Prove an inclusion of open balls given an inequality.

Let $$d_1$$ and $$d_2$$ be two metrics on a space $$X$$ such that $$d_1(x_1,x_2) \leq d_2(x_1,x_2)$$ for all points $$x_1,x_2\in X$$. Prove the inclusion $$B_{d_2}(x,r) \subseteq B_{d_1}(x,r)$$ of balls.

So, I understand that the interval that we will get from $$B_{d_1}(x,r)$$ will be larger than that of $$B_{d_2}(x,r)$$ because of the inequality given, but how would you prove this formally?

• What does it mean that $y \in B_{d_1} (x, r)$ or $y \in B_{d_2} (x, r)$? Can you see some implication between those conditions? – Martin R Mar 16 at 13:00

You prove it as you'd expect: by supposing $$y$$ is some arbitrary element of $$B_{d_2}(x, r)$$, and proving that $$y \in B_{d_1}(x, r)$$.
Suppose $$y \in B_{d_2}(x, r)$$. Then $$d_2(x, y) < r$$. Given $$d_1(x, y) \le d_2(x, y)$$, we have $$d_1(x, y) < r$$, i.e. $$y \in B_{d_1}(x, r)$$.