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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.

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    $\begingroup$ Why say "there exists open set $U$ such that..." since you can just use $U=V$ itself is an open subset? $\endgroup$ – Thomas Andrews Apr 24 at 19:09
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    $\begingroup$ Also, "$x\notin A\text{ or }x\notin B\implies x\notin A\cup B$" is a non-sequitur $\endgroup$ – Hagen von Eitzen Apr 24 at 19:19
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    $\begingroup$ 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.) $\endgroup$ – Thomas Andrews Apr 24 at 19:27
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    $\begingroup$ 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. $\endgroup$ – Thomas Andrews Apr 24 at 19:31
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    $\begingroup$ 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. $\endgroup$ – Thomas Andrews Apr 26 at 15:13
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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 $U\subseteq V$.

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  • $\begingroup$ +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 $\cap \emptyset =X.$ $\endgroup$ – DanielWainfleet Apr 25 at 7:20
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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 $U=X.$

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

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