Intersection of topologies: two topologies, finite number of topologies and topologies indexed by index set – why we need distinct proofs

general-topology

I'm self learning topology by reading "Topology without tears" by Sidney A. Morris. There are three exercises related to intersection of topologies:

  1. Let $\tau_1$ and $\tau_2$ be topologies on $X$. Prove that $\tau_1 \cap \tau_2$ is a topology on X
  2. If $\tau_1, \tau_1, …, \tau_n$ are topologies on $X$, prove $\bigcap_{i=1}^{n} \tau_{i}$ is a topology on X
  3. If for each $i \in I$, for some index set $I$, each $\tau_i$ is a topology on the set $X$, prove $\bigcap_{i \in I} \tau_{i}$ is a topology on X

Why do we need three distinct proofs for 2 topologies, countable number of topologies and (possibly infinite number of?) topologies indexed by index set? Why we cannot naturally extend the proof from 1. to 2. and 3? Is there other similar construct in mathematics that is true for two operands but not when taking account more ?

Best Answer

There is clearly no actual requirement for three separate theorems, since the last implies the others and is easily proved without appealing to either of the others. The reason for having all three exercises is probably pædagogical: even though the arguments for (1) and (3) are essentially the same, many students find it easier to think about two topologies than to think about an arbitrary family of topologies. As for (2), it can of course be proved by the same argument, but it can also be proved by induction on $n$, using (1) to carry out the induction step, much as one proves that the intersection of any finite family of open sets is open.

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