While reading Arkhangel'skii's *[Topological function spaces](http://books.google.co.uk/books?id=nl3mFq7FzNcC&lpg=PP1&dq=Arkhangel'skii%20topological%20function%20spaces&hl=pl&pg=PP1#v=onepage&q=Arkhangel'skii%20topological%20function%20spaces&f=false)*, I encountered an unexpected application of Martin's Axiom. This is Theorem II.5.20:



> 
> Assume $\mathsf{MA}+\neg \mathsf{CH}$. Let $X$ be a compact Hausdorff space. If every discrete subset of $C\_p(X)$ is countable, then $X$ is metrisable.
> 
> 
> 


Here $C\_p(X)$ stands for the space of all real-valued continuous functions on $X$ with the topology of pointwise convergence.


Is it just a matter of proof or there is a consistent counter-example to this statement? If there is a counter-example, $X$ must be necessarily [hereditarily separable](https://dantopology.wordpress.com/tag/hereditarily-separable/).



Discrete subsets in the topology of pointwise convergence vs. metrisability
Yes, that statement is independent of the axioms of set theory.


Let $X$ be a compact *strong S-space*, that is a compact space such that $X^n$ is hereditarily separable but not hereditarily Lindelof for every $n \in \mathbb{N}$ (De La Vega and Kunen constructed a homogeneous space with these features in <http://www.sciencedirect.com/science/article/pii/S0166864103002153>)


If $C\_p(X)$ contained an uncountable discrete set, this would yield an uncountable discrete set in $X^n$ for some $n$. But this is impossible, because every subspace of $X^n$ is separable. Now, $X$ is not metrizable, because separable metrizable spaces have a countable base, and hence are hereditarily Lindelof.


EDIT: It remains to prove that if $C\_p(X)$ contains an uncountable discrete set then some finite power of $X$ also does. I believe this is due to Tkachuk.


Let $D=\{f\_\alpha: \alpha <\omega\_1\}$ be a discrete set in $C\_p(X)$ of size $\omega\_1$. Let $\mathcal{B}$ be a countable base for $\mathbb{R}$. Since $D$ is discrete, for every $\alpha < \omega\_1$, we can find a positive integer $n\_\alpha$, a sequence $(x^\alpha\_1, \dots x^\alpha\_{n\_\alpha})$ of elements of $X$ and a sequence $(B^\alpha\_1, \dots B^\alpha\_{n\_\alpha})$ of elements of $\mathcal{B}$ such that $[x^\alpha\_1, \dots x^\alpha\_{n\_\alpha}, B^\alpha\_1, \dots, B^\alpha\_{n\_\alpha}] \cap D=\{f\_\alpha\}$, where $[x\_1, \dots x\_n, B\_1, \dots B\_n]$ denotes the following basic open subset of $C\_p(X)$: $\{f \in C\_p(X): (\forall i \leq n)(f(x\_i) \in B\_i)\}$. 


By the pigeonhole principle, we can find an uncountable set $E \subset \omega\_1$, a positive integer $n$ and a finite sequence $(B\_1, \dots B\_n)$ of elements of $\mathcal{B}$ such that $n\_\alpha=n$ and $(B^\alpha\_1, \dots B^\alpha\_{n\_\alpha})=(B\_1, \dots B\_n)$, for every $\alpha \in E$. 


We claim that $S=\{(x^\alpha\_1, \dots, x^\alpha\_n): \alpha \in E \}$ is a discrete subset of $X^n$. Indeed, by definition of the $B\_i$s and the $x^\beta\_i$s, we have that $f\_\beta$ is the only element of $D$ such that $f\_\beta(x^\beta\_i) \in B\_i$ for every $i \leq n$. But this implies that the open set $\prod\_{i=1}^n f\_\alpha^{-1}(B\_i)$ intersects $S$ in the single point $(x^\alpha\_1, \dots x^\alpha\_n)$, and hence $S$ is discrete in $X^n$.


