Show that there exists a set $U$ which is both open and closed and $x \in U \subseteq V$.

Let $$X$$ be a compact topological space. Suppose that for any $$x, y \in X$$ with $$x \neq y$$, there exist open sets $$U_x$$ and $$U_y$$ containing $$x$$ and $$y$$, respectively, such that $$U_x \cup U_y = X\quad \text{and}\quad U_x \cap U_y = \varnothing.$$ Let $$V \subseteq X$$ be an open set. Let $$x \in V$$ . Show that there exists a set $$U$$ which is both open and closed and $$x \in U \subseteq V$$.

My Try:

$$\forall \;x\neq y\quad U_x \cap U_y = \varnothing$$, $$X$$ is Hausdorff hence a $$T_1$$-space. $$X$$ is compact Hausdorff hence normal subsequently normal $$T_1$$-space i.e. $$T_4$$-space. As a consequence of $$T_1$$-space, $$\{x\}$$ is closed in $$X$$.

Given $$V$$ is open and $$x\in V$$. So $$\bbox[5px,border:1px solid red]{\text{there exists open set U such that x \in U \subseteq V.}}$$ $$U^c$$ is closed. By normality of $$X$$ there exists disjoint open sets $$W_1$$ and $$W_2$$ such that

$$\{x\}\subseteq W_1\subseteq U \;\text{and}\; U^c\subseteq W_2\implies W_2^c\subseteq U\tag 1$$

$$W_2^c\subseteq U\implies X=W_2\cup W_2^c\subseteq U \cup W_2 \tag 2$$ Claim: $$U\cap W_2=\varnothing$$. For suppose $$x\in U\cap W_2$$ then, $$x\in U \;\text{and}\; x\in W_2\implies\bbox[5px,border:1px solid red]{{ x\not\in U^c \;\text{or}\; x\not\in W^c_2\implies x\not\in U^c\cup W^c_2}}$$ Also from $$(1)$$ and $$(2)$$ we see that $$U^c\cup W^c_2\subseteq U\cup W_2\implies x\not\in U^c\cup W^c_2\subseteq U\cup W_2$$ a contradiction since $$x\in U\cap W_2$$ but $$x\not\in U \cup W_2$$. Hence the claim subsequently $$U$$ is both open and closed such that $$x \in U \subseteq V$$.

Is there anything incorrect or missing in my proof ? Are there any other alternative proofs?

Update: I figured the two highlighted portions are the incorrect parts of the proof. Thanks to @Thomas Andrews and @Hagen von Eitzen for clarifying my doubts and for their answers.

• Why say "there exists open set $U$ such that..." since you can just use $U=V$ itself is an open subset? – Thomas Andrews Apr 24 at 19:09
• Also, "$x\notin A\text{ or }x\notin B\implies x\notin A\cup B$" is a non-sequitur – Hagen von Eitzen Apr 24 at 19:19
• If your proof fails to work when $U=V,$ then something is wrong with your proof, because your only assumption about $U$ is that $x\in U\subseteq V$ and $U$ is open. $V$ satifies that condition, which means that $V$ must be clopen, if your proof is corrent (and thus every open subset of $X$ is clopen.) – Thomas Andrews Apr 24 at 19:27
• That misunderstands a basic piece of logic, @YadatiKiran. You've said the rest of your proof works if we find a $U$ which fits a basic criterion. But $V$ fits that criterion. If it doesn't work for $V,$ your proof fails. – Thomas Andrews Apr 24 at 19:31
• The problem in your proof is in the next section. You assume that if $x\notin A$ and $A\subseteq B$ then $x\notin B.$ That is false - you can just set $B=A\cup\{x\}$ to show that. – Thomas Andrews Apr 26 at 15:13

We are given (with improved notation) that for $$x,y\in X$$ with $$x\ne y$$, there exist open sets $$U_{(x,y)}$$ and $$U_{(y,x)}$$ such that $$x\in U_{(x,y)},\quad y\in U_{(y,x)},\quad U_{(x,y)}\cup U_{(y,x)} =X,\quad U_{(x,y)}\cap U_{(y,x)} =\emptyset.$$ Now assume $$x\in X$$ and $$V\ni x$$ is open. Then $$V$$ together with all $$U_{(y,x)}$$, $$y\in V^\complement$$ form an open cover of $$X$$. Pick a finite sub-cover consisting of $$V$$ and some $$U_{(y_i,x)}$$, $$i=1,2,\ldots, n$$. Let $$U=\bigcap_{i=1}^nU_{(x,y_i)}.$$ Then $$U$$ is a finite intersection if clopen sets, hence clopen. Clearly $$x\in U$$. And as the $$U_{(y_i,x)}$$ cover $$V^\complement$$, it follows that $$U\subseteq V$$.

• +1.. It covers the case $V^C=\emptyset$ (...which is a trivial case as then we can let $U=V=X$...) only if in your def'n of $U$ we put $n=0$ and allow that $\cap \emptyset =X.$ – DanielWainfleet Apr 25 at 7:20

The problem in your proof is after you conclude: $$x\not\in U^c\cup W^c_2\subseteq U\cup W_2$$

From which you deduce, incorrectly, that $$x\notin U\cup W_2.$$ But if $$x\notin A$$ and $$A\subseteq B$$ you cannot conclude $$x\notin B.$$ Just take, for example, $$B=A\cup \{x\}.$$

A proof.

For each $$y\notin V,$$ we have a both closed and open set $$U_y$$ with $$x\notin U_y.$$ Then the set of all $$U_y$$ is an open cover of $$V^c,$$ a closed subset of compact space, so there must be $$y_1,\cdots, y_n$$ which cover $$V.$$

But then $$U=U_{y_1}^c\cap U_{y_2}^c\cap \cdots \cap U_{y_n}^c$$ is a finite intersection of sets that are both closed and open, and hence $$U$$ is both closed and open. Also, since $$x\in U_y^c$$ for every $$y,$$ $$x\in U$$, and finally, since the $$U_{y_i}$$ cover $$V^c,$$ we have that $$U=\left(U_{y_1}\cup U_{y_2}\cup \cdots U_{y_n}\right)^c\subseteq V.$$

As noted in a comment to another proof, you have to take some liberties with the argument when $$V=X,$$ when necessarily $$n=0.$$

If we think of $$U$$ as: $$U=X\setminus\left(U_{y_1}\cup \cdots \cup U_{y_n}\right)$$ it is more obvious that when $$V=X$$, and hence $$n=0,$$ we get $$U=X.$$

• Very nice proof. I also don't like to remember the results. – Unknown x Apr 26 at 14:51
• You are right. I realise I have complicated the argument with too much with sets and their compliments. – Yadati Kiran Apr 26 at 18:12