Let $(X,d)$ be a metric space, let $x\in X$, let $\delta <\epsilon$. Then, $\overline{B(x, \delta)} \subset B(x,\epsilon)$.

Question

Let $(X,d)$ be a metric space, let $x\in X$, and let $\delta <\epsilon$. Then, $\overline{B(x, \delta)} \subset B(x,\epsilon)$.

My attempt:- Let $y\in \overline{B(x, \delta)}$ iff every open set containing $y$ has a non-empty intersection with $y$. Let $U_y$ be the open set containing $y$. Let $t \in U_y \cap \overline{B(x, \delta)}$, then $d(x,t)<\delta < \epsilon$. So, $y\in \overline{B(x, \epsilon)}$. I am not getting $y\in {B(x, \epsilon)}$. Please help me.

Answer 1

If $z\in \overline{B(x, \delta)}$, then for each $n\in \mathbb N$, $B(z,1/n)\cap B(x, \delta)\neq \emptyset, $ so we may choose $y_n\in B(z,1/n)\cap B(x, \delta)$ and use the triangle inequality to write $d(z,x)\le d(z,y_n)+d(y_n,x)<1/n+\delta.$ As this is true for all integers $n$, we must have $d(z,x)\le \delta.$

To finish, choose $\delta<r<\epsilon$, and observe that $\overline{B(x, \delta)}\subset B(x,r)\subset B(x,\epsilon).$

Answer 2

Show that $y \in \overline{B(x,\delta)}$ implies $d(y,x) \le \delta$. So then $d(y,x) < \varepsilon$ and you've got the inclusion.

To see the implication, go for the contrapositive: $d(y,x) > \delta$ implies $y \notin \overline{B(x,\delta)}$, which follows as when $d(y,x) > \delta$, the open ball $B(y, d(y,x)-\delta)$ misses $B(x,\delta)$ by the triangle inequality (check this!) showing that $y \notin \overline{B(x,\delta)}$.