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In subjects like Differential Geometry/ General Topology one often constructs for each $x$ in a space $X$ a set $U_x$ satisfying certain properties. Examples where one does constructions like this:

  1. Deducing that every open cover of a second countable space has a countable subcover.
  2. Proving that any covering space of a topological $n$ - manifold is second countable.

I am interested in whether there are any set theoretic issues like say use of the axiom of choice. In particular, how can we say "for each $x$ construct a set $U_x$.." simultaneously for all $x \in X$?

Also, my set theory is not so deep but I know that assuming that the set of all sets is a set leads to a contradiction in the style of Russell's paradox. However somehow it still makes sense to speak of a "maximal smooth atlas containing a given atlas" of a smooth manifold, or all closed sets containing a given set in a topological space.

Why is there no problem in doing this and we don't get to a situation like Russell's paradox?

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This is the same as defining a function on X. Why do you say it wouldn't be possible for a computer to do? –  Adeel Mar 14 '13 at 12:28
    
You should be aware by now that mathematics is a huge mind game. When you say "construct ..." you hardly ever mean that you are going to give a computer this task and it will solve it. Heck, I don't know how a computer can even grasp something as large as the number of dimensions of $\Bbb R^n$ for $n=A(100,100)$ (where $A$ is the Ackermann function). And yet, mathematically speaking we work with this space plenty (not this one in particular, but with more general ones, and the results apply to this one as well). –  Asaf Karagila Mar 15 '13 at 1:58
    
@AsafKaragila My last line was just one of the philosophical questions that I ask myself about maths. Also sometimes I get confused - if maths is all in the mind then how is it real? Maybe such questions are taboo. –  user38268 Mar 15 '13 at 5:15
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Taboo? I won't go that far. But insisting on doing only things which are "real" would only limit you to positive numbers smaller than $2^{100}$ or so. Enjoy. –  Asaf Karagila Mar 15 '13 at 5:20
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1 Answer

up vote 9 down vote accepted

In general, one needs the axiom of choice to choose $U_x$'s as in your question, but in certain cases one can be more explicit and thus avoid the axiom of choice.

Your second example requires some form of the axiom of choice. Specifically, it is consistent with ZF to have a countable sequence of pairwise disjoint $2$-element sets $P_0,P_1,\dots$ with no selector (i.e., no set consisting of exactly one element from each of the pairs $P_n$). In this situation, let $X$ be the union of the $P_n$'s, with the discrete topology, also give $\mathbb N$ the discrete topology, and let $f:X\to\mathbb N$ send the two elements of any $P_n$ to $n$. Then $f$ is a $2$-to-$1$ covering map, $\mathbb N$ is obviously second countable, but $X$ is not second countable (any basis for its topology would have to include all the singletons and would thus give an enumeration of $X$, contrary to the non-existence of a choice function for the pairs $P_n$).

I suspect that your first example also requires some form of the axiom of choice, but I don't immediately see a counterexample.

EDIT: Now I see a counterexample for the first one. Work in Cohen's first model for the failure of choice. This model contains a set $A$ of mutually generic Cohen subsets of $\mathbb N$ such that $A$ has no countably infinite subset. It follows from genericity that $A$ is a covering of $\mathbb N$ but any countable subfamily, being finite, fails to cover $\mathbb N$. Giving $\mathbb N$ the discrete topology, we get an open cover $A$ with no countable subcover, even though $\mathbb N$ is trivially second countable.

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Note that, in both of my counterexample, all the spaces involved are discrete. I take that as meaning that the questions are purely set-theoretic, not topological. –  Andreas Blass Mar 14 '13 at 12:59
    
For lots of information about the role of the axiom of choice in topology, you should look at the papers of Norbert Brunner and at the topology part of the book "Consequences of the Axiom of Choice" by Paul Howard and Jean Rubin. –  Andreas Blass Mar 14 '13 at 13:00
    
Thanks for your answer. I mistyped 2; what I meant was that the covering space of a topological $n$ - manifold is always second countable. –  user38268 Mar 14 '13 at 13:42
    
@BenjaLim: $\mathbb{N}$ with discrete topology is a toplogical $n$-manifold (with $n = 0$). –  Jason DeVito Mar 14 '13 at 13:46
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@BenjaLim: I don't understand - all this means is that this fact uses AC somewhere. Looking at Lee's proof (in the older edition of the book), in the last line of the first full paragraph on pg. 43, he unions together countably many countable sets and concludes the result is countable. This fact, in particular, uses AC. (I'm not sure if he uses choice anywhere else, I didn't read it that deeply.) –  Jason DeVito Mar 14 '13 at 14:01
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