Exercise 1155Logic, sets and proofs › Quantifiers and negationL1Added 3 Nov 2026

A false $\exists\forall$ statement

#exists forall#false#counterexample

Statement

Disprove: $\exists\delta>0,\ \forall x\in(0,\delta),\ \sqrt x>1$ .

Hint

Find $x$ close to $0$ with $\sqrt x\le1$ .

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Negation. $\forall\delta>0,\ \exists x\in(0,\delta),\ \sqrt x\le1$ .

Proof. For any $\delta>0$ , choose $x=\min(\delta/2,\ 1/4)$ . Then $x\in(0,\delta)$ and $\sqrt x\le1/2\le1$ .

Conclusion. The original statement is false.