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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

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    $\begingroup$ If $A$ are the integers, and $B$ are the even integers, can we find a bijection between $A$ and $B$? $\endgroup$ – Thomas Andrews Apr 2 at 17:43
  • $\begingroup$ 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? $\endgroup$ – Paras Khosla Apr 2 at 17:46
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    $\begingroup$ 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. $\endgroup$ – SMM Apr 2 at 17:47
  • $\begingroup$ Thanks for clarifying @SMM $\endgroup$ – Paras Khosla Apr 2 at 17:48
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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, $f:A\mapsto B \mid f(n)=2n$.

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    $\begingroup$ Well, it is a definition of an infinite set. However, there may be infinite sets that don't fit this definition. For more information, look into Dedekind-infinite sets. $\endgroup$ – Cameron Buie Apr 2 at 18:27
  • $\begingroup$ @CameronBuie Sure that may be, but considering the tag attached to the problem, this answer suffices. Such advanced topics are best covered in "set-theory", not in "elementary-set-theory" and are likely to confuse an introductory student. Cheers :) $\endgroup$ – Paras Khosla Apr 2 at 18:43
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    $\begingroup$ 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. $\endgroup$ – Cameron Buie Apr 2 at 18:51
  • $\begingroup$ When you put it like that, it certainly is insightful. Thanks :)) $\endgroup$ – Paras Khosla Apr 2 at 18:53

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