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I'm undergrad Applied Math Student and I'm writing my thesis about Non-Standard Analysis. I find a Spriger Book named Lectures On Non-Standard Analysis which are based on short courses of lectures delivered by Machover (first part) and the Master of Science thesis of J. Hisrchfeld.

So, the construction starts by considering some set which has certain desirable closure properties. This set needs the smallest transitive set such that contains any set $X$, sometimes called the transitive closure of $X$. Here is a fragment of the construction.

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I may be wrong, but I think the right set $U_0$ has to be defined as the UNION the set $\{V_i\}_{i=0}^\infty$, i.e $U_0=\bigcup\{V_i\mid i=0,1,2,\ldots\}$.

Can someone tell me if I'm right or I'm missing something?

Sergio Ivan
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    I'm not familiar with the book you're using, but just as a sidenote, the best non-standard analysis textbook to this day is Robinson's classic "Non-standard Analysis" – Chubby Chef Dec 07 '20 at 00:49

2 Answers2

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Not having read the aforementioned source, this looks like a simple but unfortunate typo to me. Of course $U_0$, as defined above, would not be the transitive closure.

For example, define $\downarrow$ as the function satisfying $\downarrow\! n = \{ y \in \mathbb{N} \:|\: y < n \}$ for each natural number $n$.

Then the set $V = \{ \downarrow\! x \:|\: x \in \mathbb{N}\}$ is clearly transitive. Thus, $V_{i+1} = V$ for each $i$, and $U_0 = \{ \emptyset, V \}$. But $U_0$ is not transitive, since e.g. $\downarrow\!2 \in V$, and $V \in U_0$ both hold, while $\downarrow\!2 \in U_0$ does not.

As you noted, one can easily fix this by writing $U_0 = \bigcup \{V_i \:|\: i \in \mathbb{N} \}$ instead.

Z. A. K.
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You're right. You can compare with the construction of the transitive closure of a set on English Wikipedia to check.

Mark S.
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