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. $\endgroup$ Jan 27, 2013 at 20:06

1 Answer 1


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? $\endgroup$ Jan 27, 2013 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, 2013 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. $\endgroup$ Jan 27, 2013 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, 2013 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. $\endgroup$ Jan 27, 2013 at 20:35

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