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I'm working on a tutorial question. The question asks whether the following claim is true or false, if it is true: one is supposed to provide a proof or counter-example otherwise if it's false.

Let $d:X \times X \rightarrow \mathbb{R}$ be a metric and X be a metric space. Let $A,B \subset X$.

Claim: $A \cap B \not= \emptyset \Rightarrow dist(A,B)=0$

I believe that it's true. My attempt at the proof is as follows

Proof:

  1. Let $A\cap B \not= \emptyset$.
  2. By definition, we know that $dist(A,B) = inf${$d(x,y):x \in A, y\in B$}.
  3. Let Z = {$d(x,y):x \in A, y\in B$}. We note that since $d$ is metric, $\forall$ $\mu \in Z, \mu \geq0$.
  4. Also, since $A\cap B \not= \emptyset$ then $\exists \pi \in A $ s.t $\pi \in B$.
  5. We then see that $d(\pi,\pi)=0 \in Z$.

[Problem start here -- I'm unsure of the following part]

We now we'll show that $0=inf${$Z$}. We know from $(3)$ that $0$ is a lower bound. Let $a$ be another lower bound for $Z$. This means that $a \leq z, \forall z\in Z$, in particular $a\leq 0$ since $0\in Z$. We then conclude that $inf${$Z$}$=0=dist(A,B)$

Most of the numbered list is just interpretation or presentation of what we know already, is it advisable to communicate it because I've seen most people do not include it?

Adeeb
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    I don't understand step 4: How can $\pi$ lie in both $A$ and $B$ if $A \cap B = \varnothing$? (Also, LaTeX has a \inf command.) – Aeolian Feb 17 '13 at 10:56
  • Typo: The claim is corrected. It's not supposed to be $A \cap B = \emptyset$ – Adeeb Feb 17 '13 at 11:02

2 Answers2

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You know that $Z \subset [0,\infty)$ by the definition of a metric. You state this in 3. We also know that $0 \in Z$, by 5. So $Z$ contains it minimum, and the minimum of a set if always its infimum. Done.

Proof of the last statement: let $A$ be a set such that $\inf(A)$ exists. Suppose $m = \min(A)$, then $\inf(A) = m$: $m$ is a lower bound for $A$ by definition of being its minimum, and if $x$ is any lower bound for $A$, $x \le m$, as $x \le$ all elements of $A$, including in particular $m$. So $m$ is the largest lower bound for $A$, and so $m = \inf(A)$.

Henno Brandsma
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Generally, a proof needs to include enough steps and enough verbal direction so that it should be easy for the reader to fill in the rest.

When one as a homework exercise, the condition changes to needing enough detail that the reader (i.e. your professor or grader) is confident that you are able to fill in the rest.

Trying not to change anything, I would have written your proof as follows:

Let $Z = \{ d(a,b) \mid a \in A \text{ and } b \in B \}$.

Because $A \cap B \neq \emptyset$, we may choose $\pi \in A \cap B$.

Because $d(\pi,\pi)=0$, we have $0 \in Z$, and therefore $\text{inf}(Z) \leq 0$.

Because $0 \leq \text{inf}(Z) \leq 0$, we have $\text{dist}(A,B) = 0$.

In particular, I've omitted:

  • I've omitted the initial assumption that $A \cap B \neq \emptyset$. Whether I include it or not would mostly depend in my mood. It may be useful to let the reader know immediately that you are writing a direct proof of the theorem. Upon reflection, I think I would advise including this assumption to a student, and possibly combined it with the definition of $\pi$. e.g. "By the hypothesis, we can choose $\pi \in A \cap B$."
  • Restating the definition of dist. Although I might have included it immediately after defining $Z$ to emphasize the reason I've defined $Z$.
  • I've omitted the proof that $0 \leq \text{inf}(Z)$, on the principle that it's sufficiently obvious.

Alas, I did change one thing in your proof: the details of how to reason about infima. Even knowing that I wanted to change as little as possible, I find it difficult to make myself make an argument from 'first principles' when the clean algebraic approach is available.