![]() The actual content of the proof now is to show that these two competing definitions of $>$ agree! Of course we first need to define addition, but once we have done that, it is a fairly simple matter of induction.įirst we prove that $|m+1|>|m|$ (as cardinals) for all $m$. This would be trivial (and pointless) if we're using the cardinality definition of what $m>n$ means - so even though that is the most common choice, let's assume that we want to define $m>n$ to mean "there is a natural $a$ such that $n+a+1=m$. Then all we have to do is to prove this from a definitions of "has $m$ elements" and "$mn$ and $f:m\to n$, then $f$ is not injective. If $m$ and $n$ are naturals, and $m>n$, and $A$ has $m$ elements and $B$ has $n$ elements, and $f$ is a function $A\to B$, then $f$ is not injective. Of course, once we define natural numbers, we might want to prove a "finite pigeonhole principle": So if we use that as our definition, the pigeonhole principle is not a matter of proof - instead it is part of the definition of what it means for one set to be larger than the other. ![]() There is an injective function $B\to A$, but there is no injective function $A\to B$. In particular, what does "$A$ has more elements than $B$" mean? In the usual development of set theory we use this phrase to mean the same as "$A$ has larger cardinality than $B$", which again means If $A$ and $B$ are sets, and $A$ has more elements than $B$, and $f$ is a function $A\to B$, then $f$ is not injective.Ĭan it be proved? That depends. A fairly precise statement of the pigeonhole principle would be: ![]()
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