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Does an inverse limit of compact metric spaces need to be metrizable? When it is an inverse limit of a countable inverse system I know it is metrizable (even without compactness). But what if the system is not countable?

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Is an uncountable product of copies of $[0,1]$ metrizable? –  egreg Apr 29 '13 at 10:34
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2 Answers 2

For each ordinal $\alpha<\omega_1$ let $X_\alpha=\alpha+1$ with the order topology; each $X_\alpha$ is then compact and metrizable. If $\alpha\le\beta<\omega_1$ let

$$\pi^\beta_\alpha:X_\beta\to X_\alpha:\xi\mapsto\begin{cases} \xi,&\text{if }\xi\le\alpha\\ \alpha,&\text{if }\alpha<\xi\le\beta\;. \end{cases}$$

Let $X=\varprojlim X_\alpha$. Suppose that $x=\langle x_\alpha:\alpha<\omega_1\rangle\in X$. If $x_\alpha<\alpha<\beta<\omega_1$, $\pi^\beta_\alpha(x_\beta)=x_\alpha<\alpha$, so $x_\beta=x_\alpha$. Thus, if $\eta(x)=\min\{\alpha<\omega_1:x_\alpha<\alpha\}$, then $x_\alpha=x_{\eta(x)}$ for $\eta(x)\le\alpha<\omega_1$. Note that since $x$ is non-decreasing, $\eta(x)$ must be a successor ordinal; let $\gamma(x)\in\omega_1$ be such that $\eta(x)=\gamma(x)+1$. Thus, $x_{\gamma(x)}=\gamma(x)$, and $\gamma(x)\le x_{\eta(x)}=x_{\gamma(x)+1}<\gamma(x)+1$, so $x_{\eta(x)}=\gamma(x)$. If $x_\alpha=\alpha$ for all $\alpha<\omega_1$ let $\gamma(x)=\omega_1$.

Define

$$h:X\to\omega_1+1:x\mapsto\gamma(x)\;;$$

it’s not hard to check that $h$ is a homeomorphism and hence that $X$ is not metrizable.

Alternatively, and with even less effort, just observe that $X$ is not first countable at the point $\langle\alpha:\alpha<\omega_1\rangle$.

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Brian Scott's answer and egreg's comment give counterexamples, but here's an even simpler one. The product of uncountably many copies of a $2$-point discrete space is not metrizable (not even first-countable). But it is the inverse limit of an inverse system of finite products, each of which is a finite discrete space and therefore metrizable.

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