4
$\begingroup$

Let $\mathcal A$ be a collection of subsets of $\Bbb R$.

"$\forall n\in \Bbb N$ and $\forall \varepsilon>0$, $\exists A\in\mathcal A$ such that $n<|a|<n+\varepsilon$ for infinitely many $a\in A$."

What is the negation of above statement?

Is this the correct negation? "$\exists n\in \Bbb N$ and $\exists\varepsilon>0$ such that for each $A\in \mathcal{A}, n<|a|<n+\varepsilon$ holds for finitely many $a\in A$."

Some of my friends say that negation is: "$\exists n\in \Bbb N$ and $\exists\varepsilon>0$ such that for each $A\in \mathcal A, n\ge|a|$ or $|a|\ge n+\varepsilon$ holds for infinitely many $a\in A$."

Which one of these is correct?

$\endgroup$
2
$\begingroup$

As @Omnomnomnom said in his answer (provided that we intend "finitely many" as "only finitely many"), your statement

$\exists n\in \Bbb N$ and $\exists\varepsilon>0$ such that for each $A\in \mathcal A, n<|a|<n+\varepsilon$ holds for only finitely many $a\in A$

is the correct negation. Let us see why (it could be useful for negating similar statements).

The statement you want to negate is

$\forall n\in \Bbb N$ and $\forall \varepsilon>0$, $\exists A\in\mathcal A$ such that $n<|a|<n+\varepsilon$ for infinitely many $a\in A$

which can equivalently be reformulated as

$\forall n\in \Bbb N$ and $\forall \varepsilon>0$, $\exists A\in\mathcal A$ such that $\exists \, \alpha \subseteq A$ infinite such that $n<|a|<n+\varepsilon$ for all $a\in \alpha$.

It is clear that the negation of the last statement is the following:

$\exists n\in \Bbb N$ and $\exists \varepsilon>0$ such that $\forall A\in\mathcal A$ and $\forall \alpha \subseteq A$, if $\alpha$ is infinite then "$n<|a|<n+\varepsilon$" does not hold for some $a\in \alpha$

which is clearly equivalent to your negation, because it says that it is impossible to find an infinite subset of $A$ such that "$n<|a|<n+\varepsilon$" holds for all $a$ in this subset.

Now, your friend's negation

$∃𝑛 \in ℕ$ and $∃𝜀>0$ such that for each $𝐴∈\mathcal{A}$, $𝑛≥|𝑎|$ or $|𝑎|≥𝑛+𝜀$ holds for infinitely many $𝑎∈𝐴$

is not equivalent to your negation because, according to your friend's negation, it is still possible that "$n<|a|<n+\varepsilon$" does not hold for infinitely many $a \in A$, and "$n<|a|<n+\varepsilon$" holds for infinitely many $a \in A$ as well.

$\endgroup$
  • $\begingroup$ That was an awesome approach! $\endgroup$ – Silent Aug 15 at 5:43
1
$\begingroup$

Your statement "$\exists n\in \Bbb N$ and $\exists\varepsilon>0$ such that for each $A\in \mathcal A, n<|a|<n+\varepsilon$ holds for finitely many $a\in A$." is the correct negation.

This is equivalent to the statement "$\exists n\in \Bbb N$ and $\exists\varepsilon>0$ such that for each $A\in \mathcal A, n\ge|a|$ or $|a|\ge n+\varepsilon$ holds for all but finitely many many $a\in A$." This is close to (but not quite the same as) your friends' statement.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.