Question

If S and T are equvialent sets then,
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Answer 1

Answer:

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Step-by-step explanation:

Definition Let be sets. We say that Eß F E is equivalent to F iff there exists

a bijection . If is equivalent to , we write or or or 0ÀEÄF E F E¸FÐ E¶F EµF

something similar: the . notation varies from book to bookÑ

It is intuitively clear that for sets finite E¸F E F iff and have the same number of elementsÞ

Therefore Ö+ß ,× ¸ Ö"ß #× Ö"ß #ß $× ¸ Ö"ß #×Þ ( ) but assuming that +Á, Î

Theorem The relation is an among sets. ¸ equivalence relation

Proof Let , be sets. E Fß G

a) The identity mapping is a bijection . Therefore , so the 0 ÐBÑ œ B 0 À E Ä E E ¸ E

relation is . reflexive

b) If , then there must exist a bijection Then the function E ¸ F 0 À E Ä FÞ

is also a bijection, so Therefore the relation is . 1 œ 0 À F Ä E F ¸ EÞ " symmetric

(Therefore, to show two specific sets B a E +8. re equivalent, it doesn't matter whether

you show that there is a bijection or that there is a bijection . from EF FE to from to )

c Suppose and . Then there are bijections and . Ñ E¸F F¸G 0ÀEÄF 1ÀFÄG

Then is a bijection ( ) so Therefore the relation is

Answer 2

Answer:

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Step-by-step explanation: