$X=\prod_{j\in J}X_j$ is Hausdorff(regular) iff each $X_j$ is Hausdorff(regular) on box-topology. Then what about the $T_1$ and product-topology?

Question

The box-topology proof is quite obvious as $\prod{U_j}$ is always open in $X$, where the product-topology may not stand.

By the way, if regularity stands, does this mean $T_1$ axiom stands first? Then, will "$X=\prod_{j\in J}X_j$ is of $T_1$ iff each $X_j$ is of $T_1$?" stand?

Answer 1

Yes, the product is Hausdorff/ regular /$T_1$ iff all factors are is a fact that holds in both the box topology and the product topology. The proofs are trivial.