Universal and Existential Quantifiers

A game on the number line teaches the quantifiers. Swap the order and watch the same game collapse.

One side moves first and the other side answers. 'For all' demands an answer to every challenge. 'There exists' needs just one good answer. Swapping the order changes the claim itself. This pattern becomes the epsilon-delta definition later.

For example, for every 🧑 classmate, someone taller exists. But one person taller than everybody does not exist. A single swap changes the whole meaning.

Examples & Exercises

Concept check

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)

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

Show explanation

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

The negation flips every quantifier ( $\forall$ becomes $\exists$ , $\exists$ becomes $\forall$ ) and rewrites the final implication via $\lnot(P \to Q) \equiv P \land \lnot Q$ . The first option fails to flip the $\delta$ quantifier, the third keeps the implication instead of turning it into a conjunction, and the fourth never flips the outer $\forall\varepsilon$ and merely reverses the inequalities.

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 $\forall M\ \exists x \in (0,1]\ |f(x)| > M$ , 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. $\forall M\ \exists x \in (0,1]\ (|f(x)| > M)$ 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.