Negation of Cauchy Criterion

Question

What is the negation of the Cauchy criterion for sequences? I initially believed that the negation of the Cauchy criterion for a sequence ${p_n}$ in $\mathbb{R}$ is

$\exists \epsilon>0:\forall N\in \mathbb{N}, \exists (n\geq N \lor m\geq N):|p_n - p_m| \geq \epsilon$

due to the effect of DeMorgan's Law on the part $\forall(n \geq N \land m\geq N)$ ... of the Cauchy criterion statement.

However, I am unsure if this negation is correct (if, for example, there only exists $n\geq N : |p_n - p_m| \geq \epsilon$, what is the significance of $m$ in $|p_n - p_m|$?).

Answer 1

The cauchy criterion states $\forall \epsilon>0 \exists N\in\mathbb{N}\forall n\geq N \forall m\geq N(|p_n-p_m|)<\epsilon)$. The negation should be $\exists \epsilon > 0 \forall N\in \mathbb{N} \exists n\geq N \exists m\geq N (|p_n -p_m|\ge\epsilon)$.

I think your mistake can be attributed to the following: the cauchy criterion is often written $\forall \epsilon>0 \exists N\in\mathbb{N}\forall n,m\geq N (|p_n-p_m|)<\epsilon)$, which people pronounce as "for all epsilon greater than zero, there existn $N$ in $\mathbb{N}$ such that, for all $n$ and $m$ greater than $N$, $|p_n-p_m|<\epsilon$". However, the "$n$ and $m$" is misleading. This does not mean logical and, but rather just suggest you quantify over $n$ and you then quantify over $m$.

Answer 2

1. $\exists \epsilon>0:\forall N\in \mathbb{N}, \exists (n\geq N \lor m\geq N):|p_n - p_m| \geq \epsilon$

   Expanding out a part of the quantification shorthand in Cauchy's criterion gives $$\forall \varepsilon{>}0\:\;\exists N{\in} \mathbb{N}\:\;\forall n{,}m{\in}\mathbb{N}\:\;\Big(n,m \geq N \implies |p_n - p_m|< \varepsilon\Big),$$ so its negation is $$\exists \varepsilon{>}0\:\;\forall N{\in} \mathbb{N}\:\;\exists n{,}m{\in}\mathbb{N}\:\;\Big(n,m \geq N \quad\text{and}\quad|p_n - p_m| \geq \varepsilon\Big).$$ You can then recompress it as $$\exists \varepsilon{>}0\:\;\forall N{\in} \mathbb{N}\:\;\exists n{,}m{\geq} N\quad|p_n - p_m| \geq \varepsilon,$$ where it is implicit that the indices $n,m$ are elements of $\mathbb N.$

2. In general, due to the long forms of these quantifier notational shorthand, the correct negation of, say, $$\forall x{\in}(4,7) \; P(x) \quad\text{and}\quad \exists x{\le}3 \; P(x)$$ is $$\exists x{\in}(4,7) \; \lnot P(x) \quad\text{or}\quad \forall x{\le}3 \; \lnot P(x),$$ rather than $$\exists x{\in}\color\red{\mathbb R{\setminus}}(4,7) \; \lnot P(x) \quad\text{or}\quad \forall x{\le}3 \; \lnot P(x).$$

3. Notice that the commas and colons in your proposed formula are unnecessary. As they are not typically meant as delimiters, including them is generally a stylistic choice.