Prove that for any $x_0 \in X$ and any $r>0$, the open ball $B_r(x_o)$ is open.
My attempt: Let $y\in B_r(x_0)$. By definition, $d(y,x_0)<r$. I want to show there exists an $r_1\in\mathbb{R^+}$ s.t. $B_{r_{1}}(y)\subseteq B_r(x_0)$. Let $a\in B_{r_{1}}(y)$. Then, $d(a,y)<r_1$. For $a\in B_{r}(x_0)$, $d(a,x_0)<r$. I want to show $d(a,y)<r_1$ implies $d(a,x_0)<r$. By triangle inequality, $d(a,y)\leq d(a,x_0) + d(y,x_0) \rightarrow$ $d(a,y)<r_1\leq d(a,x_0)+d(y,x_0)<2r…$
I'm a little stuck after this point.
Best Answer
You need to specify $r_1$.
For $y\in B_r(x_0)$, let $r_1=r-d(y,x_0)$. Then if $x\in B_{r_1}(y)$ $$ d(x,x_0)\le d(x,y)+d(y,x_0)< r_1+ d(y,x_0)=r. $$ So $ B_{r_1}(y)\subseteq B_r(x_0)$. This shows that $B_r(x_0)$ is open.
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