EXERCISE Show that $F$ is closed set $\iff$ if for all ball centered at $x$ contains points in $F$, then $x\in F$.

Thanks by your replies.

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.

  • 4
    $\begingroup$ What definition of closed set are you using? $\endgroup$ – checkmath Aug 18 '12 at 22:03
  • $\begingroup$ a set is closed if that complement is open $\endgroup$ – juaninf Aug 18 '12 at 22:03
  • 4
    $\begingroup$ @Juan What is your definition of an open set? $\endgroup$ – Alex Becker Aug 18 '12 at 22:06
  • $\begingroup$ I do not understand the statement in the question (after if for). $\endgroup$ – Did Aug 18 '12 at 22:09
  • 1
    $\begingroup$ Dear Juan look at the definitions of open and closed sets you are using and you will easily figure it out. Don't mind the downvoted you had on your post. People are a bit bitter here. $\endgroup$ – checkmath Aug 18 '12 at 22:21

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 $x \notin F$.

  • 1
    $\begingroup$ @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. $\endgroup$ – William Aug 18 '12 at 22:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.