Product of Closed Sets is Closed

Theorem
Let $\left \langle{ \left({S_i, \tau_i}\right)}\right \rangle_{i \mathop \in I}$ be a family of topological spaces where $I$ is an arbitrary indexing set.

Let $\displaystyle S = \prod_{i \mathop \in I} S_i$.

Let $T = \left({S, \mathcal T}\right)$ be the Tychonoff product of $\left \langle{ \left({X_i, \tau_i}\right) } \right \rangle_{i \mathop \in I}$.

Suppose we have an indexed family of sets $\left \langle{C_i}\right \rangle_{i \mathop \in I}$, where each $C_i$ is closed in $ \left({S_i, \tau_i}\right)$.

Then $\displaystyle \prod_{i \mathop \in I} C_i$ is closed in $\left({S, \mathcal T}\right)$.

Proof
First note that:

Thus by Intersection of Closed Sets is Closed, our result is proven if we can show that $\forall i \in I: \left \{{ x \in S: x_i \in C_i }\right \}$ is closed in $T$.

Let $y \in I$. We see that:

Since $C_y$ is closed, $\left({S_y \setminus C_y}\right)$ is open in $\left({S_y, \tau_y}\right)$ by definition.

Thus:
 * $\operatorname {pr}_y^{-1} \left({S_y \setminus C_y}\right) \in \mathcal S$

where $\mathcal B$ is the natural sub-basis of $T$:
 * $\mathcal B = \left \{{ \operatorname {pr}_i^{-1} \left({U}\right) : i \in I, U \in \tau_i }\right \}$

Therefore $\operatorname{pr}_y^{-1} \left({S_y \setminus C_y}\right) = S \setminus \left \{{x \in S : x_y \in C_y}\right \}$ is open in $T$.

This implies that $\left \{ {x \in S : x_y \in C_y } \right \}$ must be closed in $T$ by definition.