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).


Be more precise, the argument is subtly different:

$\tau_1 \times \tau_2$ 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.

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