# Proving that if $S$ has an infinite subset then $S$ is infinite

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?

• Looks good to me. The only thing I would change is to use the identity on $S\setminus T$ as $\delta$. – A.P. Apr 22 '15 at 17:33
• I find it perfect. – ajotatxe Apr 22 '15 at 17:34
• @A.P. I'm not sure what you mean exactly. What do you mean by "use the identity"? – fancynancy Apr 22 '15 at 17:34
• I mean the map $x \mapsto x$. – A.P. Apr 22 '15 at 17:34
• @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. :) – fancynancy Apr 22 '15 at 17:38

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}