Prove:If (X,$\tau$) is a normal space which has an uncountable closed discrete subspace ,then there are at least $2^c$ distinct continuous functions f:(X,$\tau$)-->[0,1].Hint: Urysohn lemma. Stuck for days.Please help. card(R)=c
If (A,$\tau1$) is an uncountable closed discrete subspace of X,then (A,$\tau1$) is normal. F={{{$ai$}{$aj$}}:$ai$,$aj$ $\in$ A and $ai$$\neq$$aj$}.Card($I^F$) $\ge$ $c^{\omega*(\omega-1)}$ = $c^\omega$.That is all I got.
If $A$ is a closed discrete subspace of $X$, then all subsets $S$ of $A$ are closed in $A$ (because $A$ is discrete in the subspace topology), and hence closed in $X$ (because $A$ is closed in $X$) and the same holds for $A \setminus S$.
So the Urysohn lemma says that (as $X$ is normal), for all such $S$ there is a continuous function $f_S: X \to [0,1]$ that is $0$ on all points of $S$ and $1$ on all points of $A \setminus S$.
If the subsets $S_1, S_2$ are different, then its's clear that $f_{S_1}$ and $f_{S_2}$ are different too.
So there are at least as many continuous functions on $X$ as subsets of $A$, so $|C(X,[0,1])| \ge |2^{|A|}$.
You cannot conclude there are at least $2^\mathfrak{c}$ from just uncountability of $A$, it is consistently possible that $|A| = \aleph_1 < \mathfrak{c}$ and $2^{\aleph_1} = \mathfrak{c}$. For a general conclusion, it is best to assume that $A$ is not merely uncountable but at least size $\mathfrak{c}$. (This is only automatic under CH.)