The first approach is less constructive (as it operates with the family of all topologies containing every topology $T_{\alpha}$ without specifying how to obtain it), but it is more general, as whenever we need to find the smallest set satisfying property $P$ that contains every set $T_{\alpha}$ , the first thing to do is to check whether the intersection of all sets satisfying $P$ and containing every $T_{\alpha}$ , satisfies $P$ . If it does (as in our case, where $P$ is “being a topology”), then the intersection is automatically the smallest such set.

The second approach, at the same time, is more constructive, as it prescribes exactly how to obtain a subbasis for the smallest topology, but it requires additional proof to show that the resulting topology is indeed the smallest (see Exercise 5).

Note, that if we used the other method to obtain the smallest topology containing both given topologies, we would have to consider all 29 three-element topologies, including those that are not present in the picture above, check which ones contain both topologies given, and then take their intersection. With three elements this at least looks manageable, but the number of topologies grows as $e^{n^{2}/4}$ with the number of elements $n$ , and the exact number of topologies is unknown for even small values of $n$ . Imagine if our topologies were infinite, or if there were an infinite number of them. This is exactly what I meant when I said that the first method is less constructive.