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In Stillwell's "Real Numbers," a set $Z$ is said to be inaccessible if it satisfies the following three conditions:

1) $Z$ has infinite members

2) $X\in Z$ implies the power set $P(X)\in Z$

3) $X\in Z$ implies the range of any function with domain $X$ and values in $Z$ is a member of $Z$.

I would appreciate help with Ex. 3.8.6 which asks for an example of a set that satisfies the first two conditions, but not the third.

Also, in that context, how can I show that $V_{\omega}$ does satisfy the third condition.

($V_{\omega}$ is defined to be the union of all the $V_n$ where $V_0= \emptyset$ and $V_{n+1}=V_n\cup P(V_n)$

Thanks

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    Superstructure over any infinite set satisfies first two conditions. As for the third, it is simply a matter of unwinding how functions are represented as sets of ordered pairs, and checking that those sets belong to $V_{\omega}$ by definition. – Conifold Jul 03 '17 at 23:29
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    The set $Z={\mathbb N,P(\mathbb N),P(P(\mathbb N)),\dots}$ satisfies the first two conditions but not the third. – bof Jul 03 '17 at 23:53
  • @bof - Thanks. I can see the first two. If you would, I would appreciate help as to how to approach the third condition. –  Jul 04 '17 at 00:07
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    Since $\mathbb N\in Z$ and $Z\notin Z,$ if you can find a function $f$ with domain $\mathbb N$ and range $Z,$ that will show that the third condition is not satisfied. – bof Jul 04 '17 at 00:45
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    Let $Z=V_A$ where $A$ is a non-zero limit ordinal and $A$ is not a regular cardinal. (E.g. $A=\omega +\omega.$) Then (1) and (2) hold but $cf(A)=B\in V_A$ and there is a function $f:B\to A$ with $\sup {f(x):x\in B}=A$ so $f\not \in V_A.$ – DanielWainfleet Jul 04 '17 at 12:44
  • @bof I've thought for some time to come up with such a function. (No excuses for this attempt.) There are no surjective functions from $\mathbb{N}$ to $P(\mathbb{N})$; can I let $f(n)=n$ with $n\in \mathbb{N}$? Thanks, –  Jul 04 '17 at 14:11
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    If $$\mathbb N={1,2,3,\dots}$$ and $$Z={\mathbb N,P(\mathbb N),P(P(\mathbb N)),\dots}$$ then the most obvious function with domain $\mathbb N$ and range $Z$ is the function $f$ such that $f(1)=\mathbb N,\ f(2)=P(\mathbb N),\ f(3)=P(P(\mathbb N)),$ etc. – bof Jul 04 '17 at 22:21
  • @bof Thanks you. Maybe you would please post your comment as an answer so I may accept it. With regards, –  Jul 04 '17 at 23:02
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    You can accept the answer posted by Athar Abdul-Quader. – bof Jul 04 '17 at 23:45

1 Answers1

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Let $Z = \{ \mathbb{N}, P(\mathbb{N}), P(P(\mathbb{N})), \ldots \}$. Notice that this set is countable -- you should be able to find a surjection from $\mathbb{N}$ to $Z$ (hint: find a function $f$ so that $f(0) = \mathbb{N}, f(1) = P(\mathbb{N})$, etc.). It's not hard to see that 1 and 2 are satisfied by this $Z$, and by finding such a surjection, you can get a counterexample for 3.

Athar Abdul-Quader
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