Let $x\_n$ be the continuous function on $[0, 1]$ that is $1$ at $2^{-n}$ and $0$ outside $[2^{-n-1}, 2^{-n+1}]$. For an arbitrarily given $\epsilon >0$, is there a finite dimensional subspace $V$ of $C([0, 1])$ such that the distance from $x\_n$ to $V$ is $<\epsilon$ for all $n$?



Uniform finite dimensional approximation of continuous function sequence
No, there isn't.


Note first that $\|x\_n\|=1$ and $\|x\_n - x\_m\| = 1$ for all $n \ne m$.


Suppose there were, so that for each $n$ there is $y\_n \in V$ with $\|x\_n - y\_n\| < \epsilon$. Let's say $\epsilon < 1/3$. Then $\|y\_n\| < 4/3$ for all $n$, and $$\|y\_n - y\_m\| \ge \|x\_n - x\_m\| - \|y\_n - x\_n\| - \|y\_m - x\_m\| \ge 1/3$$ for all $n \ne m$. In particular, the sequence $y\_n$ has no convergent subsequence. But since $V$ is finite dimensional, the closed ball of $V$ with radius $4/3$ is compact, so this is absurd.


