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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?

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That works. If you have it, you could also appeal to Riesz' theorem that a topological vector space is locally compact if and only if it is finite-dimensional. – Daniel Fischer May 22 '14 at 20:23
As an aside, the Baire space $\mathbb{N}^{\mathbb{N}}$ is also a non-locally-compact Polish space. (It is not too difficult to show that all compact subsets of it have empty interior.) – arjafi May 23 '14 at 2:31
up vote 4 down vote accepted

Indeed, the example works.

More generally, an infinite dimensional separable complete normed space would do the job (Riesz theorem prevents local compact).

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Is it necessary that the normed space is infinite-dimensional? – Petite Etincelle May 22 '14 at 20:28
Yes, otherwise it is locally compact. It's added now. – Davide Giraudo May 22 '14 at 20:31

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