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I am presently studying the Natural Numbers section of Enderton's "Elements of Set theory" and I wonder if I have once taken the right approach with with the following question

Let $A = \{1\}$

(a) What is $A^+$? (b) What is $\cup \{A^+\}$?

Is (a): $\{1\}^+ = \{1\} \cup\{\{1\}\}$?

$\implies$

(b) $\cup \{A^+\} = 1, \{1\}$?

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  • $\begingroup$ As far as I know, your notation is not standard. Could you explain what $A+$ and $U[A+]$ mean? $\endgroup$ – Zach L. Nov 11 '13 at 3:40
  • $\begingroup$ Sorry, I'm having trouble finding the right notation on my keyboard - I mean "A to the power of +" (the successor of A) and "U" is meant to imply a large union cup (I hope that helps!)$\bigcup X$ $\endgroup$ – Stumped philosopher Nov 11 '13 at 8:23
  • $\begingroup$ I have tried to rewrite what you wrote using LaTeX, but you were inconsistent in your use of brackets {} [] () so I may have confused things - you may want to check. $\endgroup$ – Henry Nov 12 '13 at 7:56
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For any set $\alpha$, Enderton defines the successor of $\alpha$, denoted by $\alpha^+$, to be the set $\alpha \cup \{\alpha\}$ (Enderton, Elements of Set Theory, p. 68). To answer part (a), we compute $A^+$ as follows:

$$ A^+ = A \cup \{A\} = \{1\} \cup \{\{1\}\} = \{1,\{1\}\}$$

To answer part (b), we compute $\bigcup \{A^+\} = \big \{\{1,\{1\}\}\}$. From the definition of set union, we know that $x \in \bigcup \{A^+\}$ if and only if $x \in B$ for some $B \in \{A^+\}$. Now $A^+$ is the only member of $\{A^+\}$, so the problem reduces to that of finding the members of $A^+$. But $A^+ = \{1,\{1\}\}$, so we must have have $\bigcup \{A^+\} = \{1,\{1\}\}$. So $\bigcup \{A^+\}$ equals neither $1$ nor $\{1\}$, but their (unordered) pair.

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