what is the difference between ∀x∈ℝ, ∀ε>0, (|x|≤ε ⇒ x=0) and ∀x∈ℝ, ((∀ε>0, |x|≤ε) ⇒ x=0)

Question

I am supposed to Prove or disprove: ∀x∈ℝ, ∀ε>0, (|x|≤ε ⇒ x=0) and: ∀x∈ℝ, ((∀ε>0,|x|≤ε) ⇒ x=0)

I think I understand how to prove the second one (by contradiction) but I don't understand what makes it different from the first one.

Answer 1

The statement says that, if we take any real number $x$ and any $\varepsilon > 0$ with $|x| \leq \varepsilon$, then we have $x = 0$. But it is far from being true; just too many counterexamples are available; take $x := 1$ and $\varepsilon := 2$, for example.

However, the second statement says that, if we take any real number whose absolute value is less than or equal to every given number $> 0$, then we have the fact that the real number is $=0$.

Answer 2

The second is a statement about all real numbers $x$. Namely that if $|x|$ is smaller than all $\varepsilon > 0$, then $x$ is 0.

The first is a statement about all pairs of numbers $x \in \mathbb{R}$ and $\varepsilon > 0$. Namely that whenever the $|x| \le \varepsilon$, we can conclude that $x = 0$. But we can find many pairs $x$ and $\varepsilon > 0$ where $x \neq 0$ where we have $|x| \le \varepsilon$. So this disproves the simple implication.

In the second the $\forall \epsilon>0$ statement is as a whole the assumption of the implication which is much stronger.

Answer 3

The first sentence, $$ (\forall x\in \Bbb R)(\forall \varepsilon > 0)\,(\lvert x\rvert\le \varepsilon \Rightarrow x=0) $$ is equivalent to $$ (\forall x\in \Bbb R)\,((\exists \varepsilon > 0)\,(\lvert x\rvert\le \varepsilon) \Rightarrow x=0) $$ because

• $\varepsilon$ doesn't occur in the conclusion of the implication,
• $p\Rightarrow q$ is equivalent to $\neg p\lor q$,
• $(\forall v)\,(r\lor s)$ is equivalent to $(\forall v)\,r\lor s$ if $v$ doesn't occur in $q$,
• $\forall v\,\neg\,p$ is equivalent to $\neg\,\exists v\,p$.

So the sentence says: if there is some $\varepsilon > 0$ such that $\lvert x\rvert \le \varepsilon$, then $x=0$. This is clearly false.

The second sentence, however, is true.