Definition: Let X be a metric space, and E $\subseteq X$, E is closed if it is equal to its closure.

Definition: A metric subset U of X is open if for every point in U there exists an open ball centered around the point that is contained within U.

Theorem: Let X be a metric space. A Subset $Y \subseteq X$ is closed iff $Y^c$ is open.


Assume Y is closed so $Y= \bar{Y}$ (so it is equal to is closure). Which means that $y\in Y \iff \forall$ $r>0 B(y,r) \cap Y \neq \emptyset$.

Let $a \in Y^c$ be arbitrary. Since Y is closed $\exists r>0 : B(a,r) \cap Y = \emptyset$. This means that $B(a,r) \subseteq Y^c$. Since a was arbitrary, $Y^c$ is open.

Suppose $Y^c$ is open. So $y\in Y^c$ $\iff$ $\exists r>0$ : $B(y,r) \subseteq Y^c$. Let x $\in \bar{Y}$ be arbitrary. Note that it suffices to show that $\bar{Y} \subseteq Y$. So for any possible $r>0$ we have $B(x,r) \cap Y \neq \emptyset$. Since $Y^c$ is open, it follows that $x \in Y$.

Is the proof correct? I would very much love feedback, I'd be very very thankful.

  • 2
    $\begingroup$ in general, it would be helpful to include your definitions of open and closed, because there are many equivalent ways of approaching a subject. In fact, the theorem you wrote is sometimes taken as the definition. Even though we might be able to hazard a guess as to which definition you're using based on what you've written, including the definitions can help us to provide better answers (so that we don't assume facts unknown to you etc) $\endgroup$ – peek-a-boo Jun 4 at 2:59
  • $\begingroup$ @peek-a-boo I just did. Thanks for telling me. $\endgroup$ – topologicalmagician Jun 4 at 3:11
  • $\begingroup$ A metric space is closed by definition. $\endgroup$ – copper.hat Jun 4 at 4:09
  • $\begingroup$ Closed is usually defined as the complement of open, so you may want to specify what you mean by closed. $\endgroup$ – copper.hat Jun 4 at 4:11
  • $\begingroup$ @copper.hat I did. A subspace E of X is closed if $E=\bar{E}$, where $\bar{E}$ is defined to be the closure, i.e. the set of all limit/adherent points. $\endgroup$ – topologicalmagician Jun 4 at 4:12

I'm not concinced by the last part of the $Y^\complement$ open implies $Y$ closed part. Why "Since $Y^\complement$ is open it follows that $x \in Y$" is unclear: but can this be expanded:

Assume $x \notin Y$ then $x \in Y^\complement$ and by openness there is some $r>0$ such that $B(x,r) \subseteq Y^\complement$ or equivalently, $B(x,r) \cap Y=\emptyset$, but that contradicts $x \in \overline{Y}$. So $x \in Y$ must hold.

  • $\begingroup$ Suppose $Y^c$ is open. So $y\in Y$ $\iff$ $\forall r>0$ : $B(y,r) \cap Y \neq \emptyset$. Why isn't that enough? $\endgroup$ – topologicalmagician Jun 4 at 4:53
  • $\begingroup$ Your A appeared while I was removing my usual infinite series of typos from mine. $\endgroup$ – DanielWainfleet Jun 4 at 4:53
  • $\begingroup$ @topologicalmagician I see no (logical direct) link between the two statements $Y^\complement$ open and that equivalence. $\endgroup$ – Henno Brandsma Jun 4 at 5:55
  • $\begingroup$ @Henno Brandsma, what about the contrapositive of $y\in Y^c$ iff $\exists r>0 $ : $B(y,r) \subseteq Y^c$? $\endgroup$ – topologicalmagician Jun 4 at 13:36
  • $\begingroup$ P iff Q is equivalent to not P iff not Q $\endgroup$ – topologicalmagician Jun 4 at 13:46

The last sentence is unclear. I think you should add a little, like: $$x\in \bar Y \implies \forall r>0\,(B(x,r)\cap Y\ne \emptyset) \implies$$ $$\implies (\neg (\exists r>0\, (B(x,r)\subset Y^c)\,))\implies$$ $$ \implies (\neg (x\in Y^c))\; \text {...[because } Y^c \text { is open...] }\implies$$ $$ \implies x \in Y.$$

  • $\begingroup$ Suppose $Y^c$ is open. So $y\in Y$ $\iff$ $\forall r>0$ $B(y,r) \cap Y \neq \emptyset$. Isn't that enough? $\endgroup$ – topologicalmagician Jun 4 at 4:54
  • $\begingroup$ I would very much appreciate your help $\endgroup$ – topologicalmagician Jun 4 at 5:03
  • $\begingroup$ It basically is the contrapositive of "So $y\in Y^c \iff ….."$ $\endgroup$ – topologicalmagician Jun 4 at 5:08
  • 1
    $\begingroup$ It's not that you made a mistake. It's just that it's not totally obvious how your last sentence follows from the 2nd last. $\endgroup$ – DanielWainfleet Jun 4 at 5:10
  • $\begingroup$ but is using the contrapositive logically correct? Is what I used in fact the contrapositive? $\endgroup$ – topologicalmagician Jun 4 at 13:52

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.