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

1,015

Solution 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 $=0$.

Solution 2

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.

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Weeeezard
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Updated on August 01, 2022

Comments

  • Weeeezard
    Weeeezard over 1 year

    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.

    • Matt Dickau
      Matt Dickau almost 8 years
      Hint: in the second one you can't exchange the quantifiers because one of them is inside an implication statement. But in the first one you can exchange the quantifiers, since they are adjacent.