Universal and Existential Quantifiers

A game on the number line: the challenger names

x

, and you answer with the bigger point

y = x + 1

—

\forall x \,\exists y \,(y > x)

. Swap the quantifier order and the same game collapses. The

\varepsilon

\delta

line is exactly this pattern, with

\delta = \varepsilon/2

answering

f(t) = 2t

Examples & Exercises

Negate completely the $\varepsilon$ - $\delta$ definition of continuity of $f$ at $a$ — "for all $\varepsilon > 0$ there is $\delta > 0$ such that for all $x$ , $|x-a| < \delta$ implies $|f(x)-f(a)| < \varepsilon$ ". State precisely what the negation says about $f$ .

\lnot\big(\forall\varepsilon{>}0\ \exists\delta{>}0\ \forall x\,(|x-a|<\delta \to |f(x)-f(a)|<\varepsilon)\big)

\exists\varepsilon{>}0\ \forall\delta{>}0\ \exists x\,(|x-a|<\delta \;\land\; |f(x)-f(a)|\ge\varepsilon)

The

\varepsilon

\delta

statement (

\varepsilon = 2.5

\delta = \varepsilon/2 = 1.25

), negation chain shown: final step

\exists\varepsilon{>}0\ \forall\delta{>}0\ \exists t\,(|t| < \delta \land |2t| \ge \varepsilon)

Show proof

Push the outer negation inward one step at a time, flipping each quantifier ( $\lnot\forall = \exists\lnot$ , $\lnot\exists = \forall\lnot$ ). Step 1: $\exists\varepsilon > 0\ \lnot\big(\exists\delta > 0\ \forall x\,(|x-a| < \delta \to |f(x)-f(a)| < \varepsilon)\big)$ . Step 2: $\exists\varepsilon > 0\ \forall\delta > 0\ \lnot\big(\forall x\,(|x-a| < \delta \to |f(x)-f(a)| < \varepsilon)\big)$ .
Step 3: $\exists\varepsilon > 0\ \forall\delta > 0\ \exists x\ \lnot\big(|x-a| < \delta \to |f(x)-f(a)| < \varepsilon\big)$ . Finally use the negation of an implication: $\lnot(P \to Q) \equiv P \land \lnot Q$ (scene 1, problem 3). The final negation is $\exists\varepsilon > 0\ \forall\delta > 0\ \exists x\,\big(|x-a| < \delta \land |f(x)-f(a)| \ge \varepsilon\big)$ .
What this says about $f$ : there is a fixed gap $\varepsilon > 0$ such that, no matter how close to $a$ we look (however small $\delta$ is taken), there is always a point $x$ with $f(x)$ at least $\varepsilon$ away from $f(a)$ — i.e. $f$ stays "torn" at $a$ . This is exactly the definition of " $f$ is discontinuous at $a$ ".

Prove $\forall x \in \mathbb{R}\ \exists y \in \mathbb{R}\ (y > x)$ (witness $y = x+1$ ) and disprove $\exists y \in \mathbb{R}\ \forall x \in \mathbb{R}\ (y > x)$ . Conclusion: quantifier order matters.

\forall x\ \exists y\ (y > x)\ \text{is true},\qquad \exists y\ \forall x\ (y > x)\ \text{is false}

The statement

\forall x\,\exists y\,(y > x)

(true), challenge

x = 2

: witness

y = 3

, and

3 > 2

holds.

Show proof

Prove the first statement. Let an arbitrary $x \in \mathbb{R}$ be given. Set $y = x + 1$ ; then $y \in \mathbb{R}$ and $y - x = 1 > 0$ , i.e. $y > x$ . Since $x$ was arbitrary, such a $y$ exists for every $x$ — $\forall x\ \exists y\ (y > x)$ is true. At the scene's challenge $x = 2$ the witness is $y = 3$ and $3 > 2$ holds.
Disprove the second statement. Whatever candidate $y_0 \in \mathbb{R}$ is proposed, set $x = y_0 + 1$ ; then $x \in \mathbb{R}$ and $y_0 > x$ is equivalent to $y_0 > y_0 + 1$ , i.e. $0 > 1$ , which is false. Hence no $y_0$ satisfies " $y_0 > x$ for all $x$ " — $\exists y\ \forall x\ (y > x)$ is false. In the scene the candidate $y = 2$ falls to $x' = 3$ .
Conclusion: quantifier order completely changes the meaning. In $\forall x\ \exists y$ the choice of $y$ may depend on $x$ ( $y = x+1$ ), but in $\exists y\ \forall x$ a single $y$ must work for every $x$ . This ordering is exactly why $\delta$ in the $\varepsilon$ - $\delta$ definition is allowed to depend on $\varepsilon$ .

Negate " $f$ is bounded on $[0,1]$ " — i.e. $\exists M\ \forall x \in [0,1]\ |f(x)| \le M$ — to obtain $\forall M\ \exists x\ |f(x)| > M$ , then use that form to prove $f(x) = 1/x$ is unbounded on $(0,1]$ .

\lnot\big(\exists M\ \forall x \in [0,1]\ (|f(x)| \le M)\big)\ \equiv\ \forall M\ \exists x \in [0,1]\ (|f(x)| > M)

x = \frac{1}{|M|+2} \in (0,1],\qquad f(x) = |M|+2 > M

The statement

\exists y\,\forall x\,(y > x)

(false), negation chain shown: final step

\forall y\,\exists x\,(y \le x)

— the same quantifier pattern as the negation of boundedness.

Show proof

Negate the definition of boundedness: $\lnot\big(\exists M\ \forall x \in [0,1]\ (|f(x)| \le M)\big)$ . Pushing the negation inward gives $\forall M\ \lnot\big(\forall x \in [0,1]\ (|f(x)| \le M)\big)$ , then $\forall M\ \exists x \in [0,1]\ \lnot(|f(x)| \le M)$ , i.e. $\forall M\ \exists x \in [0,1]\ (|f(x)| > M)$ . In words: "whatever bound $M$ is proposed, there is a point $x$ exceeding it" — that is the definition of unboundedness.
Now apply it to $f(x) = 1/x$ ( $x \in (0,1]$ ). Let an arbitrary $M \in \mathbb{R}$ be given. Set $x = \frac{1}{|M|+2}$ ; the denominator is positive so $x > 0$ , and $|M|+2 \ge 2$ gives $x \le \frac12 < 1$ — hence $x \in (0,1]$ . Then $|f(x)| = f(x) = |M|+2$ , and since $|M| \ge M$ we have $|M|+2 > M$ . So $|f(x)| > M$ holds at this $x$ .
Since $M$ was arbitrary, for every $M$ there is an $x \in (0,1]$ with $|f(x)| > M$ — i.e. $\forall M\ \exists x\ (|f(x)| > M)$ holds, so $f$ is unbounded on $(0,1]$ . The key is the quantifier order: because $M$ is given first ( $\forall M$ ), the witness $x$ may be chosen afterwards ( $\exists x$ ), which is why the $M$ -dependent choice $x = 1/(|M|+2)$ is legitimate.