Why is the union of two topologies not in general a topology?
(Note: this question is based on this question from the other website, written by user390960. The answer was previously written by me.)
I read that the intersection of two topologies $\mathcal{T}_1$ and $\mathcal{T}_2$ is always a topology, which makes sense; but I also read that the union of two topologies is not in general a topology, because the intersection of an open in $\mathcal{T}_1$ and an open in $\mathcal{T}_2$ is not necessarily open.
How can this be? Shouldn't the two open sets in $\mathcal{T}_1$ and $\mathcal{T}_2$ be disjoint and hence have an open (empty) intersection?
2 answers
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.1
Once this is grasped, it becomes somewhat intuitive that the union of two topologies isn't in general another topology (while the intersection is).2
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:)
1 By contrast, intersections of two opens in disjoint topological spaces are always empty. Hence the OP's confusion.
2 (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.)
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Your last sentence makes me think you haven't understood your own question. There is no reason to think that the intersection of an open set in one topology and an open set in another must be empty.
Consider the set $(0,1)\times(0,1).$ Let $\mathcal T_1$ be the topology whose basic open sets are all sets of the form $U\times\{\,x\,\}$ for some open $U\subseteq(0,1)$ and some $x\in(0,1),$ and let $\mathcal T_2$ be the topology whose basic open sets are all sets of the form $\{\,x\,\}\times U$ for some open $U\subseteq(0,1)$ and some $x\in(0,1).$ Both $\mathcal T_1$ and $\mathcal T_2$ are topologies. In $\mathcal T_1,$ every horizontal slice through the square is a connected component of the space, and in $\mathcal T_2$ the same is true of every vertical slice. But the intersection $\Big( (0.3,0.4)\times \left\{~0.35~\right\} \Big) \cup \Big( \left\{ ~0.35~\right\}\times(0.3,0.4) \Big)$ contains only the one point $(0.35,0.35),$ which is not an open set in either topology.
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