I'm reading this article and others about the same topic and I found that all autors assume the next theorem. First, a definition. Here $Y^{*}:=\beta{Y}\setminus Y$, i.e., $Y^{*}$ is the remainder of the Stone-Cech compactification.
Let $Y$ a topological space. We say that $p\in Y^{*}$ is a butterfly point in $\beta Y$ if there exist $F,G\subseteq Y^{*}\setminus\{p\}$ such that $\{p\}=\text{cl}_{\beta Y}(F)\cap\text{cl}_{\beta Y}(G)$ and $\text{cl}_{\beta Y}(F\cup G)\subseteq Y^{*}$
Next, the theorem
If $Y$ is a non compact topological space and $z\in\beta Y\setminus Y$ is a butterfly point in $\beta Y$, then $\beta Y\setminus\{z\}$ is not normal.
The say that the proof is trivial but I can't figure out. My attempt: as $\text{cl}_{\beta Y}(F)$ and $\text{cl}_{\beta Y}(G)$ are closed sets in $\beta Y$ (also compact sets) then $$F_1=\text{cl}_{\beta Y}(F)\cap (\beta Y\setminus\{z \})=\text{cl}_{\beta Y}(F)\setminus\{ z\}$$ and $$F_2=\text{cl}_{\beta Y}(G)\cap (\beta Y\setminus\{z \})=\text{cl}_{\beta Y}(G)\setminus\{ z\}$$are disjoint closed sets in $\beta Y\setminus\{z \}$. My intuition says that $F_1$ and $F_2$ are two disjoint closed sets that cannot be separated by disjoint open sets and, therefore, $\beta Y\setminus\{z\}$ is not normal. For this, take $U_1$ and $U_2$ open sets of $\beta Y\setminus\{z\}$ such that $F_1\subseteq U_1$ and $F_2\subseteq U_2$. We need to prove that $U_1\cap U_2\neq\emptyset$. But, we can take $V_1,V_2\subseteq\beta Y$ two open sets such that $U_1=V_1\cap(\beta Y\setminus\{z \})$ and $U_2=V_2\cap(\beta Y\setminus\{z \})$. We have the next cases.
- If $V_1=U_1\cup\{z \}$ and $V_2=U_2\cup\{z \}$. From here, $V_1\cap V_2\neq\emptyset$ and therefore $V_1\cap V_2\cap \text{cl}_{\beta Y}(F_1)\neq\emptyset$ and thus $U_1\cap U_2\neq\emptyset$
But I don't know how to proceed with the other cases (when some $U_i$ is open in $\beta Y$ and when both $U_i$ are open in $\beta Y$). Am I doing correctly? At this point, is my proof correct? Any hint? Thanks.
*Edit: I noticed that $U_1$ and $U_2$ are open sets in $\beta Y\setminus\{ z\}$ and $\beta Y\setminus\{z \}$ is open in $\beta Y$. So, my proof is incorrect. Any suggestion? Thanks in advance.
I find these things easier to think about in terms of continuous functions. By Urysohn's lemma, if $\beta Y\setminus\{z\}$ were normal, there would be a continuous function $f:\beta Y\setminus\{z\}\to[0,1]$ which is $0$ on $F_1$ and $1$ on $F_2$. Now by the universal property of $\beta Y$, $f|_Y$ extends continuously to $\beta Y$ and this extension must agree with $f$ on $\beta Y\setminus\{z\}$ (since $Y$ is dense in it). But now since $z$ is in the closure of both $F_1$ and $F_2$, the continuous extension is forced to map $z$ to both $0$ and $1$, which is a contradiction.
(Note that this argument does not require $F$, $G$, or their closures to be contained in $Y^*$; it only needs them to be subsets of $\beta Y\setminus\{z\}$ such that $\{z\}=\text{cl}_{\beta Y}(F)\cap\text{cl}_{\beta Y}(G)$.)