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

Stub grade: A*

This page is a stub

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

• Related claims:
  1. For any two open balls sharing the same centre point then one or both of them contains the other
  2. Given two open balls sharing the same centre the one defined to have a strictly smaller radius is contained in the other
• Demote to grade B once this is linked to by a few pages.

[ilmath]\newcommand{\ball}[1]{B_{r_#1}(x_#1)} [/ilmath]

Statement

Let [ilmath](X,d)[/ilmath] be a metric space. Let [ilmath]x_1,x_2\in X[/ilmath] be given and let [ilmath]r_1,r_2\in\mathbb{R}_{>0} [/ilmath] be given also (so that [ilmath]B_{r_1}(x_1)[/ilmath] and [ilmath]B_{r_2}(x_2)[/ilmath] are open balls of [ilmath](X,d)[/ilmath]). Let [ilmath]B_i:\eq B_{r_i}(x_i)[/ilmath] (for brevity). Then:

• [ilmath][B_1\cap B_2\neq\emptyset]\implies\big[\forall x_3\in B_1\cap B_2\exists r_3\in\mathbb{R}_{>0}[x\in B_3\wedge B_3\subseteq B_1\cap B_2]\big][/ilmath]
  • In full this is: [ilmath]\underbrace{\forall x_1\in X\forall r_1\in\mathbb{R}_{>0}\forall x_2\in X\forall r_2\in\mathbb{R}_{>0} }_{\text{any two open balls} } [/ilmath][ilmath]\overbrace{\big[\underbrace{(\ball{1}\cap\ball{2}\neq\emptyset)}_{\text{they overlap} }\implies\underbrace{\forall x_3\in \ball{1}\cap\ball{2} }_{\text{forall }x_3\text{ in that overlap} }\underbrace{\exists r_3\in\mathbb{R}_{>0} }_{\text{there is a radius for the ball at }x_3}\overbrace{[x_3\in \ball{3}\wedge\underbrace{\ball{3}\subseteq\ball{1}\cap\ball{2} }_{\text{this third ball is contained in the overlap} }]}^{\text{such that} }\big]}^{\text{such that} } [/ilmath]

In words:

• If [ilmath]B_1\cap B_2[/ilmath] is non-empty (the open balls overlap) then for any point in their intersection there exists an open ball

Proof

• Let [ilmath]x_1,x_2\in X[/ilmath] and [ilmath]r_1,r_2\in\mathbb{R}_{>0} [/ilmath] be given. (i.e. let [ilmath]\ball{1} [/ilmath] and [ilmath]\ball{2} [/ilmath] be given)
  • Suppose they are disjoint
    • By the nature of logical implication, the result is true in this case regardless of the truth or falsity of the RHS of the implication symbol
  • Suppose they are not disjoint.
    • Let [ilmath]x_3\in\ball{1}\cap\ball{2} [/ilmath] be given
      • We must now pick [ilmath]r_3[/ilmath] such that [ilmath]\ball{3}\subseteq\ball{1}\cap\ball{2} [/ilmath]
      • By An open ball contains another open ball centred at each of its points
        • We see that:
          1. [ilmath]\exists t_1\in\mathbb{R}_{>0}[B_{t_1}(x_3)\subseteq\ball{1}][/ilmath], and
          2. [ilmath]\exists t_2\in\mathbb{R}_{>0}[B_{t_2}(x_3)\subseteq\ball{2}][/ilmath]
        • Now we have [ilmath]t_1,t_2\in\mathbb{R}_{>0} [/ilmath] such that the above hold.
      • Suppose [ilmath]t_1\eq t_2[/ilmath], then obviously [ilmath]B_{t_1}(x_3)\eq B_{t_2}(x_3)[/ilmath]
        • We may write [ilmath]B_{t_1}(x_3)\subseteq B_{t_2}(x_3)[/ilmath] and [ilmath]B_{t_2}(x_3)\subseteq B_{t_1}(x_3)[/ilmath] and not be wrong in either case
      • Suppose [ilmath]t_1\neq t_2[/ilmath], then:
        • By 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 we see that if:
        • [ilmath]B_{\text{min}(t_1,t_2)}(x_3)\subseteq B_{t_2}(x_3)[/ilmath] and [ilmath]B_{\text{min}(t_1,t_2)}(x_3)\subseteq B_{t_1}(x_3)[/ilmath]
          • One of these cases will be equality the other will be a [ilmath]\subseteq[/ilmath] from the linked statement.
      • Choose [ilmath]r_3:\eq\text{min}(t_1,t_2)[/ilmath]
        • Now we have:
          1. [ilmath]\ball{3}\subseteq B_{t_1}(x_3)\subseteq\ball{1} [/ilmath] giving [ilmath]\ball{3}\subseteq\ball{1} [/ilmath], and
          2. [ilmath]\ball{3}\subseteq B_{t_2}(x_3)\subseteq\ball{2} [/ilmath] giving [ilmath]\ball{3}\subseteq\ball{2} [/ilmath]
        • By the implies-subset relation we see that [ilmath]\forall y\in\ball{3}[y\in\ball{1}][/ilmath] and [ilmath]\forall y\in\ball{3}[y\in\ball{2}][/ilmath]
          • Combining these we see: [ilmath]\forall y\in\ball{3}[y\in\ball{1}\wedge y\in\ball{2}][/ilmath]
          • This is easily seen to be the same as [ilmath]\forall y\in\ball{3}[y\in \ball{1}\cap\ball{2}][/ilmath] (by the definition of intersection)
        • By the implies-subset relation again we see that:
          • [ilmath]\forall y\in\ball{3}[y\in \ball{1}\cap\ball{2}][/ilmath] if and only if [ilmath]\ball{3}\subseteq\ball{1}\cap\ball{2} [/ilmath] - as required
        • We still have to show that [ilmath]x_3\in\ball{3} [/ilmath].
          • As [ilmath]y\in\ball{3} [/ilmath] if and only if [ilmath]d(y,x_3)<r_3[/ilmath] and as [ilmath]d(x_3,x_3)\eq 0[/ilmath] we see:
            • [ilmath]d(x_3,x_3)<r_3[/ilmath] thus [ilmath]x_3\in\ball{3} [/ilmath] as required
      • We have shown the result holds for our choice of [ilmath]r_3[/ilmath]
    • Since [ilmath]x_3\in\ball{1}\cap\ball{2} [/ilmath] was arbitrary we have shown it for all.
  • We have shown the implication holds in either case of its LHS
• Since the open balls [ilmath]\ball{1} [/ilmath] and [ilmath]\ball{2} [/ilmath] were arbitrary we have shown it for all

This completes the proof.