Product of Closed Sets is Closed

Theorem
Let $\family {\struct {S_i, \tau_i} }_{i \mathop \in I}$ be a family of topological spaces where $I$ is an arbitrary indexing set.

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

Let $T = \struct {S, \TT}$ be the product space of $\family {\struct {S_i, \tau_i} }_{i \mathop \in I}$ with Tychonoff topology $\TT$.

Suppose we have an indexed family of sets $\family {C_i}_{i \mathop \in I}$, where each $C_i$ is closed in $\struct {S_i, \tau_i}$.

Then $\displaystyle \prod_{i \mathop \in I} C_i$ is closed in $\struct {S, \TT}$.

Proof
First note that:

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

Let $y \in I$.

We see that:

Since $C_y$ is closed, $\paren {S_y \setminus C_y}$ is open in $\struct {S_y, \tau_y}$ by definition.

Thus:
 * $\pr_y^{-1} \sqbrk {S_y \setminus C_y} \in \BB$

where $\BB$ is the natural sub-basis of $T$:
 * $\BB = \set {\pr_i^{-1} \sqbrk U: i \in I, U \in \tau_i}$

Therefore $\pr_y^{-1} \sqbrk {S_y \setminus C_y} = S \setminus \set {x \in S: x_y \in C_y}$ is open in $T$.

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