Let X be a metric space. suppose cl(Z)=X. Show that if $Y\subseteq X$ is nonempty and open, then $Z\cap Y\neq \emptyset$ 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!
 A: 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 $Z$.
A: $X=\bar Z$ 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:
$y\cap Z\neq \emptyset$.
