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First, the statement of the question:

Let A,B be two non-empty sets. Show that there exists an injective map of A into B, or there exists a bijective map of a subset of A onto B.

My idea is to use Zorn's Lemna (which is covered in the chapter this question is in, next chapter is on cardinality) on the family of injective maps of subsets of A into B, thus determining that there is a maximal element, but I'm not quite sure how this would imply the existence of a bijection. If S is the set of injective maps of subsets of A into B, my idea would be to define a partial ordering $f<g$ iff $f(A')\subset{g(A'')}$, where A' and A'' are the subsets for f and g respectively. It is clear that S is thus inductively ordered so there must therefore exists a maximal element of S by Zorn's Lemna however Im not sure as stated above what this would mean in this context. All help and feedback appreciated. Thank you

Pianoman1234
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3 Answers3

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Let $\mathcal S$ be the set of maps $f\colon A'\to B$ where $A'\subseteq A$ and $f$ is injective. If $f\colon A'\to B$ and $g\colon A''\to B$ are elements of $\mathcal S$, say $f<g$ if $A'\subsetneq A''$ and $f=g|_{A'}$. This set is inductively ordered, so let $m\colon M\to B$ be a maximal element. If $M=A$, we are done. So assume $M\ne A$ and let $a\in A\setminus M$. Assume there exists $b\in B\setminus m(M)$. Then we can extend $m$ to $M\cup\{a\}$ by mapping $a\mapsto b$, contradicting maximality of $M$. Therefore, no such $b$ exists, i.e., $m$ is onto.

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Consider partial functions $f$ from $A$ to $B$, that is $f:A'\to B$ where $A'$ is a subset of $A$. Order these by $f\le g$ if the domain of $f$ is a subset of the domain of $g$, and also $g$ restricted to the domain of $f$ is $f$. Now apply Zorn.

Angina Seng
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Assuming the axiom of choice, given any two sets $A$ and $B$ either there is an injective function from $A$ into $B$ or there is an injective function from $B$ into $A$. In the later case, if $f$ is such a function, then $f^{-1}\colon f(B)\longrightarrow B$ is a bijection from a subset of $A$ onto $B$.