I want to find an example of Polish space which is not locally compact. I am thinking about the space of all continuous function from $[0,1]$ to $R$, endowed with the metric $d(f,g) = \sup_{x\in [0,1]}|f(x)-g(x)|$.
I know this space is complete. And by Weierstrass Approximation Theorem, all the polynomials with rationals coefficients are a countable dense subset of it, so it is Polish.
Then suppose the function $f=0$ has a compact neighbourhood, then there exists $r >0$ such that all the continuous functions bounded by $r$ are in the neighbourhood. But then we can define a sequence of functions such as $g_n(x) = \begin{cases} 0, x<a_n\\ r,x>a_{n+1}\\r \frac{x-a_n}{a_{n+1} – a_n}, a_n \leq x\leq a_{n+1}\end{cases}$, where $(a_n)_n$ increases to(but never reaches) 1. Then for $m,n$ different, we have $d(g_n, g_m) = r$, so the function $f=0$ has no compact neighbourhood. Therefore this space is not locally compact.
Did I miss something?
Best Answer
Indeed, the example works.
More generally, an infinite dimensional separable complete normed space would do the job (Riesz theorem prevents local compact).