Product topology on $X \times Y$ is the coarsest topology on $X \times Y$ that makes the projection maps continuous Show:Product topology on $X \times Y$ is the coarsest topology on $X \times Y$ that makes the projection maps $\pi_x,\pi_y$ on $X \times Y$ continuous.
Attempt: Let $\pi_x:X \times Y \rightarrow X$ and $\pi_y:X \times Y \rightarrow Y$ be the projection maps on $X \times Y$. Let $\tau$ be the product topology on $X \times Y$, and let $\tau'$ be any topology on $X \times Y$ that makes the projection maps continuous. Need to show $\tau \subset \tau'$.
Let $\pi_x,\pi_y$ be continuous. Then by definition of continuity, for any sets $U,V$ open in $X,Y$ respectively,$\pi_x^{-1}(U)=U \times Y,\pi_y^{-1}(V)=X \times V $  are in open in the topology on $X \times Y$. But since $\pi_x^{-1}(U),\pi_y^{-1}(V)$ are necessarily in $\tau$ and in order for the projection functions to be continuous for any topology $\tau'$ on $X \times Y$, $\pi_x^{-1}(U),\pi_y^{-1}(V) \in \tau'$, $\tau \subset \tau'$ and $\tau$ is the coarsest topology on $X\times Y$ making the projection functions continuous.
Comment: Is this all I needed to show for this proof? Is this completely correct, because it seems the proof is mostly trivial, so I am unsure.
The product topology is generated by the subbase $$\mathcal{S}:= \{\pi_X^{-1}[O]\mid O \in \mathcal{T}_X\} \cup \{\pi_Y^{-1}[O]\mid O \in \mathcal{T}_Y\}$$
by definition and if $\tau'$ makes $\pi_X,\pi_Y$ continuous, by definition of contituity it must contain $\mathcal{S}$ and thus $\tau$ (which is the minimal topology containing $\mathcal{S}$..). Ça c'est tout.
Yes, it is that easy. What do you think of the following way of writing the proof?

Let $\tau$ be the product topology on $X \times Y$, and $\tau'$ be any topology on $X \times Y$ such that $\pi_X : (X \times Y,\tau') \to X$ and $\pi_Y : (X \times Y,\tau') \to Y$ are continuous. Let $O_X$ be an open subset of $X$ and $O_Y$ be an open subset of $Y$. Then, $\pi_X^{-1}(O_X) \in \tau'$ and $\pi_Y^{-1}(O_Y) \in \tau'$ by assumption, and so $O_X \times O_Y = \pi_X^{-1}(O_X) \cap \pi_Y^{-1}(O_Y) \in \tau'$. Thus, if $U \in \tau$, $U$ is a union of sets of the form $O_X \times O_Y$, and therefore $U \in \tau'$ by the above argument.