Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|cite|improve this question
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.) –  Arthur Fischer May 23 '14 at 2:31

1 Answer 1

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).

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.