Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The union of two topologies on some set may or may not be a topology. When is it a topology?

share|cite|improve this question
Here's an obvious sufficient condition: when one is contained in the other. But that's not a necessary condition. – Zhen Lin Sep 27 '11 at 11:31
I'd very much doubt there's any non-trivial condition, you would have to check these kinds of things on a case by case basis. – Ragib Zaman Sep 27 '11 at 11:34
up vote 12 down vote accepted

Let $\tau_1$ and $\tau_2$ be two topologies on a point set $X$. That is, $\tau_1$ and $\tau_2$ are subsets of the powerset of $X$, each indicating which subsets of $X$ should be called open. Let $\tau = \tau_1 \cup \tau_2$. In general $\tau$ is not a topology on $X$, but let's see which conditions are satisfied and which could fail:

(1) $\emptyset \in \tau$. True since $\emptyset \in \tau_1$ (and $\tau_2$).

(2) $X \in \tau$. True since $X \in \tau_1$ (and $\tau_2$).

(3) Arbitrary unions of open sets are in $\tau$, that is,

$$\bigcup_{\alpha \in \mathscr{I}} U_{\alpha} \in \tau.$$

Not true in general. At best, I can partition the sets into those in $\tau_1$ and those in $\tau_2$ so that we have:

$$\bigcup_{\alpha \in \mathscr{I}} U_{\alpha} = \bigcup_{\alpha \in \mathscr{I} \cap \tau_1} U_{\alpha} \cup \bigcup_{\alpha \in \mathscr{I} \cap \tau_2} U_{\alpha} = V \cup W,$$

where $V \in \tau_1$ and $W \in \tau_2$ because each $\tau_i$ is a topology. But here's where we get stuck. I have no guarantee that $V \cup W$ is in either $\tau_1$ or $\tau_2$, hence I can't place $V \cup W \in \tau$.

(4) Finite intersections of open sets are in $\tau$, that is,

$$\bigcap_{i=1}^n U_i \in \tau.$$

Again, I could partition the sets according to $\tau_1$ and $\tau_2$ so that

$$\bigcap_{i=1}^n U_i \in \tau = V \cap W,$$

for $V \in \tau_1$ and $W \in \tau_2$. But unless $V \cap W \in \tau_1$ or $\tau_2$, I cannot find $V \cap W \in \tau = \tau_1 \cup \tau_2$.

So this leads to the necessary (and sufficient) conditions: $\tau = \tau_1 \cup \tau_2$ is a topology if every pairwise union $U \cup V$ and intersection $U \cap V$ of open sets $U \in \tau_1$ and $V \in \tau_2$ lies in either $\tau_1$ or $\tau_2$. Of course, this does not strike me as a very useful or easy condition to check.

share|cite|improve this answer
Thanks for the reply. I was looking for an useful necessary and sufficient condition, but this makes me think there is no such thing. – spin Sep 27 '11 at 14:31

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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