# Given two open balls sharing the same centre but with differing radius then the one defined to have a strictly smaller radius is contained in the other

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## Statement

Let [ilmath](X,d)[/ilmath] be a metric space. Let [ilmath]x\in X[/ilmath] be given and let [ilmath]r_1,r_2\in\mathbb{R}_{>0} [/ilmath] be given such that [ilmath]r_1<r_2[/ilmath]. Then:

• [ilmath]B_{r_1}(x)\subseteq B_{r_2}(x)[/ilmath] (where [ilmath]B_\epsilon(x)[/ilmath] denotes the open ball of radius [ilmath]\epsilon>0[/ilmath] centred at [ilmath]x\in X[/ilmath])

## Caveat

It would perhaps be better phrased as follows:

• [ilmath]\forall x\in X\forall r_1,r_2\in\mathbb{R}_{>0}[(r_1<r_2)\implies B_{r_1}(x)\subseteq B_{r_2}(x)][/ilmath]
TODO: Adapt the proof to the caveat
- this basically means saying "suppose [ilmath]r_1\ge r_2[/ilmath], then by the nature of logical implication the statement holds regardless of the truth or falsity of the right hand side" then proceeding to the other case. As usual.

## Proof

• Let [ilmath]x\in X[/ilmath] be given.
• Let [ilmath]r_1,r_2\in\mathbb{R}_{>0} [/ilmath] be given such that [ilmath]r_1<r_2[/ilmath]
• We need to show: [ilmath]B_{r_1}(x)\subseteq B_{r_2}(x)[/ilmath]
• Proof of: [ilmath]\forall p\in B_{r_1}(x)[p\in B_{r_2}(x)][/ilmath]
• Let [ilmath]p\in B_{r_1}(x)[/ilmath] be given
• Then by definition of [ilmath]p[/ilmath] we see [ilmath]d(p,x)<r_1[/ilmath]
• But [ilmath]r_1<r_2[/ilmath]
• By transitivity of a strict partial order we see:
• [ilmath]d(p,x)<r_1<r_2[/ilmath] so just: [ilmath]d(p,x)<r_2[/ilmath]
• But [ilmath][d(p,x)<r_2]\iff[p\in B_{r_2}(x)][/ilmath]
• So we see that [ilmath]p\in B_{r_2}(x)[/ilmath] - as required.
• Since [ilmath]p\in B_{r_1}(x)[/ilmath] was arbitrary we have shown it for all
• We have now shown [ilmath]B_{r_1}(x)\subseteq B_{r_2}(x)[/ilmath]
• Since [ilmath]r_1[/ilmath] and [ilmath]r_2[/ilmath] were arbitrary, we have shown it for all radiuses
• Since [ilmath]x\in X[/ilmath] was arbitrary we have shown it for all open balls of strictly differing radiuses.