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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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    $\begingroup$ Here's an obvious sufficient condition: when one is contained in the other. But that's not a necessary condition. $\endgroup$ – Zhen Lin Sep 27 '11 at 11:31
  • $\begingroup$ 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. $\endgroup$ – Ragib Zaman Sep 27 '11 at 11:34
  • $\begingroup$ If $X$ has cardinality one or two then this result is true, otherwise it is false. proofwiki.org/wiki/… $\endgroup$ – MathCosmo Sep 5 '18 at 14:13
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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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  • $\begingroup$ Thanks for the reply. I was looking for an useful necessary and sufficient condition, but this makes me think there is no such thing. $\endgroup$ – spin Sep 27 '11 at 14:31
  • $\begingroup$ I understand completely the failure of condition (4), but I don't understand the failure of condition (3). Do you think you could explain it? By definition $V \in \mathscr T_1$ and $W \in \mathscr T_2$, which seems like it should by definition of unions imply that $V \cup W \in \mathscr T_1 \cup \mathscr T_2 \equiv \mathscr T$...? Or am I mistaken? $\endgroup$ – Sam Blattner Dec 22 '16 at 17:13
  • $\begingroup$ @SamBlattner The set $\mathscr{T}_{1} \cup \mathscr{T}_{2}$ consists of all \textit{sets} in $\mathscr{T}_{1}$ or $\mathscr{T}_{2}$. When one considers whether $V \cup W$ belongs to this union, we evaluate whether one of $\mathscr{T}_{1}$ or $\mathscr{T}_{2}$ contains $V \cup W$. You are mistakenly thinking (methinks) that one must evaluate whether, if $\mathscr{T}_{1} = \{ T_{1}, T_{2}, \dots \}$ and $\mathscr{T}_{2} = \{ T'_{1}, T'_{2}, \dots \}$, $V \cup W \in \{ T_{1} \cup T'_{1}, T_{1} \cup T'_{2}, \dots, T_{2} \cup T'_{1}, T_{2} \cup T'_{2}, \dots \}$ $\endgroup$ – Muno May 22 '17 at 10:42
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Years late to answer this but:

There is actually a problem in Munkres related to this: prove for a family of topologies $\{\tau_\alpha \}$ on a set $X$ that there is a unique smallest topology containing all of the topologies in the family.

If you set $S = \{ U : U \in \tau_\alpha \text{ for some } \alpha \}$ and consider the topology generated by $S$ (as a subbasis) I think this does the trick. It also provides a less wordy sufficient and necessary condition to your question.

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