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

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?

• Infinite-dimensional spaces are a lot different from finite-dimensional ones. I picture an infinite row of fence-posts, each a copy of $[0,1]$ and I picture $x\in [0,1]^{\omega}$ as a sequence of spots, one spot on each post. And I picture a nbhd of $x$ as something that covers an interval of length $r>0$around each spot (Or the sequences of spots that stay in these intervals,) – DanielWainfleet Jul 27 '17 at 19:17

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.

• 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

(1). Generalities. For any space $X:$ If $Y$ is a closed discrete subspace of $X$ then every $Z\subset Y$ is closed: $\quad Cl_Y(Z)=Z$ because $Y$ is a discrete space, so $Cl_X(Z)=Cl_X(Z)\cap Cl_X(Y)=Cl_X(Z)\cap Y=Cl_Y(Z)=Z.$

Let $Y$ be an infinite closed discrete subspace of $X.$ Let $[Y]^{<\omega}$ be the set of finite subsets of $Y.$ (This is a standard notation in set-theory.) For each $W\in [Y]^{<\omega}$ the set $(X$ \ $Y)\cup W$ is open (...because its complement $Z=Y$ \ $W$ is closed). Now $\{(X$ \ $Y)\cup W: W\in [Y]^{<\omega}\}$ is an open cover of $X$ with no finite sub-cover, so $X$ is not compact.

(2). Specifics. Let $x=(x_j)_{j\in \omega}\in N\subset [0,1]^{\omega}$ where $N$ is a nbhd of $x.$ Take $r\in (0,1)$ such that the open ball $B_p(x,r)\subset N.$ For $n\in \omega$ let $y(n)=(y_{n,j})_{j\in \omega}$ where $y_{n,j}=x_j$ when $n\ne j$, and $y_{n,n}\in [0,1]$ such that $|y_{n,n}-x_n|=r/2.$

Then $Y=\{y(n): n\in \omega\}\subset N.$ And $Y$ is an infinite closed discrete subspace of $[0,1]^{\omega}$ because $p(y(m),y(n))=r/2$ whenever $m\ne n.$ So $Y$ is (a fortiori) an infinite closed discrete subspace of $\overline N,$ so $\overline N$ is not compact.

(3). Remarks. A non-compact metrizable space has an infinite closed discrete sub-space. But some other spaces (e.g. $\omega_1$ with the $\epsilon$-order topology) are non-compact and have no infinite closed discrete subspaces.

The set of all bounded real sequences with the metric $p((x_n)_n,(y_n)_n)=\sup_n|x_n-y_n|$ is called $l^{\infty}.$ ("El-infinity"). It is a complete metric space. The topology generated by $p$ is called the topology of uniform convergence: Members of $l^{\infty}$ are bounded functions from $\omega$ to $\mathbb R$, and a sequence $(f_n)_n$ of members of $l^{\infty}$ converges (with respect to the metric $p$) to $f\in l^{\infty}$ iff $(f_n)_n$ converges unformly to $f$ (as a sequence of functions).