# Is a set containing only ONE element equal to the element itself?

I know this may sound like a silly question, but suppose $\space S \space$ is a set such that:

$$S= \{ 2 \}$$

Does this mean $\space S \space$ is just equal to the number $\space 2$?

OR

Is $\space 2 \space$ just an element in the set $\space S \space$ and not actually equal to the set itself?

• This: "Is $\space 2 \space$ just an element in the set $\space S \space$ and not actually equal to the set itself?". You have $2 \in \{2\}$ but not $2 = \{2\}$. – StackTD Apr 5 '16 at 14:25
• @StackTD - Ah I see, thank you :) – Max Echendu Apr 5 '16 at 14:25
• A candy bar in a plastic wrapper is not the same thing as just a candy bar. I would want to eat the latter, but not the former :) – MPW Apr 5 '16 at 14:27
• @Max You should keep in mind that Quine atoms aren't 'mainstream'. So your default assumption, when communicating about set theory, should be that regularity is assumed and therefore Quine atoms provably don't exist in our background universe. Also: Apparently Quine atoms make for awful candy bars... – Stefan Mesken Apr 5 '16 at 14:38
• Although they are not the same, there is a "natural" mapping from $\mathbb{N}$ to the power set of $\mathbb{N}$, taking $a$ to $\{a\}$. Sometimes it is convenient to deliberately blur the distinction between $a$ and $\{a\}$. – André Nicolas Apr 5 '16 at 14:43

$S$ is not equal to $2$. As another example, $\{\emptyset\}$ has exactly one element (teh empty set) and therefore differs form its only element $\emptyset$ (which has no elements at all).
• It might be worth noting that ZF set theory defines $0$ as the empty set $\emptyset$ and $1$ as the set $\{ \emptyset \}$ as part of its set-theoretic construction of $\mathbb{N}$. – DylanSp Apr 5 '16 at 14:42