}{2\|v\|}v\in B(0,r)\subset W,$$ and thus since $W$ is closed under scalar multiplication, $v\in W$. It follows that $V \subseteq W$, so $V = W$.

froggie
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