Question about proof of the product of a finite number of T1 spaces is a T1 space. Show that the product of a finite number of T1 spaces is a T1 space.
Since each $X_i$ is a $T_1$ space, $$\prod_{i \in I} X_i \backslash \{x_i\} = \{(a_1, a_2, ..., a_n):a_i \in X_i \land (a_1, a_2, ..., a_n) \neq (x_1, x_2, ..., x_n)\} $$ is open, and  $$(\prod_{i \in I} X_i) \backslash \{(x_1, x_2, ..., x_n)\} = \{(a_1, a_2, ..., a_n):a_i \in X_i \land (a_1, a_2, ..., a_n) \neq (x_1, x_2, ..., x_n)\}$$
has the same definition as the first product, which implies that it's open.
$\therefore$ The two are equal and finite product of $T_1$ spaces is a $T_1$ space.
Is this correct to assume that $\prod X_i \backslash \{x_i\} = (\prod X_i) \backslash \{(x_1, x_2, ..., x_n)\}$ because the definitions are the same?

To make things more simple you can focus on closed sets instead of open sets.
If $X$ denotes the product of $X_1,\dots,X_n$ and $\pi_i:X\to X_i$ the projections then a singleton $\{x_1,\dots,x_n\}\in X$ equals $\bigcap_{i=1}^n\pi_i^{-1}(\{x_i\})$. The singletons in every $X_i$ are closed (since $X_i$ is $T_1$) and the $\pi_i$ are continuous so we are dealing with a finite intersection of closed sets. So it is closed itself. We conclude that singletons are closed in $X$, or quivalently that $X$ is $T_1$.
