Question

Please show that, if {s^k} is bounded, then, for any element c in the metric space, there is a constant t>0 with d(c,s^k)<=t, for all k.

Answer 1

Task. Show that, if $\{s^k\}$ is bounded, then, for any element $c$ in the metric space, there is a constant $t > 0$ with $d(c,s^k) \leq t$, for all $k$.

Proof. By definition, the set $\{s^k\}$ is bounded if there exists some $A > 0$ such that

$$d (s ^ {k}, s ^ {l}) \leq A$$

for all $k,l\geq 0$. Put

$$t = d (c, s ^ {0}) + A.$$

Then by triangle inequality for any $k$ we have that

$$d (c, s ^ {k}) \leq d (c, s ^ {0}) + d (s ^ {0}, s ^ {k}) \leq d (c, s ^ {0}) + A = t.$$