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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?

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  • $\begingroup$ 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? $\endgroup$ – Martin R Mar 16 at 13:00
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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. $y \in B_{d_1}(x, r)$.

There are no tricks to it.

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