1

In the proof of Urysohn's lemma, Rudin uses Theorem 2.7 (seemingly) infinitely many times:enter image description here

He uses it to create a set for every rational number between 0 and 1. Why is this allowed? It seems to me like this is a proof with infinitely many steps.

For clarity, theorem 2.7 states that for an open set $V$ and a compact set $K\subseteq V$, there exists an open set $V_0$ such that $K\subseteq V_0\subseteq\overline{V_0}\subseteq V$

John Doe
  • 994
  • 6
  • 16
  • check out Infinitary logics and this question on math overflow – powerline Sep 08 '24 at 08:37
  • I'm aware of infinitary logic, I was just wondering why it's possible here since ZFC is built on finitary logic (if I'm not mistaken), and I think it's safe to assume that's what the book is assuming as well – John Doe Sep 08 '24 at 08:39
  • I’m kind of guessing here, but could it be that this is proof by induction (somewhat in disguise) and he uses the phrase “Continuing” in a misleading way, but it’s actually meant “And now that we’ve shown that the statement holds for r+1, we continue and truly do obtain this collection…” – powerline Sep 08 '24 at 08:47
  • I don't think that's the case, because he goes on to say "we obtain a ocllection ${V_r}$ of open sets, one for every rational," which is a collection with infinitely many sets. An argument by induction would only permit the statement for some $n$ we have a collection of $n$ sets, each corresponding to a (unique) rational number. – John Doe Sep 08 '24 at 08:49
  • I’m not sure I understand exactly what you said in the last part, but he does enumerate the rationals at the start and then inducts over them, as I see it – powerline Sep 08 '24 at 08:58
  • If I’m correct, this seems relevant: How can I expand mathematical induction to rational numbers – powerline Sep 08 '24 at 09:10
  • This is still a statement of the form: given a rational number $p$, then $P(p)$ is true, but what it seems like Rudin is saying here, is that there exists an infinite object which for every rational number has a corresponding element. At least that's what it seems like to me. – John Doe Sep 08 '24 at 09:13
  • What is wrong with the $P(p)$ here being: "For every rational there's a collection of open sets ${V_r}$ with the stated properties" ? He doesn't say it explicitly but that's what I think he's doing – powerline Sep 08 '24 at 09:24
  • 2
    You know the rationals are countable, so the list $(r_i)_{i \in \Bbb{N}}$ is a complete listing of the rationals. He then proves given a sequence of appropriately nested sets for each initial segment of this list, the list can be extended (at least) once by Thm. 2.7. But this completes the induction, so the entire list gives a list-length sequence of appropriately nested sets and the length of the list is as long as all the rationals... – Eric Towers Sep 08 '24 at 09:26

1 Answers1

2

It is a proof with infinitely many steps, but is generally considered unproblematic. In particular, one of the usual axioms of set theory, the Axiom of Choice, allows us to make an appropriate choice for all of the steps at once provided that at each stage an appropriate next step can be made. To be specific, I'll state an equivalent of the Axiom of Choice called Zorn's Lemma.

Zorn's Lemma. Let $\leq$ be a partial order on a set $Z$. Suppose that given any subset $A \subseteq Z$ with the property that given distinct $x , y \in A$ either $x \leq y$ or $y \leq x$ (such a sebset is called a chain in $(Z , \leq )$), there is a $z \in Z$ such that $x \leq z$ for all $x \in A$ ($x$ is an upper bound of $A$). Then there is a $z \in Z$ such that there is no $x \in Z \setminus \{ z \}$ such that $z \leq x$ ($z$ is a maximal element of $Z$).

For the proof of Urysohn's Lemma, we can take $Z$ to be the set of all (finite or infinite) sequences $( V_i )_i$ which satisfy the conditions in Rudin's proof of Urysohn's Lemma up to its length. We define a partial order relation $\leq$ on $Z$ by $( V_i )_i \leq ( W_j )_j$ if $( V_i )_i$ is an initial segment of $( W_j )_i$ (i.e., $( W_j )_j$ is not shorter than $( V_i )_i$ and $V_i = W_i$ for all possible $i$). It is easy to show that every chain in $( Z , \leq )$ has an upper bound by just taking the "union" of the sequences in the chain. Then by Zorn's Lemma $( Z , \leq )$ has a maximal element $( V_i )_i$. If this maximal element happened to be finite, we could use the construction in Rudin's proof to make it longer, contradicting that it is maximal! Therefore the maximal sequence must be infinite, as required.

Note that use of a principal like the Axiom of Choice is necessary to prove Urysohn's Lemma. In

Läuchli, H., Auswahlaxiom in der Algebra, Comment. Math. Helv. 37, 1-18 (1962). ZBL0108.01002.

the author constructs a model of ZF (where the Axiom of Choice fails) in which there is an infinite normal (Hausdorff) space $X$ such that every continuous function $X \to \mathbb{R}$ is constant. This implies that Urysohn's Lemma fails in this model (since all continuous real-valued functions on $X$ are constant, you can't even separate points by continuous functions in $X$, and by Hausdorffness singleton sets are closed in $X$).

As a final aside, note that in

Good, C.; Tree, I. J., Continuing horrors of topology without choice, Topology Appl. 63, No. 1, 79-90 (1995). ZBL0822.54001.

it is proved that Usysohn's Metrization Theorem does not require the Axiom of Choice (or anything like it). Essentially, they prove that in a second-countable regular space $X$ one can use a fixed countable base $\mathcal{B}$ to define for each pair $A , B$ of disjoint closed subsets of $X$ a specific open set $U \subseteq X$ satisfying $A \subseteq U \subseteq \overline{U} \subseteq X \setminus B$. One can then prove that Urysohn's Lemma holds for this space by following, e.g., Rudin's proof of Urysohn's Lemma, but instead of choosing an intermediary set at each stage, use a previously fixed countable base to define it. Since each step was determined, one then uses the Principle of Recursive Definitions instead of the Axiom of Choice to justify that it could all be done at once.

Lääne-Viru
  • 6,918