Prove that $A=\{y\in M:d(x, y)\le r\}$ is closed in $M$

Question

Let $(M, d)$ be a metric space. $A=\{y\in M:d(x, y)\le r\}$ is called the closed ball about $x$ of radius $r$. Prove that $A$ is closed in $M$.

My attempt:

I prefer to use the sequences definition: Suppose the sequence $(y_n)\subset A$ converges to some point $y\in M$. We must show that $y\in A\Leftrightarrow d(x, y)\le r$.

Since $(y_n)\subset A$, $d(x, y_n)\le r\ \forall n\in\mathbb{N}$.

Let $\epsilon>0$. $\exists N\in\mathbb{N}$ s.t. $d(y_n, y)<\epsilon$ whenever $n\ge N$.

By the triangle inequality,

$d(x, y)\le d(x, y_n)+d(y_n, y)\ \forall n\in\mathbb{N}$.

For $n\ge N$, $d(x, y_n)+d(y_n, y)<r+\epsilon$

Thus, $d(x, y)<r+\epsilon$ for all positive values of $\epsilon$.

How do I prove that $d(x, y)\le r$?

Answer 1

Use another definition of closedness: e.g. $A$ is closed iff for all $x \notin A$ there is some $s>0$ such that $B(x,s) \cap A = \emptyset$.

Then note that if $y \notin M$ we have by definition that $d(y,x) > r$ and then we can use $s = d(x,y) - r > 0$ in this condition. (triangle inequality!).