Update with Direct Question
Based on Asaf's comments, here is a related question:
Prove that the mapping $n \mapsto n \cup \{n\}$ on the set $\Bbb N$ is injective without the axiom of foundation.
The wikipedia $\text{ZF}$ article under axiom 7 contains the text
(It must be established, however, that these members are all
different, because if two elements are the same, the sequence will
loop around in a finite cycle of sets. The axiom of regularity
prevents this from happening.)
ORIGINAL QUESTION
Without the axiom of foundation (axiom 2 in the wikipedia $\text{ZF}$ article)
can any infinite sets be constructed?
By an infinite set we mean a set that is not Kuratowski finite.
I suspect that without it, the axiom of infinity (axiom 7) might be better described as
$\quad$ The formula of finitary frustration.
My work
I saw the axiom of foundation mentioned inside parentheses in the paragraph for axiom 7 allowing us to construct the natural numbers. So apparently, the familiar program of constructing the natural numbers $\Bbb N$ can't be carried out without axiom 2.
Best Answer
It is true, if $x=\{y\}$ and $y=\{x\}$ and $x\neq y$, then $x\cup\{x\}=\{x,y\}=y\cup\{y\}$. So without assuming the axiom of regularity the map $x\mapsto\{x\}$ is not provably injective.
But just because it is not necessarily injective on the entire universe does not mean that we cannot find a set on which it is injective.
Setting $\omega$ to be the intersection of all the inductive sets, we can now prove that $x\mapsto\{x\}$ is in fact injective on $\omega$. The reason being is that $\omega$ is in fact well-founded, so in particular the above situation does not happen inside $\omega$, it it happens at all.
The quickest, dirtiest, hackiest way to see this is to simply note that $\omega$ belongs to the von Neumann universe which is an inner model of the universe (working in $\sf ZF-Reg$). But this can be shown by hand, through what is tantamount to an ad-hoc proof that $\omega$ is well-founded.