I'm aware of the existence of this question: Surjectivity implies injectivity
However, the question is regarding a finite set $S$. I was wondering, though: What happens when $S$ is an infinite set? Zhen Lin addresses this cases in his answer, by saying that it ceases to be necessarily true, for example $f:\mathbb{N}\rightarrow \mathbb{N}$ defined by $x \mapsto x+1$ is injective but not surjective.
My question is: what happens if $S = \mathbb{R}$? Constructing a counterexample for $S=\mathbb{N}$ seems simple enough, but I'm struggling to find a function $f:\mathbb{R}\rightarrow \mathbb{R}$ that's injective but not surjective. Does such a function even exist? If so, how to construct it?
And perhaps a more general question (maybe too broad): For which infinite sets $S$ there is a function $f:S\rightarrow S$ such that $f$ is injective but not surjective?
Best Answer
The function $f(x) = \arctan x$ is injective but clearly not surjective on $\mathbb{R}$.
In general, it is true that every Dedekind-infinite set has this property.