Alternate Axiom of Infinity

Question

The Axiom of Infinity states that there is a set $S$ containing $\varnothing$ such that if $x$ is an element of $S$ then so is $x\cup\{x\}$.

Is the following variant equivalent? There exists a nonempty set $S$ such that if $x$ is an element of $S$ then so is $\{x\}$.

This one also guarantees an infinite set, and it's shorter. The fact that we use the longer, more complicated one makes me think that the shorter version isn't as strong. Is this correct?

EDIT: To be clear, I'm wondering if they're equivalent if I leave the other axioms of ZFC unchanged.

Answer 1

I'm not sure it is shorter. To say a non-empty set you need $\exists x (x \in S)$ which adds to your statement. To prove equivalence, you define a function $f(x)=\emptyset, f(\{x\})=\{\emptyset \}\dots$ and use replacement on your $S$ to show the existence of the canonical $\omega$.

Answer 2

In the absence of the Axiom of Foundation they are not equivalent. Without Foundation it is consistent that x={x} for some x. The Axiom of Foundation is that any non-empty x has a member y which is disjoint from x.