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Is my prove correct?

  • Thread starter Ka Yan
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27
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There is a simple problem, and I gave my simple prove. Could anybody help me check whether it is correct:

Show that if B is not finite and [tex]{B}\subset{A}[/tex], then A is not finite.

My prove:
Since B is not finite, there exists a bijection of B into one of its proper subset, C, say, and denote the function to be f: [tex]{B}\rightarrow{C}[/tex].
Since B is a proper subset of A, as an extension of f to A,there exist a (some) bijiection(s), say g, such that g: [tex]{A}\rightarrow{C}[/tex]. And thus g maps A onto its proper subset, A cannot be finite. Hense A is infinite.

Thks!
 
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Answers and Replies

HallsofIvy
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What, precisely, do you mean by "an extension of f to A". In other words, what, exactly, is g?
 
Dick
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You should also notice that there doesn't generally exist a bijection of A to C. Which is ok, because to show A is infinite you don't need a bijection with C, you only need one with 'some subset' of A.
 
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can't you simply say if [tex]{B}\subset{A}[/tex]

n [tex]\le[/tex] cardinality of (B) [tex]\leq[/tex] cardinality of (A) for all n in N.

So A is not finite.
 
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Dick
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can't you simply say if [tex]{B}\subset{A}[/tex]

n [tex]\le[/tex] cardinality of (B) [tex]\leq[/tex] cardinality of (A) for all n in N.

So A is not finite.
Yes, if you have cardinality theorems available. Ka Yan appears to want to prove it using the definition that a set is infinite if there exists a bijection with a proper subset of itself.
 
27
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Yes, thanks gentlemen.

That, I was originally considered, if f is a function on a subset B of A, then an "extension" of f from B to A is that, exist a function g on A, such that g(x)=f(x) for all x in B.

But B and A are seem neccessary to be closed, and I didnt quite sure if those A and B are closed or not to asure the concept "extension" be apply.

Besides, my classmate gave me another way:
construct a sequense {an} by: take a0[tex]\in[/tex]B, a1 [tex]\in[/tex] B-{a0}, a2 [tex]\in[/tex] B-{a0, a1}, and so on. And this sequense is infinite, and all belong to A, thus A is infinite.
I don't quite well remember the constructing process, and I showed just what he meant.
 
Dick
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Here's another way. Consider the subset D=(A-B) union C. It's a proper subset of A since C is a proper subset of B. Define g(x)=f(x) for x in B and g(x)=x for x in A-B. g is a bijection of A with a proper subset D. So A is infinite. I think that's the 'extension' you were after.
 
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