A union of two *topological spaces* is not the same thing as a union of two *topologies* on the *same* set. This is why intersections of two opens in different topologies can be nonempty.<sup>1</sup>
Once this is grasped, it becomes somewhat intuitive that the union of two topologies isn't in general another topology (while the intersection is).<sup>2</sup>
For a pictorial example, take $T_1 := I_2×D_2$ and $T_2 := D_2×I_2$ where I_2 and D_2 are respectively the indiscrete and discrete topologies on $\mathbf{2} = \{0,1\}$:
The regions enclosed by rounded rectangles are the (nonempty) opens. There are clearly three such regions.
The union of these topologies contains five nonempty sets (besides the empty set).
It is visually evident that this is not a topology. Indeed, the intersection of horizontal regions (opens of $T_1$) and vertical regions (opens of $T_2$) is always a single point, but single points are not open in either $T_1$ or $T_2$.
Nevertheless a union $T_1 \cup T_2$ can always be *extended* to a topology by adding more open sets (in this case we get the discrete topology, which has $2^4=16$ open sets, hence I only show a basis below:)
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<sup>1</sup> By contrast, intersections of two opens in disjoint *topological spaces* are always empty. Hence the OP's confusion.
<sup>2</sup> (A topology on $X$ is just a subset of $2^X$ closed under unions and finite intersections, much as a group is closed under composition. Thus, just as the union of two subgroups is not in general a subgroup but the intersection is, so also the union of two topologies is not in general a topology but the intersection is.)
[1]: https://i.sstatic.net/Z4pApQmS.png
[2]: https://i.sstatic.net/KkrFVsGy.png
[3]: https://i.sstatic.net/lQj6K5Y9.png
clemens
·
2026-03-07T05:40:02Z (6 months ago)