Topology under separation and countability

51 Views Asked by At

A nonempty product of spaces is $T_0$ if and only if each factor space is $T_{0}$.

This is my solution:

If $X_\alpha$ is a $T_0$ space for each $\alpha$ that belongs to $A$ and $x$ is not equal to $y$ in $\pi X_\alpha$ then for some coordinate $\alpha$ we have $x_\alpha$ not equal to $y_\alpha$ so there exists a neighborhood $U_\alpha$ containing, say $x_\alpha$ but not $y_\alpha$, and since the $\pi^{-1} U_\alpha$ is open in $\pi X_\alpha$ containing $x$ but not $y$, thus $\pi X_\alpha$ is $ T_0$. Please is it ok or there is another work to add to it? please help out