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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.

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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?

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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.

(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).

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