Vtyjkyuk

Given two vector subspaces $V_1,V_2$ of an infinite-dimensional normed space $X$, we say that $V_1$ is $epsilon$-close to $V_2$,

if for every $v_1 in V_1$, $|v_1|=1$, there exist $v_2 in V_2$ such that $|v_1-v_2|<epsilon$.

Now, suppose that $(V_n)_n ge 0$ is a sequence of $k$-dimensional subspaces ($k < infty$) of $X$, such that $V_n$ is $epsilon_n$-close to $V_0$, where $epsilon_n to 0$.

Is it true that $V_0$ becomes arbitrarily close to $V_n$ when $n to infty$?

In other words, I am asking whether the closeness relation is "asymptotically-symmetric".

The "equal dimensions" assumption is necessary: Otherwise $V_n$ could be strict subspaces of $V_0$.

Edit:

gerw proved that the answer is positive whenever $X$ is a Hilbert space.

The question remains whether or not the answer stays positive if the norm is not induced by an inner product.

After giving a flawed counterexample, I give a proof that works (at least) in Hilbert spaces.

Let us assume that $V_n$ is $varepsilon$-close to $V$ with $varepsilon < 1$. Let us denote by $S$ the unit ball in our Hilbert space $X$. Moreover, let $P$ be the orthogonal projection onto $V$.

By $varepsilon$-closedness, we have $| v - P v| le varepsilon$, hence, $| P v | ge 1-varepsilon > 0$ for all $v in V_n cap S$, i.e., the map $Q : V_n cap S to V cap S$ defined via $Qv := Pv/|Pv|$ is continuous and odd. Since $V_n cap S$ and $V cap S$ are essentially $n-1$-spheres, $Q$ should be surjective, this should follow, e.g., from Borsuk-Ulam theorem. Now, its inverse $Q^-1$ gives you a nice map from $V cap S$ to $V_n$. This shows that $V$ is $delta$-close to $V_0$ for some $delta$. Moreover, we get $delta to 0$ as $varepsilon to 0$.

Here is some explanation for the should. I was not able to come up with a reference, so I might miss something obvious:
Let $f : S^n to S^n$ be a continuous, odd map, i.e., $f(-x) = -f(x)$. Here, $S^n$ is the $n$-sphere. We claim that $f$ is surjective. Otherwise, there is $x in S^n$ which is not in the image of $S^n$. Since $f$ is odd, $-x$ does not belong to the image, too. Now, let $g : S^n to mathbb R^n$ be the orthogonal projection onto the orthogonal complement of $x$ within $mathbb R^n+1$. Then, $gcirc f : S^n to mathbb R^n$ is a continuous, odd map and $0$ does not belong to the image of $g circ f$. This is a contradiction to Borsuk-Ulam.