I don't understeand the following proof of Jones lemma, that is presented in "Elementos de Topología general" by Angel Tamariz and Fidel Casarrubias.
Theorem (F.B. Jones)
Let be $X$ a normal and separable. Well if $X$ contain a discrete and closed subspace $Y$ such that $|Y|=k$, then $2^k\le 2^{\aleph_0}$.
Proof. Let $Y$ be a discrete and closed subspace of $X$ such that $|Y|=k$ and let $D$ be a dense and numerable subspace of $X$. Since it result that any $A\subseteq Y$ is a closed set in $X$ and so for any $A\subseteq Y$ there exist two disjoint open sets $U_A$ and $V_A$ of $X$ such that $A\subseteq U_A$ and $Y\setminus A\subseteq V_A$. For any $A\subseteq Y$ we define $C_A=U_A\cap D$. So we observe that if $A,B\subseteq Y$ are such that $A\neq B$ then $\overline{U_A}\neq\overline{U_B}$: indeed since it is $A\neq B$
without loss of generality we can suppose that $A\setminus B\neq\varnothing$ (the case $B\setminus A\neq\varnothing$ it is analogous) and so $\overline{U_A}\cap V_B\neq\varnothing$, since $A\setminus B\subseteq\overline{U_A}\cap V_B$; and so from this we can conclude that $\overline{U_A}\neq\overline{U_B}$, since $\overline{U_B}\cap V_B=\varnothing$. Now since $D$ is dense we can claim that for any $A\subseteq Y$ it result that $\overline{U_A}=\overline{U_A\cap D}=\overline{C_A}$: so we can conclude that if $A,B\subseteq Y$ are such that $A\neq B$, then $C_A\neq C_B$, since otherwise it was result that $\overline{U_A}=\overline{U_B}$. Well we can define an injective function $\psi:\mathcal{P}(Y)\rightarrow\mathcal{P}(D)$ defined by the condiction $\psi(A)=C_A$ for any $A\in\mathcal{P}(Y)$: so we can conclude that $2^k=|\mathcal{P}(Y)|\le|\mathcal{P}(D)|=2^{\aleph_0}$.
Well I don't understand why any $A\subseteq Y$ is closed in $X$: since $X$ is normal, it results that $X$ is $T_1$ and so every finite subset of $Y$ is closed in $X$ but unfortunately I can't claim anyting about an infinite subset of $Y$.
Anyways, I could perhaps solve the question in this way: since $Y$ is a set of isolated points then for any $A\subseteq Y$ it results that $\mathscr{der}(A)=\varnothing$, since $A\subseteq Y\Rightarrow\mathscr{der}(A)\subseteq\mathscr{der}(Y)=\varnothing$.
Could someone help me, please?
Best Answer
A set $Y$ is discrete in $X$ iff all points of $Y$ are isolated in $Y$. So for $y\in Y$ there is an open set $O$ of $X$ such that $O \cap Y =\{y\}$, so all singletons of $Y$ are open in $Y$.
Because all subsets of $Y$ are unions of singletons, all subsets of $Y$ are open in $Y$. It follows that all subsets of $Y$ are closed in $Y$ too. And as in the lemma situation $Y$ is not only discrete but also closed in $X$, all subsets $A$ of $Y$ are also closed in $X$. ($A = C \cap Y$ for some closed $C$ in $X$, but this is an intersection of two closed sets in $X$, so closed in $X$). That’s why we have that many sets of disjoint closed subsets in $X$, every pair of disjoint sets in $Y$ will give such pairs, in particular $A$ and $Y\setminus A$ for any $A \subseteq Y$, as is used here.