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Definition$\quad$ A set $S$ can be defined as infinite if there exists a mapping from $S$ to $S$ that is one-to-one but not onto. Otherwise, $S$ is finite.

Problem: Using the definition of infinite above, prove that if a set $S$ has an infinite subset, then $S$ is infinite.

My attempt: Suppose $T\subseteq S$, where $T$ is infinite. By the supplied definition of infinite, there exists a mapping $\eta\colon T\to T$ that is one-to-one but not onto. That is, for all $x_1,x_2\in T$, we have $\eta(x_1)=\eta(x_2)\to x_1=x_2$, but there exists $\tau\in T$ such that $\eta(x)\neq\tau$ for all $x\in T$.

Consider a one-to-one and onto mapping $\delta\colon S\setminus T\to S\setminus T$. There exists a mapping $\gamma\colon S\to S$ such that $$ \gamma\colon S\to S\equiv \begin{cases} \eta\colon T\to T &\text{if $x\in T$},\\[0.25em] \delta\colon S\setminus T\to S\setminus T &\text{if $x\in S\setminus T$}. \end{cases} $$ The mapping $\gamma\colon S\to S$ is one-to-one because $x_1,x_2\in T\cup S\setminus T\to x_1,x_2\in S$ and $\gamma(x_1)=\gamma(x_2)\to x_1=x_2$ because $\eta$ and $\delta$ are both one-to-one mappings. However, $\gamma$ is not onto because there exists an element in $S$, namely $\tau$ (since $\tau\in T\to\tau\in S$ because $T\subseteq S$), that is not mapped to. Hence, there exists a mapping $\gamma$ from $S$ to $S$ that is one-to-one but not onto when $T\subseteq S$ and $T$ is infinite. Thus, $S$ is infinite. $\Box$

Question: Is this a good/correct proof? If not, where did I go wrong? If it is correct, then is there a way I can improve it or is there a more elegant approach?

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  • $\begingroup$ Looks good to me. The only thing I would change is to use the identity on $S\setminus T$ as $\delta$. $\endgroup$ – A.P. Apr 22 '15 at 17:33
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    $\begingroup$ I find it perfect. $\endgroup$ – ajotatxe Apr 22 '15 at 17:34
  • $\begingroup$ @A.P. I'm not sure what you mean exactly. What do you mean by "use the identity"? $\endgroup$ – fancynancy Apr 22 '15 at 17:34
  • $\begingroup$ I mean the map $x \mapsto x$. $\endgroup$ – A.P. Apr 22 '15 at 17:34
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    $\begingroup$ @A.P. Thanks for the input! That does it make it simpler since the identity mapping is always onto and one-to-one...I can make it more specific in that sense. :) $\endgroup$ – fancynancy Apr 22 '15 at 17:38
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Looks good to me. The only thing I would change is to define $\delta$ as the identity on $S∖T$, i.e. as $$ \begin{align} \delta \colon S \setminus T &\to S \setminus T \\ x &\mapsto x \end{align} $$

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  • $\begingroup$ I'm not looking for easy rep, but if you accept this (or any other) answer your question will not appear unanswered anymore, contributing to keep the site a bit more tidy. $\endgroup$ – A.P. Apr 22 '15 at 17:52
  • $\begingroup$ Could always make it CW $\endgroup$ – Daniel W. Farlow Apr 22 '15 at 18:00
  • $\begingroup$ @MagicMan I'm a bit ashamed about this, but I couldn't find the option. :P $\endgroup$ – A.P. Apr 22 '15 at 18:03

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