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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.

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up vote 22 down vote accepted

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.

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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 $v = t^{-1}(tv) \in S$.

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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 $V = W$.

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