EXERCISE Show that $F$ is closed set $\iff$ if for all ball centered at $x$ contains points in $F$, then $x\in F$.
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DEFINITION Given a metric space $(X,\rho)$ and a point $x$, an open ball about $x$ with radius $\epsilon$ is the set $B(x,\epsilon)=\{y\in X:\rho(x,y)<\epsilon\}$
DEFINITION Given a metric space $(X,\rho)$ and a point $x$ in $X$, a set $N$ is a neighborhood of $x$ if it contains an open ball about $x$.
DEFINITION Given a metric space $(X,\rho)$ and a subset $O$ of $X$, $O$ is open if it is a neighborhood of each of its points.
DEFINITION Given a metric space $(X,\rho)$ and a subset $F$ of $X$, $F$ is closed if it is a $X\setminus F$ is open.
Best Answer
Let $X$ be a metric space.
Suppose $F$ is closed. Suppose $x \notin F$. Then $x \in X - F$. Since $X - F$ is open, $x$ is an interior point. There exists a ball $B$ containing $x$ such that $B \subset X - F$. Hence $B$ does not contain points of $F$.
Suppose $F$ is not closed. Then $X - F$ is not open. Thus there exists a point $x \in X - F$ which is not an interior point. That is, every ball $B$ containing $x$ is not completely contained in $X - F$. Hence there is a point $x$ such that every ball containing $x$ intersects $F$ but $x \notin F$.