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I want to prove the following: Prove that $d(a, B \cup C):=inf \{d(a,z); z\in B\cup C \} $ is the smaller of $d(a,B)$ and $d(a,C)$ for a point a and subsets B, C of a metric space.

My attempt so far:

$d(a,B \cup C) \leq d(a,B)$ and $d(a,B \cup C) \leq d(a,C)$. Hence $d(a,B\cup C)\leq min \{d(a,B), d(a.C)\}$ Now I want to assume that $d(a,B\cup C)< min \{d(a,B), d(a.C)\}$ and look for a contradiction. Now I am not sure if the following is correct: $\exists$ $z \in(B\cup C)$ s.t. $d(a,z)\leq d(a,B\cup C) + \epsilon $ for all $\epsilon>0$. From there I think I can get the contradiction I want but is the last step above true? If so can anyone convince me it is?

Joogs
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  • Yes, that step is correct (as is everything you've written!). You're using the fact, which follows from the definition of the infemum of a set, that there's always an element of a set $S$ that's less than inf$(S) + \epsilon$. (assuming the infemum isn't $\pm\infty$, anyway) – Greg Martin Oct 23 '16 at 19:21
  • How to finish the proof now? I first thought I could use $min(d(a,B),d(a,C) \leq d(a,z)$ but I can not just let $\epsilon$ go to zero right? If I could do that it would also imply that there is a z which is smaller than or equal to the infimum. – Joogs Oct 23 '16 at 19:29
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    Try choosing $\epsilon$ smaller than $\min{d(a,B),d(a,C)}-d(a,B\cup C)$. – Greg Martin Oct 23 '16 at 19:31
  • In general, in what instances can you let $\epsilon$ go to zero in inequalities as these? – Joogs Oct 23 '16 at 19:34

1 Answers1

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Your proof is actually allready finished. Consider the following:

  • Because of the inf-definition of distance, there must be a sequence of points $(z_i)\subset B\cup C$ such that $\lim_{i\rightarrow\infty}d(a,z_i)=d(a,B\cup C)$.
  • Assuming $d(a,B\cup C)< \min\{d(a,B),d(a,C)\}$, for each $\epsilon$ such that $\min\{d(a,B),d(a,C)\}-d(a,B\cup C)>\epsilon>0$, there must exists an $n$ such that

$$ d(a,B\cup C)\le d(a,z_i) < \min\{d(a,B),d(a,C)\} \quad\text{for all }i>n. $$

The contradiction now is that $z_i\in B\cup C$ by assumption, while $z_i\notin B, z_i\notin C$ by the right-hand side of the last equation.

Sebastiaan
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