# Is it true $\operatorname{st}(^\ast A) = \overline{A}$, when $A$ is a bounded subset of $\mathbb{R}$?

Given a bounded subset of $\mathbb{R}$, let $\operatorname{st}(^\ast A)$ be the set of the standard parts of all elements in $^\ast$-transform of $A$, $\overline{A}$ is the closure of $A$.Is it true $\operatorname{st}(^\ast A) = \overline{A}$?

It seems to me it's true $\overline{A} \subseteq \operatorname{st}(^\ast A)$.For every $x \in \overline{A}$, there is a sequence $r = \{r_i\}_{i \in \mathbb{N}} \in ^\ast A$ such that $r_i \in A$ for all $i \in \mathbb{N}$ and $\lim_{k \to \infty} r_k = x$. $\operatorname{st}(r) = x$, because for an arbitary $\epsilon$, $\{ i \in \Bbb{N}:|r_i - x|>\epsilon\}$ is finite.

But what about the other direction?

It appears that you’re using the ultrapower construction. Let $\mathscr{U}$ be the ultrafilter. Suppose that $x\in\operatorname{st}{}^*A$. Then there is $a\in{}^*A$ that is represented by a sequence $\langle a_n:n\in\omega\rangle$ in $A$ with the property that for each $n\in\omega$, $$U_n\triangleq\{k\in\omega:|a_k-x|<2^{-n}\}\in\mathscr{U}\;.$$
For each $n\in\omega$ fix $k(n)\in U_n$; then $\langle a_{k(n)}:n\in\omega\rangle$ is a sequence in $A$ converging to $x$, so $x\in\operatorname{cl}A$.
• @Ittay: The fact that it fundamentally uses AC has nothing to do with whether it’s explicit in the way in which I think Metta World Peace is using the term. It’s explicit in the sense that it gives you relatively concrete objects with which to work; yes, they’re very ‘fuzzy’, since you can’t actually put your hands on $\mathscr{U}$, but they’re much more concrete and intuitively manipulable than things conjured up via the compactness theorem, for instance. I would call the ultrapower construction explicit in comparison with any other that I’ve seen. – Brian M. Scott May 13 '13 at 18:36