Is my prove correct?

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!

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

Dick
Homework Helper
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.

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.

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Dick
Homework Helper
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.

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$$\in$$B, 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