3

Is the following statement true or false?

Let $\{A_i,\ i\in I\}$ be a family of at most countable sets. Suppose the family is totally ordered under set inclusion, i.e. $\forall i\in I,\ j\in I$, either $A_i\subset A_j$ or conversely. Then $\bigcup_{i\in I}A_i$ is also at most countable.

I had this problem when I was trying to prove something else using Zorn's lemma. Although my original problem had a simpler solution, I am now more interested in this new question I came up with, so I won't post my original problem as it is sort of irrelevant.

The difficulty is that $I$ may not be a countable index set. Yet, I was unable to find a counterexample or a proof off the top of my head. Any ideas?

Andrés E. Caicedo
  • 79,201
trisct
  • 5,221
  • The set of roots of quadratic integer polynomials is countable. So is the set of roots of cubic integer polynomials and so forth; I strongly suspect their union over $n$ (maximal degree of integer polynomial) isn't. EDIT: Actually they are. Apparently they are called 'Algebraic numbers'. – Alexander Geldhof Nov 27 '19 at 14:26
  • @Peter Yes, I mean that each $A_i$ is at most countable, but there could be uncountably many sets $A_i$. I added a pair of braces, if that helps... – trisct Nov 27 '19 at 14:28
  • @AlexanderGeldhof No, the set of algebraic numbers is countable –  Nov 27 '19 at 14:28
  • @MatthewDaly But we need not have polynomials with algebraic coefficients. – Peter Nov 27 '19 at 14:29
  • 1
    @saulspatz Note that the family needs to be totally ordered. Your example does not satisfy this condition. – trisct Nov 27 '19 at 14:31
  • OK, and if we start with some real number, than add another real number to get another set and always add a real number and $I$ is uncountable , what then ? – Peter Nov 27 '19 at 14:33
  • Try explicitly doing that, @Peter. It isn't that evident, since there exists no bijection between $\mathbb{N}$ and $\mathbb{R}$. In particular, there is no guarantee that the $i-$th set is countable. – Alexander Geldhof Nov 27 '19 at 14:33
  • @AlexanderGeldhof If this does not work, can we have an uncountable family with the desired property then ? – Peter Nov 27 '19 at 14:35
  • @Peter Just a thought, "add another number" is the same as expanding the index set by one element, which should lead to a countable index set. – trisct Nov 27 '19 at 14:35
  • @Peter Frankly I have no clue at this moment! – Alexander Geldhof Nov 27 '19 at 14:35
  • 1
    @Peter The thing is, exactly what is meant by "an uncountable number of steps", how do we walk such steps? Naturally, steps should be one after another. – trisct Nov 27 '19 at 14:38
  • 3
    I don't think this is possible. Can't we take the index set to be the first uncountable ordinal $\omega_1$, and for $\sigma<\omega_1$, take $A_\sigma=\sigma$? Then $\bigcup_{\sigma<\omega_1}A_\sigma=\omega_1$ which is uncountable. – saulspatz Nov 27 '19 at 14:44
  • 1
    @AlexanderGeldhof Yes, but each of them is countable, which is all that matters. – saulspatz Nov 27 '19 at 14:48
  • 1
    Similarly: Given a chain of finite sets, is its union necessarily finite? – Asaf Karagila Nov 27 '19 at 14:52

0 Answers0