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I have a question concerning convergence and divergence of sequences in terms of notations from propositional calculus.

Suppose we have the following statements $P(x)$, $Q(y)$, with $\forall x \exists y [P(x) \implies Q(y)] \equiv \forall x \exists y [\neg P(x) \lor Q(y)]$ (1)

If we want to negate (1), we would have the following:
$\neg(\forall x \exists y) [P(x) \implies Q(y)] \equiv \neg(\forall x \exists y) [\neg P(x) \lor Q(y)]\equiv \exists x \forall y [P(x) \land \neg Q(y)]$ (2)

Now suppose we let $P(N,\epsilon):=n>N$ and $Q(N,\epsilon):= \mid x_{n}- l\mid < \epsilon$, and the definition of convergence is usually given in the form of (1), we have:
$\forall \epsilon > 0 \exists N>0 \forall n\in \mathbb{N} [n > N \implies \mid x_{n}- l\mid < \epsilon] \equiv \forall \epsilon > 0 \exists N>0 \forall n\in \mathbb{N} [\neg (n > N) \lor \mid x_{n}- l\mid < \epsilon]\equiv \forall \epsilon > 0 \exists N>0 \forall n\in \mathbb{N} [n \leq N \lor \mid x_{n}- l\mid < \epsilon]$ (3)
If we want to negate (3) along the lines of (2) to give the definition of divergene, won't it go as follows:

$\exists \epsilon > 0 \forall N>0 \exists n\in \mathbb{N} [\neg (\neg(n > N)) \land \mid x_{n}- l\mid \geq \epsilon] \equiv \exists \epsilon > 0 \forall N>0 \exists n\in \mathbb{N} [n > N \land \mid x_{n}- l\mid \geq \epsilon]$

Am I correct in my translation to propositional calculus notation. Thank you in advance.

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  • $\begingroup$ The three leading quantifiers are correct. $\endgroup$ – Mauro ALLEGRANZA Feb 1 at 9:33
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    $\begingroup$ Then you have the formula $\lnot [n > N \to |x_n - l| < \epsilon]$. $\endgroup$ – Mauro ALLEGRANZA Feb 1 at 9:33
  • $\begingroup$ The negation of $P \to Q$ is $P \land \lnot Q$. Thus, the result will be : $[n > N \land |x_n - l| \ge \epsilon]$ as you correctly writed. $\endgroup$ – Mauro ALLEGRANZA Feb 1 at 9:34
  • $\begingroup$ @MauroALLEGRANZA. The "$<$" in $|x_n-l|<\epsilon$ in the proposer's last line should be "$\ge$". I suspect this is merely a typo. $\endgroup$ – DanielWainfleet Feb 1 at 10:28
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    $\begingroup$ @DanielWainfleet thank you for pointing out the typo. $\endgroup$ – Seth Mai Feb 1 at 15:11

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