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I am now taking a theory course in microeconomics and my professor gave us some problems on analysis. I am having a hard time getting used to it.

This is my proof

Suppose $$ Z\cap Y=\emptyset \implies Z\subseteq X\setminus Y$$ take $cl()$ on both sides $$cl(Z)\subseteq cl(X\setminus Y)$$ by using $cl(Z)=X$, $$X\subseteq cl(X\setminus Y)$$ it is a contradiction because $$Y\neq \emptyset$$ So, $$Z\cap Y\neq \emptyset$$

I am not sure whether I can take cl() on both sides I feel like my proof definitely have problems since I haven't used openness Thank you in advance!

Mr.Gandalf Sauron
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slowpoke
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  • I don't see the contradiction in $$X\subseteq cl(X\setminus Y)$$ – geetha290krm Sep 10 '22 at 06:25
  • @geetha290krm I thought X cannot be a subset of itself - a nonempty set. Could you please tell me which part has the problem? – slowpoke Sep 10 '22 at 06:26
  • $X \subseteq X\setminus Y$ would be a contradiction but you don't have that. You only have $X\subseteq cl(X\setminus Y)$. – geetha290krm Sep 10 '22 at 06:28
  • Why the heck are these questions getting downvoted?. The op clearly has a doubt in his question and has shown whatever he/she has attempted in the question. – Mr.Gandalf Sauron Sep 10 '22 at 07:54
  • @geetha290krm can you please give me the hint where to start? I am so confused.. :( Thank you so much for your comments – slowpoke Sep 10 '22 at 07:54
  • @Mr.GandalfSauron I think it is because when I first uploaded the question I forgot to add $$ and looked very messy – slowpoke Sep 10 '22 at 07:56

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Suppose that $Z\cap Y=\phi$. Then it means that for all $y\in Y$, you have an open ball say $B(y,r)\subset Y$ and that $B(y,r)\cap Z=\phi$. But what does that mean?. By definition, then $y$ cannot be a limit point of $Z$. This contradicts that $cl(Z)=X$ as $y\notin cl(Z)$.

Alternatively : following your idea, you have $cl(Z)\subset cl(X\setminus Y)$ . But $Y$ is an open set , hence $X\setminus Y$ is closed and hence $cl(X\setminus Y)= X\setminus Y$ . Thus you have $X =cl(Z)\subset X\setminus Y$ which is an impossibility as $Y$ is non-empty.

The problem with your proof is that you have not shown that $cl(X\setminus Y)$ is a proper subset of $X$. So if it is inded the whole of $X$, then there is no contradiction as $X\subset X$ is always true. The reason for that is you are not using the "openness" of $Y$. If $Y$ were any arbitrary non-empty set then $cl(X\setminus Y)$ might equal $X$. Say for example if $X=\Bbb{R}$ with usual euclidean metric and $Y=\{0\}$. Then $cl(X\setminus Y) = X$.

Recall the definition of limit point of a set $A$. A point $x\in X$ (a metric space) is said to be a limit point of a set $A$ if for any open neighbourhood $N_{x}$ of $x$ , you have $(N_{x}\setminus\{x\})\cap A\neq \phi$ . Equivalently , for all $r>0$, you have $(B(x,r)\setminus\{x\})\cap A\neq\phi$ .

The closure of a set $Z$ is the union of $Z$ and the set of limit points of $Z$.

Mr.Gandalf Sauron
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    I fully understood your proof. Taking a cl() on both sides like taking log was a dumb idea. My professor gave us the question while also teaching about a concept 'separable', so I first thought it was something that I should use 'separable' and I was completely lost. Thank you so much your detailed answer! – slowpoke Sep 10 '22 at 08:10
  • Anytime! . But I don't want to classify your "taking closure like log" as a dumb idea. It indeed works out sometimes. You just have to check whether everything makes sense and there aren't any logical holes . For example, in your case, as $Y$ is open you have $X\setminus Y$ is closed. Hence the $cl(X\setminus Y)= X\setminus Y$ which contradicts that $cl(Z)=X$ . You know what...let me add that in the answer as well. Btw separable is when the $Z$ in the question becomes countable. That is a topological space is called separable if it has a countable dense subset. – Mr.Gandalf Sauron Sep 10 '22 at 10:44
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$X=\bar Z$ so $X$ is the smallest closed set containing $Z$.

Definition 1: In a topological space $X$ a set $A$ is closed iff it's complementary set is open.

Definition 2: A point $x$ of a set $A$ is a limit point of $A$ if every open set $Y$ containing $x$ contains other points in $A$.

claim 1: A closed set $A$ contains all it's limit points. proof: suppose $x$ is a limit point of $A$ not in $A$, then $x\in A^c$ which is open. hence there exists a ball $B_r(x)\subseteq A^c$.

contradiction. $\square$

Definition 3: $\bar Z=\cap X_Z$ where every $X_Z$ is a closed set containing $Z$.

claim 2: the set $T$ containing all limit points of $Z$ is closed.

proof: let $x_0$ be a limit point of $T$. then every ball containing $x_0$ has points of $T$ let $d(x,x_0) < \frac{\epsilon}{2}$ then there exists $x_1\in Z$ s.t. $d(x_1, x)<\frac{\epsilon}{2}$. using triangle inequality: $d(x_0,x_1)<d(x_0,x)+d(x,x_1)<\epsilon$. thus $x_0$ is a limit point of $Z$. since $T$ contains all it's limit points it is closed.

$\square$.

claim 3: Let metric $X=\bar{Z}$ then every $x \in X$ is a limit point of $Z$.

proof: $T$ be the set containing all limit points of $Z$, then $X\subseteq T$ since $X$ is the smallest set closed set containing $Z$. but every closed set containing $Z$ contains all the limit points of $Z$. Thus $X=T=\bar Z$.

$\square$. Thus every point in $X$ is a limit point of $Z$. thus: $y\cap Z\neq \emptyset$.

ryaron
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