# product topology, generator

I have a question to the following topic.

The product topology, noted by $$(X, \tau_1) \times (X, \tau_2)$$ on $$X \times Y$$, is the smallest topology such that the projections $$\pi_1: X \times Y \to X$$, $$\pi_2: X \times Y \to Y$$ are continuous.

For a product of two topological spaces $$(X, \tau_1), (X, \tau_2)$$ it can be shown that $$\tau_1 \times \tau_2$$ is the generated topology by the family $$\{ U \times V : U \in \tau_1, V \in \tau_2 \}$$.

Can you tell me if my following attempt of proof seems logical?

Let $$\tau$$ be the generated topology by $$\{ U \times V : U \in \tau_1, V \in \tau_2 \}$$. We consider the projections: $$\pi_1 : X \times Y \to X, \pi_2 : X \times Y \to Y$$. Then $$\pi_1$$ and $$\pi_2$$ are cont. regarding $$\tau$$, because $$\pi_1^{-1}(U) = U \times Y \in \tau, \pi_2^{-1}(V) = X \times V \in \tau$$. Hence $$\tau_1 \times \tau_2 \subseteq \tau$$.

For the other inclusion $$\tau \subseteq \tau_1 \times \tau_2$$ it is sufficient that $$U \times V = (U \times Y) \cap (X \times V) = \pi_1^{-1}(U) \cap \pi_2^{-1}(V) \in \tau_1 \times \tau_2$$ for all $$U \in \tau_1$$, $$V \in \tau_2$$ (since $$\tau$$ is per Def the smallest topology, which contains all these sets).

$$\tau_1 \times \tau_2$$ is by definition the minimal topology that makes both projections continuous.
$$\tau$$ is the topology with as base the open rectangles.
So after showing that $$\tau$$ indeed makes both projections continuous (correctly), apply to that minimality explicitly (at this point!) to justify $$\tau_1 \times \tau_2 \subseteq \tau$$.
Now if $$\tau'$$ is any topology that makes both projections continuous, we deduce correctly, as you did that open rectangles $$U \times V$$ are in $$\tau'$$. It then follows that $$\tau \subseteq \tau'$$ (as $$\tau'$$ contains the base of $$\tau$$ it contains all of $$\tau$$), and as we can take $$\tau'= \tau_1 \times \tau_2$$ in particular (!) (because it is one of the topologies that makes both projections continuous; no minimality used here!) we can say $$\tau \subseteq \tau_1 \times \tau_2$$ and we have equality of topologies.