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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.

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    @checkmath Your *People are a bit bitter here* is odd. Please do not equate each and every downvote with the effect of a kind of moral perversity, some questions ARE badly phrased and DESERVE downvotes (ditto for answers), this is how the system works. (In case you are wondering, my only vote on this page is +1 to Alex Becker's comment.)2012-08-20

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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 $x \notin F$.

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    @tomasz He used the term ball so I assumed it was a metric space. I should probably remove the parenthetical remark, but I just wanted to mention that the result holds for general topological space if ball is replace by open set.2012-08-18