# X is infinite if and only if X is equivalent to a proper subset of itself.

Prove that a set $X$ is infinite if and only if $X$ is equivalent to a proper subset of itself.

If $X$ is finite, then suppose $|X|=n$. Any proper subset $Y$ of $X$ has size $m<n$, and so there cannot be any bijective mapping between $Y$ and $X$.

If $X$ is countably infinite, then suppose $X=\{x_1,x_2,\ldots\}$. We can map $X$ to $Y=\{x_2,x_3,\ldots\}$ by using the map $f(x_i)=x_{i+1}$.

But what if $X$ is uncountably infinite? How can we specify the mapping?

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What does uncountably infinite mean for you? That there is an injection $\mathbb N \to X$, but no bijection? – martini Jun 8 '13 at 22:01
Which definition of inifinite do you use? – Hagen von Eitzen Jun 8 '13 at 22:01

The solution was in the previous title: if $X$ is infinite, then it contains a countable infinite subset, say $X_0$. Then you gave a bijection $X_0\to X_0\setminus\{x_1\}$, that extends to a bijection $X\to X\setminus\{x_1\}$, that acts as identity on $X\setminus X_0$.
The answer is no if you do not assume the axiom of choice or a weaker version such as countable choice. If $X$ is an amorphous set, then it is Dedekind-finite, which is the negation of the property in the question.
Actually, countable choice suffices. It's okay, so long as we have just enough choice so that $\aleph_0$ is comparable to every cardinality. – Cameron Buie Jun 8 '13 at 22:54