$A$ is an open subset of $M$ $\iff$ ($x_n\to a\implies x_n\in A$ for large $n$)

My definition of an open subset $A$ of $M$ is the one that for every $x\in A$, there is an open ball contained in $A$. Now, suppose that $x_n\to a$. By definition, $\forall \epsilon>0$ there exists $n_0\in \mathbb{N}$ such that $n>n_0\implies d(x_n, a)<\epsilon$. Since we're in an open set, there exists a ball, with some $\epsilon_2$ and center $a$, contained in $A$. Pick $\epsilon = \epsilon_2$ and we have:

$n>n_0\implies d(x_n,a)<\epsilon_2$ and $B(a,\epsilon)\subset A$, therefore $x_n\in A$ for sufficiently large $n$.

Now, suppose we have the condition:

$$x_n\to a \implies x_n\in A$$

This says that there exists $n_0$ such that for $n>n_0$, $x_n\in A$ and $d(x_n, a)<\epsilon$.

I'm stuck here. I cannot simply take a ball and say that it's entirely contained in $A$.

• How about trying contraposition? Suppose there does not exist such ball. Then for any radius $1/n$ you find an $x_n$ with $d(x_n,a) < 1/n$ that does not lie in $A$. This sequence converges to $a$. – TastyRomeo May 22 '16 at 22:22

HINT: Prove the contrapositive instead: suppose that $A$ is not open, and show that there are some $x\in A$ and some sequence $\langle x_n:n\in\Bbb N$ contained entirely in $M\setminus A$ that nevertheless converges to $x$. This is fairly straightforward and logically equivalent to what you want.