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

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


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[Note 1]). 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][Note 2][Note 3]
    • 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


  • 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:
        • 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.


  1. The length of the following makes it hard to read:
    • [ilmath][\ball{1}\cap\ball{2}\neq\emptyset]\implies\big[\forall x_3\in\ball{1}\cap\ball{2}\exists r_3\in\mathbb{R}_{>0}[x_3\in\ball{3}\wedge\ball{3}\subseteq\ball{1}\cap\ball{2}]\big][/ilmath]
  2. TODO: The following is currently speculation and should be confirmed in the future

    That is why we try to find [ilmath]B_3[/ilmath] - centred at [ilmath]x_3[/ilmath] rather than any other open ball in the intersection that contains [ilmath]x_3[/ilmath] say.

    • This should all make sense once the metric topology page is completed and we have a definition of open set to show equivalence with
  3. It would be an acceptable and unambiguous abuse of notation to write:
    • [ilmath]x\in B_3\subseteq B_1\cap B_2[/ilmath] rather than [ilmath]x\in B_3\wedge B_3\subseteq B_1\cap B_2[/ilmath]