Practice question · Multiple choice
Continuity is defined with ∀ε ∃δ, and uniform continuity swaps them to ∃δ ∀x. Why is that swap the whole difference between the two ideas?
Hints
- In each version, has x already been chosen when δ is picked?
- 1/x on (0,1) is continuous everywhere and not uniformly so. Ask what goes wrong near zero.
Show the answer
B. Because in the second, one δ must work everywhere at once
Why
Later variables may depend on earlier ones, so moving δ before x forbids it from adapting. 1/x is the standard witness: continuous at every point, and no single δ survives the approach to zero. Two symbols in a different order state genuinely different theorems.
Practise Quantifiers and Predicates
The app has 5 more questions on this lesson, and keeps your place in the course. Mathematics I is free to start.
More questions on Quantifiers and Predicates
- Order the steps that negate 'for every integer x there exists an integer y with y > x'.
- If 'there exists x with P(x)' is true, then 'for all x, P(x)' must also be true.
- Sort each statement by the quantifier it genuinely needs.
- The negation of "for all x there exists y with P(x,y)" is "there exists x such that for all y, not P(x,y)".…
- The domain is {1, 2, 3, 4, 5, 6}. Let P(x) be 'x is even'. Select every x that makes P(x) TRUE.