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? $$f(x)=\begin{cases}0\quad x\in\overline U\\.5\quad x\in A \\ 1 \quad x\in \overline V\end{cases}$$

  • $\begingroup$ 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 $\endgroup$ – Mariah Oct 10 '18 at 14:25
  • $\begingroup$ 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. $\endgroup$ – memerson Oct 10 '18 at 15:40

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