Compactness of a space and metrizability 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?
 A: 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
A: 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.
A: 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: 


*

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

*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. 

*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. 
