A problem on Souslin Spaces I came across this problem, and I am having trouble with formulating a proof.
An uncountable Souslin space has the  card of the continuum.
 A: Here is an outline:
Let $X$ be an uncountable Souslin space, and consider a continuous surjection $f : \mathbb{N}^{\mathbb{N}} \to X$.  (We also endow $\mathbb{N}^{\mathbb{N}}$ with the usual metric.)  For each $x \in X$ pick some point in $f^{-1} [ \{ x \} ]$, and gather these into a subset $A$ of $\mathbb{N}^{\mathbb{N}}$.  Next consider the set $A_0$ of all condensation points of $A$.  Then $A_0$ is uncountable and has no isolated points.
Next construct a Souslin scheme $h$ mapping $2^{<\mathbb{N}}$ into the set of open balls in $\mathbb{N}^{\mathbb{N}}$ so that


*

*$\mathop{radius} ( h ( s ) ) < 2^{-\mathop{length}(s)}$;

*$h(s)$ is centred at some $z_s \in A_0$;

*$\overline{ h( s \mathop{^\smallfrown} 0 ) } , \overline{ h( s \mathop{^\smallfrown} 1 ) } \subseteq h ( s )$; and

*$\overline{ f [ h( s \mathop{^\smallfrown} 0 ) ] } \cap \overline{ f [ h( s \mathop{^\smallfrown} 1 ) ] } = \varnothing$.


This then induces an embedding $\hat{h} : 2^{\mathbb{N}} \to \mathbb{N}^{\mathbb{N}}$, defined by $\hat{h} ( \sigma ) = \lim_{n \rightarrow \infty} z_{\sigma \restriction n }$, and it is not difficult to show that $f \circ \hat{h} : 2^{\mathbb{N}} \to X$ is injective.  It follows that $2^{\aleph_0} \leq | X |$, and since the mapping $f$ above is surjective we have $| X | \leq 2^{\aleph_0}$.
A: Yes. First let us recall that a Polish space is a separable completely metrizable topological space. Such space has cardinality $\le \mathfrak c$. A Souslin space is a Hausdorff continuous image of a Polish space. As we known $f: X \to Y$, then |$Y|\le |X|$. So the caridinalty of a countable souslin space $\le \mathfrak c$. If CH holds, then it equals $\mathfrak c$. 
