closed ball always is closed in the meteric space

Q: Show that a closed ball is always closed in the meteric space?

This is my answer. Is it true?

Let t be any point in this complement of B(a;r). Compute the distance between the center of the closed ball and t; call it d.

By setting epsilon = d/2, the open neighborhood (ball) centered at t with radius epsilon is disjoint from the closed ball B(a;r).

Hence, the complement of the closed ball B(a;r) is open, thus the closed ball is a closed set.

could any one help me in this.

Re: closed ball always is closed in the meteric space

The basic idea is correct, except that $\displaystyle \epsilon=d/2$ is insufficient for the proof. For example, let $\displaystyle a=0$ and $\displaystyle r=4$. Then $\displaystyle \overline{B}(a;r)=[\pm 4]$. If you choose $\displaystyle t\in[\pm4]^c=(-\infty,-4)\cup(4,\infty)$ then you may get something like $\displaystyle t=6$, in which case $\displaystyle B(t;d/2)=B(6;3)=(3,9)$, which is not disjoint from $\displaystyle \overline{B}(a;r)=[\pm 4]$.

Instead let $\displaystyle \epsilon=d(a,t)-r$, and this will suffice for the proof.