General topology multiple choice question Let $X$ be a normal Hausdorff space. Let $A_1$, $A_2$, and $A_3$ be closed subsets of $X$ which are pairwise disjoint. Then there always exists a continuous real valued function $f$ on $X$ such that $f(x) = a_i$ if $x$ belongs to $A_i$, $i=1,2,3$


*

*iff each $a_i$ is either 0 or 1.

*iff at least two of the numbers $a_1$, $a_2$, $a_3$ are equal.

*for all real values of $a_1$, $a_2$, $a_3$.

*only if one among the sets $A_1$, $A_2$, $A_3$ is empty.

The third choice is correct. Recall Urysohn's lemma, which states that a space is normal iff disjoint closed sets can be separated by a function. Since $A_1$ and $A_2$ are disjoint closed sets, we have a continuous function $g:X\to [0,1]$ such that $g(x)=0$ for $x\in A_1$ and $g(x)=1$ for $x\in A_2$. Since $A_2\cup A_3$ and $A_3$ are closed disjoint sets, we have a continuous function $h:X\to [0,1]$ such that $h(x)=0$ if $x\in A_1\cup A_2$ and $g(x)=1$ if $x\in A_3$. Thus we have the continuous function $(g+2h):X\to \mathbb R$ which satisfies
$$(g+2h)(x)=\begin{cases}
0 &\text{if } x\in A_1\\
1 &\text{if } x\in A_2\\
2 &\text{if } x\in A_3\\
\end{cases}$$
and so composing this with your favorite function $p:\mathbb R\to\mathbb R$ which sends $0$ to $a_1$, $1$ to $a_2$, and $2$ to $a_3$ gives the desired function $f$.
