Interpreting the statement

Question

Just started quantifiers, and I'm having some trouble with interpreting this.

Here's what I understand:

For every Epsilon $> 0$, there exists a Delta $> 0$ for all $x$ in $\mathbb{R}$.

The antecedent of the implication is false, because there is no Delta that is always greater than $x$.

The precedent of the implication is also false, because not every Epsilon greater than $0$ is also greater than $x^2 -1$ for any $x$.

So the statement is false. Is my logic sound? I'm not sure if I'm reading this right.

Answer 1

You could read the implication as:

$|x^2 -1|<\varepsilon\:$ whenever $\: |x-1|<\delta$

which seems just fine.

With the quantifiers, this whole thing just says the function $x^2$ is continuous at $x=1$.

Added note: See Andreas Blass' answer. I was assuming the quantifiers applied to the entire implication. Depending on the text it's from, that's probably what is intended.

Answer 2

I believe your difficulty arises from a shortage of parentheses in the formula. I think the formula was intended to be $$ (\forall\varepsilon>0)(\exists\delta>0)(\forall x\in\mathbb R)\Big([0<|x-1|<\delta]\implies [|x^2-1|<\varepsilon]\Big). $$ That is, the quantifiers at the beginning apply to the whole implication, not just to its antecedent. This is also how the two answers preceding mine (by Ahmed and Michael Cotton) interpreted the formula.

Answer 3

$(\forall\varepsilon > 0)(∃δ > 0)(∀x ∈ R):0 <| x − 1 |< δ \implies | x^2 − 1 |< ε$

I believe you read it wrong. It should be:

For every $\varepsilon>0$ there is a $\delta>0$ such that for all real numbers $x$ it holds that
if $0<|x-1|<\delta$ then $|x^2-1|<\varepsilon$. Here $x$ is just not a real number but a real number that can be chosen that close to $1$ so that $|x^2-1|<\varepsilon$ - how close to zero you first choose $\varepsilon$.

Answer 4

This is a neat opportunity for a "mini real analysis lesson." Your formula was most certainly wrong, and Andreas kindly fixed it: $$ (\forall\varepsilon>0)(\exists\delta>0)(\forall x\in\mathbb R)(0<|x-1|<\delta\to|x^2-1|<\varepsilon). $$ What you have here is actually a particular instance of the formal definition of a limit at a real number being used. Huh? Yes, consider the following definition of a limit at a real number from James Stewart's Calculus textbook.

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Definition: Let $f$ be a function defined on some open interval that contains the number $a$, except possibly at $a$ itself. Then we say that the limit of $f(x)$ as $x$ approaches $a$ is $L$, and we write $$ \lim_{x\to a}f(x)=L $$ if for every number $\epsilon>0$ there is a number $\delta>0$ such that $$ |f(x)-L|<\epsilon\qquad\text{whenever}\qquad 0<|x-a|<\delta. $$

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Pay very close attention to how this definition is worded. Using universal quantifiers, we may represent it as follows: $$ (\forall\epsilon>0)(\exists\delta>0)(\forall x\in\mathbb{R})(0<|x-a|<\delta\to|f(x)-L|<\epsilon). $$ Now consider what Andreas wrote: $$ (\forall\varepsilon>0)(\exists\delta>0)(\forall x\in\mathbb R)(0<|x-1|<\delta\to|x^2-1|<\varepsilon). $$ What does this mean in the context of your question? By George, you are actually considering the formal statement of the claim $$ \lim_{x\to 1}(x^2-1)=0. $$ This is what your statement means, and it is not false. I will outline a formal proof below that follows the formal definition of a limit.

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Claim: $$ \lim_{x\to 1}(x^2-1)=0. $$

Proof. Given $\epsilon>0$, we need $\delta>0$ such that if $0<|x-1|<\delta$, then $$ |(x^2-1)-0|=|x^2-1|=|(x-1)(x+1)|<\epsilon. $$ Notice that if $|x-1|<1$, then $$ -1<x-1<1\Longrightarrow 1<x+1<3\Longrightarrow |x+1|<3. $$ So take $\delta=\min\left\{1,\frac{\epsilon}{3}\right\}$. Then \begin{align} 0<|x-1|<\delta &\Longleftrightarrow |(x-1)(x+1)|\\[1em] &\leq |3(x-1)|\\[1em] &= 3\cdot|x-1|\\[1em] &< 3\delta\\[1em] &\leq \epsilon. \end{align} Thus, $\lim_{x\to 1}(x^2-1)=0$ by the definition of a limit. $\Box$