Question

Let (x,d) be a metric space and A a subset of X .prove that A is closed if and only if ,if each convergent sequence of points of A converges to a point of A

Answer 1

$\text{Suppose that a ∈ S̄, then every ball B(a, 1/n) contains a point of S, so we may pick}\\\text{a sequence $(x_n )^∞ _{n=1}$ with $x_n ∈ B(a, 1/n) ∩ S$. Clearly $lim_{n→∞ x_n} = a.$} \\\text{Conversely, suppose $lim_{n→∞} x_n = a$, where $x_n ∈ S$. If $a \notin S̄$ then there must be}\\\text{some ball B(a, ε) not meeting S. But if n is large enough then $d(x_n, a) < ε$, and so}\\\text{$x_n ∈ S ∩ B(a, ε)$, by contradiction. Therefore A is closed}$