How to express "$a \leq x \leq a + \frac{y}{n} $ for every $n \geq 1$, then $x = a$" using quantifiers

Question

I've been studying the real number system from Apostol Calculus Vol I and the theorem I.31 says the following:

If three real numbers a, x and y satisfy the inequalities $a \leq x \leq a + \frac{y}{n} $ for every $n \geq 1$, then $x = a$

Now, i've tried to prove the theorem by contradiction and in my head (i think i'm wrong by the way) the statement using quantifiers is

$\forall x \in \mathbb{R} \, \forall y\in \mathbb{R} \, \forall a \in \mathbb{R} \, \forall n \geq 1 \, (a \leq x \leq a + \frac{y}{n}) \implies (x = a)$

and so the negation is

$\exists x \in \mathbb{R} \, \exists y\in \mathbb{R} \, \exists a \in \mathbb{R} \, \exists n \geq 1 \, (a \leq x \leq a + \frac{y}{n}) \wedge (x \neq a)$

But then i saw the proof in the book and Apostol uses the Archimedean property to show that if $x > a$ then there exists a number $n$ such that $n(x-a) > y \,$ and $x > a + \frac{y}{n}$. And well, this doesn't match my intuition for the statement nor the negation of the statement. What would be the correct quantified statement and its negation?

Answer 1

Be careful of operator precedence here. The statement

if $a \leq x \leq a + \frac{y}{n}$ for every $n \geq 1$, then $x = a$

is a material implication at the outermost level, so its symbolic form is $$ (\forall n \in \mathbb{Z}^+ \; (a \le x \le a + \tfrac{y}{n})) \to (x = a). $$

The full statement

if three real numbers $a$, $x$, and $y$ satisfy the inequalities $a \leq x \leq a + \frac{y}{n}$ for every $n \geq 1$, then $x = a$

is thus $$ \forall a \in \mathbb{R} \; \forall x \in \mathbb{R} \; \forall y \in \mathbb{R} \; ((\forall n \in \mathbb{Z}^+ \, (a \le x \le a + \tfrac{y}{n})) \to (x = a)), \tag{1} \label{1} $$ whose negation is $$ \exists a \in \mathbb{R} \; \exists x \in \mathbb{R} \; \exists y \in \mathbb{R} \; ((\color{red}{\forall} n \in \mathbb{Z}^+ \, (a \le x \le a + \tfrac{y}{n})) \land (x \neq a)), \tag{2} \label{2} $$ by De Morgan's laws.

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Due to the properties of quantifiers, \eqref{1} is not equivalent to $$ (\times) \quad \forall a \in \mathbb{R} \; \forall x \in \mathbb{R} \; \forall y \in \mathbb{R} \; \color{red}{\forall} n \in \mathbb{Z}^+ \, ((a \le x \le a + \tfrac{y}{n})) \to (x = a)), $$ but rather $$ (\checkmark) \quad \forall a \in \mathbb{R} \; \forall x \in \mathbb{R} \; \forall y \in \mathbb{R} \; \color{red}{\exists} n \in \mathbb{Z}^+ \, ((a \le x \le a + \tfrac{y}{n})) \to (x = a)). \tag{1’} $$ Similarly, \eqref{2} is not equivalent to $$ (\times) \quad \exists a \in \mathbb{R} \; \exists x \in \mathbb{R} \; \exists y \in \mathbb{R} \; \color{red}{\exists} n \in \mathbb{Z}^+ \, ((a \le x \le a + \tfrac{y}{n})) \land (x \neq a)), $$ but rather $$ (\checkmark) \quad \exists a \in \mathbb{R} \; \exists x \in \mathbb{R} \; \exists y \in \mathbb{R} \; \color{red}{\forall} n \in \mathbb{Z}^+ \, ((a \le x \le a + \tfrac{y}{n})) \land (x \neq a)). \tag{2’} $$

Answer 2

The statement you wrote was almost correct, but a bit unnatural. Almost the same thing, but easier to understand is:

$\forall x,y,a\in \mathbb{R}: (a\le x\le a+\frac{y}{n})\forall n\in \mathbb{N^+} \implies (x=a)$.

Important is, that the condition $(a\le x\le a+\frac{y}{n})$ is met for all natural numbers. The way you wrote it, the statement would imply, that for a fixed, the implication still holds, which is obviously untrue (x=2, y=8, a=1, n=8 gives $a\le x\le a+\frac{y}{n}$, but $1\neq 2$). Can you now work out the negation yourself?