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

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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


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.


  • 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.


  1. It is an acceptable (and unambiguous) abuse of notation to write:
    • [ilmath]p\in B_{r_2}(p)\subseteq B_{r_1}(x)[/ilmath] instead of [ilmath]p\in B_{r_2}(p)\wedge B_{r_2}(p)\subseteq B_{r_1}(x)[/ilmath].
  2. It is actually trivial that [ilmath]p\in B_\delta(p)[/ilmath] for any [ilmath]\delta\in\mathbb{R}_{>0} [/ilmath] as:
    • [ilmath][x\in B_\delta(p)]\iff[d(x,p)<\delta][/ilmath] and
    • [ilmath][d(x,p)\eq 0]\iff[x\eq p][/ilmath] (by definition of a metric)
    So we could write:
    • [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}[B_{r_2}(p)\subseteq B_{r_1}(x)][/ilmath]
    However the point of this statement is to show that for all points in an open ball there's another open ball (contained inside the first one) that contains the point; so we make it explicit
  3. Note that if we required that for all points in the open ball there is another open ball contained in that ball (but not necessarily centred about the selected point) such that this second open ball contains the selected point; then we could just choose the open ball itself!
  4. Which is certainly more easy to picture mentally as it avoids coming close to the "rim" of the initial open ball.


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