Negation of the definition of limit

Question

A sequence $(x_n) $ of real numbers converges to a real number $ x $ if

For all $\epsilon> 0 $ there exists a natural number $ n_0 $ such that for all $ n \ge n _0 $, $|x_n - x| < \epsilon $.

How to negate this statement? No matter what I try my new statement makes no sense. I am trying to negate it using propositional logic (so the negation of a universal quantifier is an existential quantifier, etc) but I'm getting lost. Please ELI5

Thanks!

Answer 1

You can write the definition of $(x_n) \to x$ as $$\forall \varepsilon > 0 \ \exists n_0 \in \mathbb{N} \ \forall n \ge n_0 \ |x_n-x| < \varepsilon$$ When you negate such a statement, the front-loaded quantifiers flip (so the $\forall \exists \forall$ becomes $\exists \forall \exists$, and the quantified proposition is negated (so the $|x_n-x|<\varepsilon$ becomes $|x_n-x| \ge \varepsilon$).

See if you can now negate the statement yourself (and, even better, re-write it in plain English).

Answer 2

In ordinary language:

For any real number $x$, there are terms $x_n$ in the sequence with arbitrarily high rank which will remain (at least) at a minimal distance from $x$.

Formally, as there's really an implication in the definition of convergence: $$\exists x\,\forall\varepsilon\,\exists n_0\,\forall n,\enspace\bigl((n\ge n_0 )\implies(\lvert x_n-x\rvert < \varepsilon)\bigr)$$ we obtain $$\forall x\,\exists\varepsilon\,\forall n_0\,\exists n,\enspace\bigl((n\ge n_0 )\wedge(\lvert x_n-x\rvert \ge \varepsilon)\bigr)$$

If $x$ is a given number, it becomes somewhat simpler:

There are terms $x_n$ in the sequence with arbitrarily high rank which will remain at least at a minimal distance from $x$.

Formally: $$\exists\varepsilon\,\forall n_0\,\exists n,\enspace\bigl((n\ge n_0 )\wedge(\lvert x_n-x\rvert \ge \varepsilon)\bigr)$$

Answer 3

$\forall \varepsilon > 0 \ \exists n_0 \in \mathbb{N}$ $\forall n$, if $ n \ge n_0$ then $ \ |x_n-x| < \varepsilon$.

Let $p: n\ge n_0$

and $q:\ |x_n-x| < \varepsilon$.

Then, $\forall \varepsilon > 0 \ \exists n_0 \in \mathbb{N} $$\forall n$, $p\implies q$.

Note that $\lnot(p\implies q)=\lnot(\lnot p\lor q)=p\wedge\lnot q$ and $\lnot\varepsilon=\forall$ etc.

Negation: $\exists \varepsilon > 0 \ \forall n_0 \in \mathbb{N} $$ \exists n$, $p\wedge\lnot q$ OR $\exists \varepsilon > 0 \ \forall n_0 \in \mathbb{N} \exists n$, ($n\ge n_0$)$\wedge (|x_n-x|\ge\epsilon$).

Answer 4

The completeness of ${\mathbb R}$ allows to formulate this negation in a way that avoids the cumbersome "for all $x$ we don't have convergence to $x\,$". It is sufficient to formulate that the sequence in question violates the Cauchy condition, as follows:

There is an $\epsilon_0>0$ such that for any $n_0$ we can find $m$, $n>n_0$ with $|x_m-x_n|\geq\epsilon_0$.

Answer 5

A sequence $(x_n) $ of real numbers does not converge to a real number $ x $ iff

$\exists \epsilon \in \mathbb{R}: [\epsilon > 0 \land \forall n_0 \in \mathbb{N}: \exists n\in \mathbb{N}:[n \ge n_0 \land |x_n-x|\ge\epsilon]]$