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Let $X$ be a Hausdorff space. Then $X$ is normal if and only if for every pair of disjoint closed sets $A,B \subset X$, there exists a continuous map $f: X \rightarrow [0,1]$ such that $f(x) = 0 $ for all $x \in A$ and $f(x)=1$ for all $x \in B$.

($\Rightarrow$) Assuming that $X$ is normal, we have for disjoint closed sets $A$ and $B$ that there exist disjoint open sets $U$ and $V$ such that $A \subset U$ and $B \subset V$. I am thinking that we need to construct a function from $X$ onto $[0,1]$ such that $f(x) = 0 $ for all $x \in A$ and $f(x)=1$ for all $x \in B$ and somehow show that it is continuous using the open sets $U$ and $V$.

($\Leftarrow$) Assume that there exists a continuous map $f: X \rightarrow [0,1]$ such that $f(x) = 0 $ for all $x \in A$ and $f(x)=1$ for all $x \in B$. I am thinking of using the fact that since $f$ is continuous, for any open set $V \subset [0,1]$, we have $f^{-1}(V)$ open in $X$. Then I want to find two open subsets $U$ and $V$ of $[0,1]$ such that $A \subset f^{-1}(U)$ and $B \subset f^{-1}(V)$ and $f^{-1}(U) \cap f^{-1}(V) = \emptyset$. But I am having trouble finding such $U$ and $V$. Any help would be appreciated, thank you!

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    Do you know Urysohn's lemma? – 2017-02-02

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For $\Rightarrow$, consider Urysohn's lemma. This is the hard part, I think.

For $\Leftarrow$, consider $f^{-1}\big([0,1/3)\big)$ and $f^{-1}\big((2/3,1]\big)$ and use the continuity of $f$.

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    Do you meant $f^{-1}([0, 1/3))$? – 2017-02-03
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    I am still stuck on the reverse direction. – 2017-02-04
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    We know that since $f$ is continuous, for every open set $V \subset [0,1]$, $f^{-1}(V)$ is open in $X$. We also know that $A \subset f^{-1}([0,1/3))$ and $B \subset f^{-1}((2/3,1])$. However $[0,1/3)$ and $(2/3,1]$ are not open sets. – 2017-02-04
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    You are correct in that it should be $f^{-1}\big([0,1/3)\big)$. $[0,1/3)$ and $(2/3,1]$ are open in $[0,1]$. – 2017-02-04
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    got it, thanks! – 2017-02-04
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    How do we know that $f^{-1}([0,1/3)) \cap f^{-1}((2/3,1]) = \emptyset $? It seems like it should be true because $f$ is continuous, but I'm not completely sure. – 2017-02-14
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    This does not have to do with continuity. $x \in f^{-1}\big([0,1/3)\big)$ iff $f(x)\in[0,1/3)$. Similarly, $x \in f^{-1}\big((2/3,1]\big)$ iff $f(x)\in (2/3,1]$. It cannot be that $f(x)\in [0,1/3)$ and $f(x)\in(2/3,1]$, simultaneously, so the intersection must be empty. – 2017-02-14
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    I thought this was true only for one-to-one functions, not necessarily in general. – 2017-02-14
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    This is true for 'single-valued' functions, in general (ie, for each $x$ in the domain, $f(x)$ is a single value). – 2017-02-14