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i'm trying to show that:

$[0,1]^{[0,1]}$ is not metrizable. My attempt so far is as follows:

Suppose it is metrizable, then, since it compact by tychonoff, it follows that it is seperable and therefore second countable. How do I reach a contradiction?

I would like a direct argument, without the use of continuums.

I'm not sure what to do next.

May I have help?

3 Answers3

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As given $Y=[0,1]^{[0,1]}$-the product of continuum number of copies of $[0,1]$. $X$ can be identified with the space of all functions $f$ acting from $[0,1]$ to $[0,1]$ equipped with the topology of pointwise convergence.This topology isn't metrizable. $X$ is compact by Tychonoff theorem. The cardinality of a compact metric space cannot be greater than continuum, but the cardinality of $X$ is greater than continuum. Moreover, see chapter $2$ section $21$ Mukres Topology

Unknown
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If $[0,1]^{[0,1]}$ were metrizable, then compactness and sequential compactness would be equivalent. You know $[0,1]^{[0,1]}$ is compact by Tychonoff, so to prove $[0,1]^{[0,1]}$ is not metrizable, we just need to find a sequence from which you cannot extract a convergent subsequence.

Here is one example - for each $n\in\mathbb{N}$ and $x\in[0,1]$, let $f_n(x)$ be the $n$-th digit in the binary expansion of $x$. If you look at the graphs of these $f_n$, they alternate between $0$ and $1$ with increasing frequency (copy and paste \operatorname{floor}\left(\operatorname{mod}\left(x,2\right)\right) into desmos graphing calculator to get an idea). Maybe you can come up with an argument as to why you cannot extract a convergent subsequence from $(f_n)$.

Alternatively, you can also show that $[0,1]^{[0,1]}$ is not first-countable (and therefore not second-countable), which may be easier. With this approach, just keep in mind that the intersection of countably many uncountable co-finite sets is, well, at least non-empty.

Seric
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  • could you elaborate on the thought process of obtaining such a sequence? –  Apr 28 '20 at 19:00
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    In general, there are 2 main approaches to construct sequences with no convergent subsequence. You can make elements of your sequence "shoot off to infinity" in some sense, or you can make elements "approach" some point which is not in the set (like in incomplete metric spaces). In this case, elements of $(f_n)$ "approach" a graph that looks like $y=0$ union $y=1$, which is clearly not a function. Note that in less intuitive topologies, this thought process probably won't be as useful. – Seric Apr 28 '20 at 19:16
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In your question here you show (with some help) that an uncountable product of Hausdorff ($T_1$ suffices) non-trivial spaces is not first countable. (Here we have $|[0,1]|$ copies of the quite non-trivial Hausdorff space $[0,1]$..).

A metrisable space is first countable, so that immediately solves it.

But to flog a dead horse:

  1. $[0,1]^{[0,1]}$ is not sequentially compact (as I showed here) but compact, while these notions are equivalent for metrisable spaces.

  2. The product $[0,1]^{[0,1]}$ is separable, but not second countable (not even first countable as we saw), which is yet another way to see it is not metrisable.

  3. By Urysohn's embedding theorem, it contains a homeomorphic copy of the Sorgenfrey plane, which is not normal so $[0,1]^{[0,1]}$ is not hereditarily normal. Also any $\{f\}$ is not a $G_\delta$ which contradicts perfect normality (again) and so metrisability.

etc etc.

Henno Brandsma
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