4
$\begingroup$

I have a very basic question on wether or not we use the axiom of choice when we prove the very simple fact that the union of open sets of $\mathbb{R}$ (defined as unions of open intervals) is an open set of $\mathbb{R}$.

Say that $(U_i)_{i\in I}$ is a family of open sets of $\mathbb{R}$. So each $U_i$ is of the form $\cup_{j\in J} I_j$ where the set $J$ may depend on $i$, and the family of open intervals $(I_j)_{j\in J}$ may also depend on $i$.

So to be able to write down $\cup_i U_i$ as the union $\cup_{i,j} I_j$ don't we have to choose (a potentially uncountable number of times) a set $J$ and a family $(I_j)_j$ for each $i$ ? Does that mean we have to apply the axiom of choice ?

$\endgroup$
  • $\begingroup$ What makes you think you are choosing anything? You include each interval which makes up any of the open sets of your choice. If you are more precise by what you mean by notations like $(U_i)_{i \in I}$, I think you'll find that you can construct the union you need easily without ever invoking AC. $\endgroup$ – Mees de Vries Sep 12 '17 at 14:33
  • $\begingroup$ Can you tell what you mean by being more precise? $(U_i)$ is a family indexed by $I$, so a function from $I$ into the set of all open sets of $\mathbb{R}$. $\endgroup$ – LCO Sep 12 '17 at 14:35
  • $\begingroup$ Can you tell me how I am not choosing the indexing set $J$ for each i ? $\endgroup$ – LCO Sep 12 '17 at 14:36
  • $\begingroup$ Ah, I'm sorry, I see what you mean now. I was a little bit dismissive before. My bad. However, you can take a canonical choice of family for each open set: namely, pick for each open set the family of all open intervals which are subsets of your open set. Thus you don't need to make any non-canonical choices. $\endgroup$ – Mees de Vries Sep 12 '17 at 14:37
  • $\begingroup$ Ah yes, I see now how to do it, thank you $\endgroup$ – LCO Sep 12 '17 at 14:43
9
$\begingroup$

Your doubt arises because we are only given for $U_i$ that there exists an index set $J$ and intervals $I_j$ such that $U_i=\bigcup_{j\in J}I_j$. But we are not given this information.

However, we can describe the union without invoking choice as $$\bigcup_{i\in I}U_i = \bigcup_{\langle a,b\rangle\in J}\left]a,b\right[ $$ where $$J=\{\,\langle a,b\rangle\in \Bbb R^2\mid \exists i\in I\colon \left]a,b\right[\subseteq U_i\,\}. $$

$\endgroup$
8
$\begingroup$

Hagen wrote correctly, that we don't need to choose, and we can take all of the intervals.

But actually, more is true. Every open set of real numbers has a unique decomposition into pairwise disjoint intervals. Simply look at the connected components of the open set.

So in this case there is absolutely no need for using choice, even if you want to have a canonical choice. But of course, the main point is that you don't need to choose anything at any point, taking everything works just fine.

$\endgroup$
0
$\begingroup$

So your definition of an open set in $\Bbb{R}$ is a set that can be written as a union of open intervals. This generalises: given a set $X$ and a system $\cal B$ of subsets of $X$ such that if $A, B \in \cal B$ then $A \cap B \in \cal B$, there is a topology $\cal T$ on $X$ whose open sets are the sets of the form $C = \bigcup_{A \in \cal I} A$, where ${\cal I}$ is a subset of $\cal B$. ($\cal B$ is called a basis for the topology$\cal T$.) You don't need the axiom of choice to show that $\cal T$ is a topology.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.