Is $\mathbb{N}$ a well-founded set?

Question

I was reading about Von Neumann's construction of $\mathbb{N}$, I understood that $\mathbb{N}=\{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\},...\} $.
I see that, with this construction, $\mathbb{N}$ has an infinite descending membership sequence. My question is, does this mean that $\mathbb{N}$ is not well founded? If yes, what are the consequences?
Thanks.

Answer 1

You need to remember that the relation on $\Bbb R$ (let alone, any set extending it) is not the $\in$ relation.

Since well-foundedness only requires $\in$ to be well-founded, there is no issue here.

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This set that you mention is usually denoted by $\omega$, the first infinite ordinal. It has nothing to do with $\Bbb R$. The "natural numbers" as we perceive them in $\Bbb R$ and the "natural numbers" as finite ordinals are two completely different arithmetical systems. One of real numbers and the other of ordinals (or cardinals, which is a third system).

The $\infty$ that we think about in calculus is not an ordinal, or a cardinal. It is a formal notion of something which exists "at the end of the line". You can use $\omega$, or any other set in your universe to represent this $\infty$, but this representation is not going to matter, since the way in which we mean that $\infty\in\overline{\Bbb R}$ is related to the topology of $\Bbb R$, rather than to the $\in$ relation, which defines the order on the ordinals (and cardinals).