# An open ball contains another open ball centred at each of its points

This page is a stub, so it contains little or minimal information and is on a to-do list for being expanded.The message provided is:
Elementary metric space theorem. Needed for proof of the metric topology (precursor to If the intersection of two open balls is non-empty then for every point in the intersection there is an open ball containing it in the intersection which is a precursor to the metric topology

## Statement

Let [ilmath](X,d)[/ilmath] be a metric space. Then we claim:

• [ilmath]\forall x\in X\forall r_1\in\mathbb{R}_{>0}\forall p\in B_{r_1}(x)\exists r_2\in\mathbb{R}_{>0}[p\in B_{r_2}(p)\wedge B_{r_2}(p)\subseteq B_{r_1}(x)][/ilmath][Note 1][Note 2]Important:[Note 3]

In words:

• For all open balls of the metric space [ilmath](X,d)[/ilmath] for every point in that open ball there exists another open ball centred at that point such that this second open ball is entirely contained in the former.

## Proof

• Let [ilmath]x\in X[/ilmath] be given
• Let [ilmath]r_1\in\mathbb{R}_{>0} [/ilmath] be given, so now we have the open ball [ilmath]B_{r_1}(x)[/ilmath].
• Let [ilmath]p\in B_{r_1}(x)[/ilmath] be given
• Thinking of open balls in the plane (i.e. open disks) we see that the closer [ilmath]p[/ilmath] is to the "rim" the smaller the ball around it would have to be.
• TODO: More explanation would be good. It's just so obvious to me now!
• Choose [ilmath]r_2:\eq r_1-d(x,p)[/ilmath] - Note that anything smaller than [ilmath]r_1-d(x,p)[/ilmath] would also work, for example [ilmath]\frac{1}{2}(r_1-d(x,p))[/ilmath][Note 4]
1. We require firstly that [ilmath]p\in B_{r_2}(p)[/ilmath], this has already been discussed. Recall [ilmath]d(p,p)\eq 0< r_2[/ilmath] so [ilmath]p\in B_{r_2}(p)[/ilmath], and in addition we require:
2. [ilmath]B_{r_2}(p)\subseteq B_{r_1}(x)[/ilmath]. Recall by the implies-subset relation this is really [ilmath]\forall q\in B_{r_2}(p)[q\in B_{r_1}(x)][/ilmath]
• Let [ilmath]q\in B_{r_2}(p)[/ilmath] be given. We must show [ilmath]q\in B_{r_1}(x)[/ilmath]
• By definition of [ilmath]q[/ilmath], we see [ilmath]d(q,p)<r_2[/ilmath]
• By definition of [ilmath]r_2[/ilmath] we see [ilmath]d(q,p)<r_1-d(x,p)[/ilmath]
• Rearranging we see: [ilmath]d(q,p)+d(x,p)<r_1[/ilmath]
• As [ilmath]d(a,b)\eq d(b,a)[/ilmath] for a metric, and by commutativity of addition of the reals we:
• [ilmath]d(x,p)+d(p,q)<r_1[/ilmath]
• Using the triangle inequality property of a metric we see:
• [ilmath]d(x,q)\le d(x,y)+d(y,q)[/ilmath] for any [ilmath]y\in X[/ilmath]
• Picking [ilmath]y:\eq p[/ilmath] we see:
• [ilmath]d(x,q)\le d(x,p)+d(p,q)[/ilmath]
• Combining these we see:
• [ilmath]d(x,q)\le d(x,p)+d(p,q)<r_1[/ilmath]
• So: [ilmath]d(x,q)<r_1[/ilmath]
• Which is equivalent to [ilmath]q\in B_{r_1}(x)[/ilmath]
• Since [ilmath]q\in B_{r_2}(p)[/ilmath] was arbitrary we have shown for all such [ilmath]q[/ilmath] that [ilmath]q\in B_{r_1}(x)[/ilmath]
• We have shown our choice of [ilmath]r_2[/ilmath] is sufficient for the hypothesis to hold
• Since [ilmath]p\in B_{r_1}(x)[/ilmath] was arbitrary we have shown it for all such [ilmath]p[/ilmath]
• Since the ball's centre and its radius were arbitrary we have shown it for all open balls.