0
$\begingroup$

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).

$\endgroup$
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.