I was just thinking if you need the axiom of choice to see if some topologies are actual topology.

$\cup_{\lambda \in \Lambda} U_{\lambda} \in \tau$

See if you have $\tau$ is uncountable and $\Lambda$ is uncountable. How do you know if you have a topology? You must need some axiom of choice to show that a set is topology or to show it's not a topology.


First let us see when $\tau\subseteq\mathcal P(X)$ is a topology on $X$.

  1. $X\in\tau$, $\varnothing\in\tau$.
  2. For every $A\subseteq\tau$, $\bigcup A\in\tau$.
  3. For every $U,V\in\tau$ we have $U\cap V\in\tau$.

The second condition is to say, that no matter how many open sets I take (finitely many, countably many, continuum many, or any other number), the union of all the open sets will be an open set.

The fact that this union is at all a set we adhere to the Axiom of Union, suppose $A$ is a set of sets, then $\{u\mid \exists y\in A: u\in y\}$ is a set, this set is often denoted $\bigcup A$ and is known as the union of $A$.

The axiom schema of replacement says that if $\Lambda$ is a set, and $X_i$ is a set for every $i\in\Lambda$, then $\{X_i\mid i\in\Lambda\}$ is also a set. This is independent of the size of $\Lambda$. We only require that it is indeed a set.

From this we have that the requirement that a topology is closed under any unions is not dependent on the axiom of choice, the closure under finite intersections is also easily definable, as the property that the empty set and the entire space are in the topology.

Note that $\mathbb R$ with the standard topology is second countable, that means that every union of open sets can be reduced to a countable union of open sets; it means that every discrete subset is countable; and it means that there are at most continuum many open sets.

  • $\begingroup$ The problem with that is that. $X=\mathbb{R}$, then you take the discrete topology on it. Then, isn't lambda by your definition countable and so can't contain an element in topology. What does arbitary mean in a topology. $\endgroup$ – simplicity Oct 12 '11 at 12:11
  • $\begingroup$ Huh? @simplicity, discrete topology just means every subset is open. You may be confusing because $\mathbb R$ is second countable and so every union of open sets can be reduced to a countable union, but in general the requirement is just to be closed under any unions, regardless to their cardinality (or even well-orderability). $\endgroup$ – Asaf Karagila Oct 12 '11 at 12:17
  • $\begingroup$ On the other hand, you can have a definition of a collection which ZFC proves to be a topology, but ZF doesn't. For example, consider the collection of wellorderable subsets of the reals. $\endgroup$ – user83827 Oct 12 '11 at 15:40
  • $\begingroup$ @ccc: Interesting. I never thought of this as an example. $\endgroup$ – Asaf Karagila Oct 12 '11 at 15:48

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