closed ball always is closed in the meteric space

**rqeeb** · August 27th 2011, 04:17 PM

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.

**hatsoff** · August 27th 2011, 06:25 PM

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

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

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