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I am asked to prove:

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.

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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 $P \implies Q.$

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  • $\begingroup$ Agh, contrapositive was the word I wanted. I'll fix that! $\endgroup$ – Decaf-Math Sep 20 '15 at 4:55
  • $\begingroup$ @pyrazolam Yes the terminology is useful, but keep in mind that the contrapositive is the same (logically) as the original implication, so if you were trying at that point to negate the implication you wouldn't want the contrapositive at that point. $\endgroup$ – coffeemath Sep 20 '15 at 4:57
  • $\begingroup$ 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 $U\cap V = \emptyset$. $\endgroup$ – Decaf-Math Sep 20 '15 at 5:12
  • $\begingroup$ @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. $\endgroup$ – coffeemath Sep 20 '15 at 5:19

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