Exercise 10, § 6

Let $X$ be a Souslin space, $f$ a continuous mapping of $X$ into a separated space $Y$. Now prove that if $f(X)$ is not countable, there exists a subspace $A$ of $X$,

¹ One can show that in the space $\mathscr{C}(I)$ of continuous numerical functions on a compact interval $I \subset \mathbf{R}$, endowed with the topology of uniform convergence (which is in fact a Polish space), the set of differentiable functions is not Souslin but has a Souslin complement (cf. S. MAZURKIEWICZ, Fund. Math., t. XXVII (1936), p. 244).

homeomorphic to the triadic Cantor set K (IV, p. 9), such that the restriction of $f$ to A is injective. (Reduce to the case where X is a complete metric space. Now prove that there exists a sieve $C = (C_n, p_n)$ such that for every $n$, $C_n$ is a set with $2^n$ elements, and for each $n$, a mapping $\varphi_n$ of $C_n$ into the set of non-empty closed subsets of X, of diameter $\leq 2^{-n}$, such that: $1^\circ \; \varphi_{n+1}(c) \subset \varphi_n(p_n(c))$ for all $c \in C_{n+1}$; $2^\circ$ for two distinct elements $c, c'$ of $C_n$, $f(\varphi_n(c))$ and $f(\varphi_n(c'))$ are disjoint. One will use I, p. 108, exerc. 17). In particular, every non-countable Souslin space contains a subspace homeomorphic to K, and consequently to the power of the continuum.