Exercise 1 Let X be any set. Consider the two conditions below: (a) Every injective function X→ X is surjective. (b) For every proper subset Y of X there is no surjection Y → X. (Note: a proper subset Y of X is a set such that y € Y ⇒ y € X and 3x € X such that x Y.) Prove that the two conditions above are equivalent. That is, prove (a) (b). A set S is said to be "finite" if either of these equivalent conditions hold. Hint: For (a) ⇒ (b) try the contrapositive. You may assume the "axiom of choice".
Exercise 1 Let X be any set. Consider the two conditions below: (a) Every injective function X→ X is surjective. (b) For every proper subset Y of X there is no surjection Y → X. (Note: a proper subset Y of X is a set such that y € Y ⇒ y € X and 3x € X such that x Y.) Prove that the two conditions above are equivalent. That is, prove (a) (b). A set S is said to be "finite" if either of these equivalent conditions hold. Hint: For (a) ⇒ (b) try the contrapositive. You may assume the "axiom of choice".
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:Exercise 1 Let X be any set. Consider the two conditions below:
(a) Every injective function X → X is surjective.
(b) For every proper subset Y of X there is no surjection Y → X.
(Note: a proper subset Y of X is a set such that y € Y ⇒ y € X and ³x € X such that x ‡ Y.)
Prove that the two conditions above are equivalent. That is, prove (a)
(b).
A set S is said to be "finite" if either of these equivalent conditions hold.
Hint: For (a) (b) try the contrapositive. You may assume the "axiom of choice".
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