Take the 2-minute tour ×
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.

On page 186 of Munkres' Topology

Show that $[0,1]^{\omega}$ is not locally compact in the uniform topology?

Uniform topology is defined as topology induced by uniform metric $p$ which is stated as follows.

For any two points $a$, $b$ in $\mathbb{R}$, $$\bar{d}(a,b) = \text{min}\{|a-b|,1\}$$ For any two points $x$, $y$ in $[0,1]^{\omega}$ $$x = \{x_i:i<\omega\}$$ $$y=\{y_j:j<\omega\}$$ $$p(x,y) = \text{sup}\{\bar{d}(x_i,y_i):i<\omega\}$$

Is there some method to gain intuition on infinite product of topological space in various topologies? Could we visualize it as the finite product case?

share|improve this question

1 Answer 1

up vote 1 down vote accepted

Use the equivalent definition of local compactness: if $x \in U \subset C$, $U$ open, $C$ compact, then there must be a ball $B_{\epsilon}(x)$ whose closure $\bar{B}$ is contained in $U$. Apply this to $x = 0$. Then the closure of such a ball is compact as it is a closed subset of a compact set in a metric space. But it is not: look at the set $A = \{ x_i = (0, \dots, 0, \epsilon, 0, \dots ), i \in \mathbb{N} \} \subset \bar{B}$, with the $\epsilon$ in position $i$. It has no limit point $x$ in $A$ ($x$ cannot contain a coordinate in $(0, \epsilon)$ or a small enough ball around $x$ will not contain any $y \in A$; and in the other case, it is at distance $0$ or $\epsilon$ from any point in $A$ - think about that!). So $A$ is not limit-point compact, which in metric spaces is equivalent to being compact. Contradiction. 

share|improve this answer
    
And no, you can't really visualize a space with this metric as a finite product. This is only sort of possible with the standard product topology. –  gnometorule Feb 3 '13 at 6:59
    
Thank you for your answer. Allow me to take the liberty to guess "$C$ compact" in the first line is a typo, which should be "$C$ hausdorff" as stated in Munkres' Topology, Theorem 29.2. If $C$ is compact, it's automatically local compact. –  Metta World Peace Feb 3 '13 at 15:26
    
@MettaWorldPeace: no it's really compact. I'm away from home for some days, and so have no access to my munkres. This is using the proposition that shows an alternative, often easier to use way to show a space is locally compact - more or less what I type at the beginning. I might have gotten a detail wrong and can't confirm for almost a week (see above), but have a look at the section. Or...maybe I misunderstand what you say, and you're right. :) But use the alternative definition. –  gnometorule Feb 3 '13 at 16:10
    
What you say is correct of course, but part of a chain of conclusions leading to a contradiction: if l.c., then for any compact set there is a closed (and so compact) set within it; but the second part shows that such a set cannot be compact. So, unrolling to the start, the space is not l.c. (at $0$). –  gnometorule Feb 3 '13 at 16:17
    
I see. It's me who misunderstood your statement. Your formulation of local compactness is different from the one I'm referring to. I thought $C$ is the underlying topological space that we try to define local compactness on. Thank you very much. –  Metta World Peace Feb 3 '13 at 16:28

Your Answer

 
discard

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.