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Let $A = \mathbb Q \cap [0,1]$. In the proof of Urysohn's lemma, we construct a family of open set such that: $\{V_q\}_{q \in A}$, where if $r_1 < r_2$, then $V_{r_1} \subset \subset V_{r_2}$.

The construction was made possible by some choice $\alpha: A \rightarrow \{\text{open set}\}$. Is there a particular reason why we chose $A = \mathbb Q \cap [0,1]$, i.e. a countable, dense set of $[0,1]$? Why can't we choose $A = [0,1]$? We still have a choice function $\alpha: [0,1] \rightarrow \{\text{open sets}\}$ such that if $r_1 < r_2$, then $\alpha(r_1) \subset\subset \alpha(r_2)$.

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    $\begingroup$ How do you know that such a function $\alpha$ exists? That's not what the axiom of choice says at all (and it is inaccurate to call this $\alpha$ a "choice function"). $\endgroup$ Jun 13, 2021 at 5:05
  • $\begingroup$ Ok, I misinterpreted AC, and the proof certainly does not use AC. However, the proof assumes that there exists a function $\alpha: \mathbb Q \cap [0,1] \rightarrow \text{ {open sets}}$. If there are finitely many rational numbers, $\{q_1, \cdots, q_n\}$ we may certainly construct $V_{q_1} \subset \cdots \subset V_{q_n}$. However, I am uncomfortable when I take the whole set $\mathbb Q \cap [0,1]$. (continued) $\endgroup$
    – James C
    Jun 13, 2021 at 5:47
  • $\begingroup$ (For simplicity, by $\mathbb Q$, I mean $\mathbb Q \cap [0,1]$) $\alpha$ corresponds to some collection of open sets such that $\mathcal V = \{V_q\}_{q \in \mathbb Q}$. So, this $\mathcal V$ is already out there in some abstract space consisting of collections of open sets, and $\alpha$ is just a function that maps \mathbb Q$ to $\mathcal V$. (continued) $\endgroup$
    – James C
    Jun 13, 2021 at 5:52
  • $\begingroup$ Here's an example. Take $\mathbb Q' := \mathbb Q \setminus \{1/2\}$. Then, $\mathbb Q'$ is a countable, dense set of $[0,1]$, and I may construct a $\mathcal V' := \{V_q\}_{q \in \mathbb Q'}$. Now, I want to choose some open set $V_{1/2}$ and include it in $\mathcal V'$. However, there's no way of properly defining $V_{1/2}$; it may be possible, but very unlikely. (continued) $\endgroup$
    – James C
    Jun 13, 2021 at 5:56
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    $\begingroup$ Side remark: Urysohn's lemma is independent of ZF + Countable Choice. To prove it, one needs some kind of choice that is stronger than countable choice. The usual one is the axiom of dependent choice, which is weaker than the full axiom of choice, but Andrew Blass commented in this MO thread that one can use an even weaker version known as "dependent multiple choice" (which I don't know what it is). $\endgroup$
    – user1551
    Jun 13, 2021 at 6:29

1 Answer 1

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The proof does not assume there is a function $\alpha$ as you describe. The construction of the $V_q$ for $q\in Q=\mathbb{Q}\cap(0,1)$ uses a choice function $C$ defined on the set of pairs $(F,U)$, where $F$ is closed, $U$ is open, and $F\subseteq U$; the codomain of $C$ is the family of open sets and $C(F,U)$ is such that $F\subseteq C(F,U)\subseteq\overline{C(F,U)}\subseteq U$. Given the disjoint closed sets $A$ and $B$ the construction is by recursion on $\mathbb{N}$ after you enumerate $Q$ as $\{q_n:n\in\mathbb{N}\}$ with $q_0=0$ and $q_1=1$. First one takes $V_{q_0}=C(a,X\setminus B)$ and $V_1=X\setminus B$.At stage $n\ge2$ look where $q_n$ is relative to $\{q_i:i<n\}$ take the largest $q_i$ and smallest $q_j$ with $q_i<q_n<q_j$ and let $V_{q_n}=C(\overline{V_{q_i}}, V_{q_j})$; as $q_0<q_n<q_1$ there are always such $i$ and $j$.
The reason we use this $Q$ is that it is countable and hance that we can do the recursion along $\mathbb{N}$.

You can then define $V_x$ for all $x\in(0,1)$ by $V_x=\bigcup\{V_q:q\in Q, q<x\}$.

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  • $\begingroup$ Thank you for a kind answer. So, it answers one of my questions why I take $\mathbb Q$ instead of $[0,1]$. I have another question. Your answer shows that "given finitely many rational points between $[0,1]$, I can always construct a family of (finitely many) open sets such that if $r_i < r_j$, then $V_i \subset \subset V_j$. With this, can I generalize the result that the same result holds for countably many points? $\endgroup$
    – James C
    Jun 13, 2021 at 22:40
  • $\begingroup$ I don't like the fact that the choice function $C$ can be arbitrarily made based on my choice. To elaborate, for $q \in \mathbb Q \cap [0,1]$, I want to "FIX" some open set $C(q)$ which does not depend on "MY CHOICE." $\endgroup$
    – James C
    Jun 13, 2021 at 22:42
  • $\begingroup$ After some search, I found out that Urysohn's Lemma depends on Axiom of Dependent Choice. As long as I figured out that the lemma indeed depends on some axiom, then I am happy. $\endgroup$
    – James C
    Jun 13, 2021 at 22:58
  • $\begingroup$ Answer to comment 1: I'm not quite sure what you mean. Every countable linear order can be embedded into $Q$, so this takes care of all countable linear orders at once. On the other hand: the finite version works without Choice (of any kind); the infinite version needs some Choice; the principle DC suffices, as you discovered. $\endgroup$
    – hartkp
    Jun 14, 2021 at 7:15
  • $\begingroup$ On comment 2: the choice function $C$ that I used was not made, it was given by an application of AC. Also without Choice it may be impossible to have a non-trivial map from $Q$ into the family of open sets. $\endgroup$
    – hartkp
    Jun 14, 2021 at 7:23

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