# Negation of Implication to Possibly Make Proof Easier

Show that if $a,b \in \mathbb{R}$, and $a\neq b$, then there exists $\epsilon$-neighborhoods $U$ of $a$ and $V$ of $b$ such that $U \cap V = \emptyset$.

I am looking to see if proving the contrapositive of the statement is easier than the given direction, or perhaps gain some insight on how the "if-then" proof works. I've written up my own proof, but I'm not "convinced" that it's correct. Thus, I began to compose the contrapositive. However, I'm not sure if I've negated parts of the statement completely.

Here's what I said:

Let $P := a,b\in\mathbb {R} \wedge (a \neq b)$ and $Q:= \exists U_\epsilon(a),V_\epsilon(b) : U\cap V = \emptyset$. Our statement is given as $$P \Rightarrow Q.$$ So, $$\neg(P\Rightarrow Q) \equiv \neg Q \Rightarrow \neg P.$$

But I got stuck when trying to negate $Q$ and $P$. I thought: $\neg P := a,b\not\in\mathbb {R} \vee (a = b)$ and $\neg Q:= \forall U_\epsilon(a),V_\epsilon(b) : U\cap V \neq \emptyset$.

A few things confused me just a little: is the proof trivial if $a,b\not\in \mathbb{R}$? Also, I've negated the quantifiers, but am I supposed to negate the "such that" condition part too? These questions assume I negated things correctly. Please let me know if my negations ended up wrong.

I don't think you want to negate the hypotheses $a,b \in \mathbb{R}$ and $a \neq b$. Keep them as they are, and negate only the conclusion, i.e. deny that there are such $\epsilon$ neighborhoods, and go for a contradiction from there.
The negation of the existence of these neighborhoods is that, no matter how small a positive $\epsilon$ is chosen, $U_\epsilon \cap V_\epsilon$ is nonempty.
That said, to me it is better to proceed directly, since if $a \neq b$ you can choose any $\epsilon$ less than $|b-a|/2$ and get it to work.
Just by the way, the negation of $P \implies Q$ is not $\lnot Q \implies \lnot P,$ actually the latter is the contrapositive and is equivalent to $P \implies Q.$
• Thanks for your answer. By the way, what's the problem with choosing $\epsilon = |b-a|/2$? The $\epsilon$-neighborhoods we'd have would be $(a-\epsilon,a+\epsilon)$ and $(b-\epsilon,b+\epsilon)$ which the midpoint of those two intervals would be $(a+b)/2$ which would be $just$ outside of each of these neighborhoods, effectively showing that there exist $U$ and $V$ such that $U\cap V = \emptyset$. – Decaf-Math Sep 20 '15 at 5:12
• @pyrazolam Yes, I was just being short about it, and choosing $\epsilon$ to be exactly equal (not necessarily less) to $|b-a|/2$ works, as you detail in your last comment. – coffeemath Sep 20 '15 at 5:19