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

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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

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up vote 10 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.

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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

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