In a metric space $(M,d)$, if $A$ is a subset of $M$, then $\bar A$ (closure of $A$) is closed.

My definition of $\bar A$ is $\{x\in M : \forall \varepsilon > 0, \; B(x,\varepsilon) \cap A \neq \emptyset\}$, where $B(x,\varepsilon)$ is the open ball with center $x$ and radius $\varepsilon$.

Partial proof: let's see that $M\setminus \bar A$ is open. If $x\in M\setminus \bar A$ then $x\not\in \bar A$ and $$ \exists \varepsilon > 0\; B(x, \varepsilon)\cap A = \emptyset \implies B(x,\varepsilon) \subset M\setminus A $$ but we need to show that $B(x,\varepsilon) \subset M\setminus \bar A$ so let's suppose that this is not true, therefore $$ B(x,\varepsilon) \cap \bar A \neq \emptyset\implies \exists y \in B(x,\varepsilon)\cap \bar A \implies \exists\delta > 0 \; B(y,\delta) \subset B(x,\varepsilon) \cap \bar A $$ so summarizing $B(x,\varepsilon)\subset (M\setminus A) $ and $B(y,\delta) \subset \bar A$ but I see no contradiction since $B(y,\delta)$ can be contained in $\bar A \cap (M\setminus A)$. I am missing something. What is wrong? (This proof may be very easy but currently I can't find the way to prove it) Thanks in advance.

  • $\begingroup$ Steady on the symbols! A few more words would make your working a lot easier to follow. $\endgroup$ – Clive Newstead Jan 27 '13 at 20:06

Take some $B(y, \delta) \subset B(x,\epsilon)$ then it contains a point of $A$ by the definition of $y\in \overline{A}$.

  • $\begingroup$ I don't get that, it could be $A\subsetneq\bar A$, right? $\endgroup$ – V. Galerkin Jan 27 '13 at 20:08
  • $\begingroup$ @V.Galerkin to be in the closure of $A$ means that every small ball around you contains a point in $A$. In this case, the small $\delta$ sized ball around $y$ must contain a point of $A$. $\endgroup$ – Deven Ware Jan 27 '13 at 20:11
  • $\begingroup$ Ok, so for every $z \in B(y,\delta)$, since $z\in \bar A$ we have that $B(z,r)\cap A \neq \emptyset \; \forall r > 0$ but this contradicts that $B(y,\delta)\subset M\setminus A$... Thank you very much. $\endgroup$ – V. Galerkin Jan 27 '13 at 20:17
  • $\begingroup$ @V.Galerkin not for every $z \in B(y,\delta)$ but the ball around $y$ itself intersects $A$. Note that you cant say that the ball around $y$ is a subset of $\overline{A}$ because that set is not open. $\endgroup$ – Deven Ware Jan 27 '13 at 20:19
  • $\begingroup$ Right, so I should say that $y\in \bar A$ and then every ball $B(y,\delta)$ intersects $A$ but also $B(y,\delta) \subset B(x,\varepsilon) \subset M\setminus A$ and this is absurd. $\endgroup$ – V. Galerkin Jan 27 '13 at 20:35

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.