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