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New update:

It seems that my question (ii) has some mistakes. It should be something like "axiom of countable dependent choice". I'm confused if there is such an axiom.

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Let's check Munkres' “Topology”, page 208, Theorem 33.1 (Urysohn lemma), Step 1.

(i) We may have countable choices to choose next $r\in\mathbb{Q}\cap[0,1]-P_{n}$. But if we arrange all the elements of $\mathbb{Q}$ as $0,1,\frac{1}{2},\frac{1}{3},\frac{2}{3},\cdots$, the choice is uniquely determined in every pick of $r$. So, I think we can avoid the Axiom of countable choice here. Am I right?

(ii) After we choose next $r\in\mathbb{Q}\cap[0,1]-P_{n}$, the normality of $X$ tells me that there exists an open set $U_{r}$ of $X$ such that $\overline{U}_{p}\subset U_{r}$ and $\overline{U}_{r}\subset U_{q}$. The normality only tells me that exists such a $U_{r}$ but we don't know what exactly this $U_{r}$ is . So, here, from $r$ to $U_{r}$, we may have infinite choices. These choices are also uncountable. So, the full Axiom of choice is applied here and we cannot avoid it. Am I right?

I conclude that the Urysohn Lemma does use full Axiom of choice (AC). Am I right?

I need your help. Thank you.

This a question from Munkres' Topology. I show you the whole context of the question below. enter image description here enter image description here

studyhard
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  • See here: https://math.stackexchange.com/questions/4171446/use-of-choice-function-in-urysohns-lemma Let me add that Munkres's text, while very good for a number of reasons, is fairly insensitive to issues surrounding choice. For example, his proof that a closed subset of a compact space is also compact involves spurious use of choice. – user43208 Feb 04 '25 at 23:38
  • @user43208 that link doesn't answer my questions (both (i) and (ii)) – studyhard Feb 04 '25 at 23:43
  • His proof doesn't use full choice, but something called the axiom of dependent choice, https://en.wikipedia.org/wiki/Axiom_of_dependent_choice. (By the way, I misremembered something in my first comment: it should be the result that a compact subset of a Hausdorff space is closed, where the proof he gives involves spurious use of choice.) – user43208 Feb 04 '25 at 23:52
  • @user43208 I think it should be something like "axiom of countable dependent choice". I'm confused if there is such an axiom. – studyhard Feb 05 '25 at 00:03
  • @studyhard Dependent choice is already "countable" in the sense that all it posits is the existence of countable dependent choice sequences $(x_n: n< \omega)$, so intuitively it's like countable choice except that each subsequent choice can depend on the ones before it. There is a generalization of dependent choice to transfinite sequences (see e.g. this question, or for much more info, chap 8 of Jech's monograph), but unqualified, "dependent choice" refers to the countable variation. – spaceisdarkgreen Feb 05 '25 at 00:30
  • No, it's known as the axiom of dependent choice, as spaceisdarkgreen said. So the point with regard to your question is that dependent choice is strictly weaker than full choice (and at the same time is stronger than countable choice); therefore full choice is not a consequence of the Urysohn lemma. – user43208 Feb 05 '25 at 00:36

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