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Consider that we define an "infinite" set as one which contains some proper subset that is in bijection with itself. Now suppose that $B$ is a set and $A \subset B$ is an infinite proper subset of $B$. Can we prove using just that definition and elementary facts about functions that $B$ must also be infinite?

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I suppose you forgot the condition that $A$ be infinite. Then yes, let $f \colon A \to C$ a bijection with $C \subsetneq A$, extend it to $g \colon B \to (B\setminus A) \cup C$ by the identity on $B \setminus A$. –  Daniel Fischer Jul 9 '13 at 12:14
    
So far I'm the only person who's up-voted this question. –  Michael Hardy Jul 9 '13 at 17:13

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You are assuming $A$ to be infinite according to your definition, right?

Let $C$ be a proper subset, and $f : A \to C$ a bijection.

Consider the proper subset of $B$ given by $D = C \cup (B \setminus A)$.

Now consider the bijection $g : B \to D$ which is the identity on $B \setminus A$, and $f$ when restricted to $A$.

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Great, thanks. I was thinking, does this argument holds if we take a concrete category, an object and a subobject and define the object to be infinite if there is an isomorphic proper subobject? So that you have this sort of "transitivity" even when keeping track of structure? –  José Siqueira Jul 9 '13 at 12:20
    
Surely some more considerations would have to be made, but still, is it possible to adapt this argument for this more general case? It seems it won't work directly even for groups. –  José Siqueira Jul 9 '13 at 12:25
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Of course this notion of categoric infinity will differ from the notion of underlying-set-inifinity. For exmaple in the category of rings (with $1$), $\mathbb Z$ has no proper subobject, hence is not infinite using this definition. –  Hagen von Eitzen Jul 9 '13 at 12:35
    
It surely will, but will you still have this sort of "transitivity" of the infinity property? –  José Siqueira Jul 9 '13 at 12:43
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I guess it's just wishful thinking, it does not work for groups (in this sense, $(\mathbb{Z},+)$ is infinite, while $(\mathbb{Q},+)$ is not, although the first one is a proper subobject of the second. –  José Siqueira Jul 9 '13 at 13:31

If you are assuming that $A$ is infinite, then yes, we can. In that case, there is a bijective function $f:A\to C$ with $C\subsetneq A$. Let $g$ be the identity function on $B\setminus A$. Then $f\cup g$ is a bijection between $B$ and a proper subset of itself, namely $(B\setminus A)\cup C=B\setminus (A\setminus C).$

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