Proving A is Not Finite: B is Not Finite

[Ka Yan]

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 {B}\subset{A}, 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: {B}\rightarrow{C}.
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: {A}\rightarrow{C}. And thus g maps A onto its proper subset, A cannot be finite. Hense A is infinite.

Thks!

 

[HallsofIvy]

What, precisely, do you mean by "an extension of f to A". In other words, what, exactly, is g?

 

[Dick]

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.

 

[fakrudeen]

can't you simply say if {B}\subset{A}

n \le cardinality of (B) \leq cardinality of (A) for all n in N.

So A is not finite.

 

[Dick]

can't you simply say if {B}\subset{A}

n \le cardinality of (B) \leq 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.

 

[Ka Yan]

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 necessary 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\inB, a1 \in B-{a0}, a2 \in 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]

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.