# If $\displaystyle \prod_{i=1}^n (X_i, \mathcal T_i)$ is a $T_1$ space, then each $(X_i,\mathcal T_i)$ is a $T_1$ space.

If $\displaystyle \prod_{i=1}^n (X_i, \mathcal T_i)$ is a $T_1$ space, then each $(X_i,\mathcal T_i)$ is a $T_1$ space.

Suppose that If $\displaystyle \prod_{i=1}^n (X_i, \mathcal T_i)$ is a $T_1$ space, then, for every $\bar{x}= (x_1, x_2,\dots , x_n)$, we have that $\left(\displaystyle \prod_{i=1}^n X_i\right)\setminus \{\bar{x}\} \in \mathcal T$, where$\mathcal T$ denotes the finite product topology.

Now notice that $$\left(\prod_{i=1}^n X_i\right)\setminus \{\bar{x}\} = \prod_{i=1}^n \left( X_i \setminus \{x_i\}\right)$$

I need now show that, for every $i=1,2,3,\dots ,n$, that $X_i \setminus \{x_i\} \in \mathcal T_i$. This is where I am stuck though.

I was thinking of trying to use the fact that $$\mathcal B=\{ O_1 \times O_2\times \dots \times O_n : O_i \in \mathcal T_i\}$$ forms a basis for $\mathcal T$ to somehow show that each

$X_i\setminus \{x_i\}$ is equal to one of those $O_i$, since this will then prove the result, however, I am not sure if that is true...

• It's not correct that $\prod_{i=1}^n X_i \setminus \{\bar x\} = \prod_{i=1}^n (X_i \setminus \{x_i\} )$. For example, $\Bbb R^2 \setminus \{(0,0)\} \neq (\Bbb R \setminus \{0\}) \times (\Bbb R \setminus \{0\})$. – user384138 Nov 12 '16 at 17:03
• Rather, one has: $\prod_{i=1}^n X_i \setminus \{\bar x\} = \bigcup_{i=1}^n \left( X_i \setminus \{x_i\} \times \prod_{j\neq i} X_j \right)$ – user384138 Nov 12 '16 at 17:12
You should have the additional hypothesis that each $X_i$ is nonempty. Otherwise, letting $X$ be any space, the product $X\times \emptyset$ is empty, and hence $T_1$, while $X$ need not be $T_1$.
So suppose we have a product space $\prod_{i=1}^n X_i$ which is $T_1$. We must show that each $X_i$ is $T_1$. By rearranging the factors, it suffices to show that $X_1$ is $T_1$. By our assumption that $X_2,\dots,X_n$ are nonempty, we can pick a point $a_i\in X_i$ in each of them.
Now given $x,y\in X_1$, we would like to show that there is an open set $U\subseteq X_1$ with $x\in U$ and $y\notin U$. Consider the points $(x,a_2,\dots,a_n)$ and $(y,a_2,\dots,a_n)$ in the product space. Since the product is $T_1$, there is a basic open set $V = V_1\times \dots \times V_n$ (with $V_i\subseteq X_i$ open for all $i$) such that $(x,a_2,\dots,a_n)\in V$ but $(y,a_2,\dots,a_n)\notin V$.
So $x\in V_1$ and $a_i\in V_i$ for all $i$. If $y\in V_1$, then $(y,a_1,\dots,a_n)\in V$, which is not the case. So $y\notin V_1$. This shows that picking $U = V_1$ witnesses the $T_1$ property for $x$ and $y$ in $X_1$.