Why is a conjunction not treated as such in the following more complicated statement?

Question

For the statement

• For all $\epsilon>0$ there is some $N\in\mathbb{N}$ such that for all $\boldsymbol{n\in\mathbb{N}}$ satisfying $\boldsymbol{n\ge N}$ and all $\boldsymbol{x\in S}$, $|f_n(x)-f(x)|<\epsilon$

and its negation

• there exists $\epsilon >0$ such that for all $N\in\mathbb{N}$, there exists $x\in S$ and there exists $n\in\mathbb{N}$ such that $n\ge N$ and $|f_n(x)-f(x)|\ge \epsilon,$

why is the boldfaced conjunction negated as two separate quantifiers, whereas the conjunction

• for all $\boldsymbol{x\in S}$ and all $\boldsymbol{n\in\mathbb{N}}$ such that $\boldsymbol{n\ge N}$

would be negated as the disjunction

• there exists $x\notin S$ or there exists $n\in\mathbb{N}$ such that $n < N,$

as expected?

Answer 1

No, the initial string of quantifiers does not include a propositional conjunction.

The logical form is $\forall x \forall y \ \varphi$:

$\forall x \in S, \ \forall n \in \mathbb N \ (n \ge N \to \text {blah})$

Thus, its negation will be:

$\exists x \in S, \ \exists n \in \mathbb N \ (n \ge N \land \lnot \text {blah})$

Answer 2

For the statement

• For all $\epsilon>0$ there is some $N\in\mathbb{N}$ such that for all $\boldsymbol{n\in\mathbb{N}}$ satisfying $\boldsymbol{n\ge N}$ and all $\boldsymbol{x\in S}$, $|f_n(x)-f(x)|<\epsilon$

why is the boldfaced conjunction negated as two separate quantifiers, whereas the conjunction

• for all $\boldsymbol{x\in S}$ and all $\boldsymbol{n\in\mathbb{N}}$ such that $\boldsymbol{n\ge N}$

The strings  "for all $n{\in}\mathbb{N}$ satisfying $n{\ge} N$"  and  "(for) all $x{\in} S$"  and  "(for) all $n{\in}\mathbb{N}$ such that $n{\ge} N$"  can neither be true nor false, so are not propositional functions, so are not propositional conjuncts. Rather, they are quantifications.

Not only is the latter bullet not a conjunction, it isn't even a sentence, but just a fragment of the full sentence "for all $\boldsymbol{x{\in} S}$ and all $\boldsymbol{n{\in}\mathbb{N}}$ such that $\boldsymbol{n{\ge} N}$, it holds that...".

would be negated as the disjunction

• there exists $x\notin S$ or there exists $n\in\mathbb{N}$ such that $n < N,$

as expected?

Correction:  "there exists $x{\notin }S$ and $n{\in}\mathbb{N}$...".

For the same reason as above, this string also contains no propositional conjunction or propositional disjunction.

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For the statement

• For all $\epsilon>0$ there is some $N\in\mathbb{N}$ such that for all $\boldsymbol{n\in\mathbb{N}}$ satisfying $\boldsymbol{n\ge N}$ and all $\boldsymbol{x\in S}$, $|f_n(x)-f(x)|<\epsilon$

and its negation

• there exists $\epsilon >0$ such that for all $N\in\mathbb{N}$, there exists $x\in S$ and there exists $n\in\mathbb{N}$ such that $n\ge N$ and $|f_n(x)-f(x)|\ge \epsilon,$

On an orthogonal note: the given statement does contain a hidden disjunction, and its negation correspondingly contains the expected conjunction (due to De Morgan's Laws): $$\forall \epsilon{>}0\; \exists N{\in} \mathbb{N}\;\forall x{\in} S \;\forall n{\in} \mathbb{N} \;(n\ge N\implies |f_n(x)-f(x)|<\epsilon)\tag1$$ $$\forall \epsilon{>}0\; \exists N{\in} \mathbb{N}\;\forall x{\in} S \;\forall n{\in} \mathbb{N} \;(n\color\red< N\;\color{cyan}\lor\;|f_n(x)-f(x)|\color\red<\epsilon)\tag1$$ $$\exists \epsilon{>}0\; \forall N{\in} \mathbb{N}\;\exists x{\in} S \;\exists n{\in} \mathbb{N} \;(n\color\red\ge N\;\color{cyan}\land\;|f_n(x)-f(x)|\color\red\ge\epsilon)\tag2$$

Answer 3

• The word "and" in "there is some $N\in \mathbb N$ such that for all $n\in \mathbb N$ satisfying $n\geq N$ and all $x\in S \ldots$" is being used as an ordinary English word that makes the English translation of the logical formula sound nice, and is not the same as a "logical conjunction".

• When you take the negation of a formula containing logical-and, you apply de Morgan's laws to the logical formula, changing $\wedge$ into $\vee$ and $\forall$ into $\exists$ and so on, as you know. When you take the negation of a formula expressed in English and containing the word "and", you've first got to figure out what logical formula the English sentence represents, and if "and" is being used as an ordinary English word that won't appear in the formula, or if it's being used as a conjunction and will show up in the formula as $\wedge$.

• You will probably use the English word "and" whenever you're trying to write a logical formula in English, when it contains two universal quantifiers in a row. For example, the math formula $\forall x \in \mathbb{N}, \forall q\in \mathbb{Q}, f(x,q) \text{ is true}$ would be written out in sentence form as "For every $x$ in $\mathbb{N}$ and every $q$ in $\mathbb{Q}$, the proposition $f(x,q)$ is true." Note that this sentence has the word "and" in it as an ordinary word— the original formula doesn't have any conjunctions $\wedge$ in it!

• A quick and dirty rule of thumb is to assume that whenever you see "and every", it just means $\forall$. It doesn't have a $\wedge$ anywhere.

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Let's talk about the formulas and their negations:

1. The sentence "For all $\epsilon > 0$ there is some $N\in\mathbb N$ such that for all $n\in \mathbb N$ satisfying $n\geq N$ and all $x\in S$, $|f_n(x) - f(x)| < \epsilon$" is an English language translation of a logical formula. That logical formula has no conjunctions in it. It is: $$\forall \epsilon > 0\quad \exists N\in \mathbb N \quad \forall n \in \mathbb N \ni n\geq N\qquad|f_n(x)-f(x)|<\epsilon.$$

2. The negation of this formula is $$\exists \epsilon > 0\quad \forall N\in \mathbb N \quad \exists n \in \mathbb N \ni n\geq N\qquad|f_n(x)-f(x)|\not\lt\epsilon.$$ which we get by applying de Morgan's laws to the original formula.

3. We can translate this negation into English as well: There is some choice of $\epsilon > 0$ where no matter which $N\in \mathbb N$ you choose, you can always find some $n\geq N$ where, unfortunately, $|f_n(x) - f(x)|$ is not less than $\epsilon$.

   This English language translation doesn't have any "and"s in it, not even as ordinary English words. This is because, unlike the original formula, the negated formula doesn't have two universal quantifiers $\forall\forall$ in a row. It also doesn't have any conjunctions.