# Equal Cardinality and Bijection

I ran into the following question:

Suppose that $$B \subset A$$ and that $$\exists$$ a bijection $$f: A\mapsto B$$. What may be reasonably deduced about $$A$$?

I think either there is something wrong with the question because the existence of a bijection is the formal condition for equal cardinality of two sets. But $$\mathrm {card}A=\mathrm {card}B$$ and $$B\subset A$$ being simultaneous conditions is rather absurd to me. Maybe $$\emptyset$$ has something to do here.

Is there another perspective that helps in understanding this better? Also, what am I missing? Thanks

• If $A$ are the integers, and $B$ are the even integers, can we find a bijection between $A$ and $B$? – Thomas Andrews Apr 2 at 17:43
• I get it because the $\mathrm{card}\mathrm {Z}$ and $\mathrm{card} {B}$ where $B=\{x: x=2n, n\in \mathrm {Z}\}$ are both equal to $\infty$ and even integers are a subset of the integers. So can we actually deduce that $A$ must be an infinite set? – Paras Khosla Apr 2 at 17:46
• You can say that $A$ is infinite. This is the definition of an infinite set: $A$ is infinite iff there exists a bijection between $A$ and its proper subset. – SMM Apr 2 at 17:47
• Thanks for clarifying @SMM – Paras Khosla Apr 2 at 17:48

This condition is the definition of an infinite set. $$A$$ is infinite iff $$\exists$$ a bijection between $$A$$ and its proper subset, that in this case is $$B$$. So it can be concluded that $$A$$ is an infinite set.
As an example, consider a bijection $$f$$ from $$A=\{x: x\in \mathbb{Z}\}$$ and $$B=\{x: x=2n, n\in\mathbb{Z}\}$$. Using simple notation, $$f:A\mapsto B \mid f(n)=2n$$.
• It certainly suffices! I would say, however, that we learn even more than that $A$ is infinite. We learn that $A$ has a countably-infinite subset, in fact. – Cameron Buie Apr 2 at 18:51