# Extremally disconnected space and separating function

Suppose $$\forall U, V \subset X$$ open, disjoint sets in $$X$$ it holds that $$\overline{U} \cap \overline{V} = \emptyset$$ as well.

I want to show that for every two disjoint open sets, $$U, V$$ there is a continuous function $$f: X \to [0,1]$$ that separates $$U, V$$. That is, $$f(U) \subset \{0\}, f(V) \subset \{1\}$$.

My attempt:

If $$U, V$$ are as above, than $$\overline{V} \subset X - \overline{U}$$ and $$\overline{U} \subset X - \overline{V}$$ and these are open sets in $$X$$.

Moreover $$X = X - \overline{U} \cup X - \overline{V}$$.

Setting $$A = X - \overline{U} \cap X - \overline{V}$$, we can define:

$$f(x) = \begin{cases} 0 & \text{if x \in X - \overline{V}} \\ \frac{1}{2} & \text{if x \in A} \\ 1 & \text{if x \in X - \overline{U}} \end{cases}$$

And it seems $$f$$ is continuous. However I feel shaky about this; what am I missing?

First, it seems like the definition of your function doesn't make sense. If $$x\in A$$, then $$x\in X-\overline U$$ and $$x\in X-\overline V$$, so is $$f(x)$$ 0, .5, or 1? Did you mean the following? $$f(x)=\begin{cases}0\quad x\in\overline U\\.5\quad x\in A \\ 1 \quad x\in \overline V\end{cases}$$
• hey, no that's not what I mean; my map is well defined as $A$ stands for the intersection of the above sets. Your map may not be continuous as $\overline{U}$ isn't open – Mariah Oct 10 '18 at 14:25
• I agree that my map isn't continuous, but I still don't see how your map is well defined. Suppose $x\in A$. Then $x\in X-\overline A$ and $x\in X-\overline B$. So what is $f(x)$. Maybe instead you meant that $f(x)=\begin{cases}0\quad x\in(X-\overline V-A)\\.5\quad x\in A\\1 \quad x\in(x-\overline U-A)\end{cases}$. This way, no element $x$ matches the cases for more than one part of the function. – memerson Oct 10 '18 at 15:40