Points in Product Spaces are Near Open Sets

Theorem
Let $X_i$ be a topological space for each $i \in I$.

Let $X$ be the topological product of the $X_i$s.

Let $U$ be nonempty and open in $X$.

Let $x$ be a point in $X$.

For each point $y$ in $X$, let $K \left({y}\right) = \left\{ {i \in I : y_i \ne x_i} \right\}$.

Then there exists a point $u$ in $U$ such that $K \left({u}\right)$ is finite.

Proof
Let $q$ be any point in $U$.

By the definition of the product topology, there exist sets $Q_i$ for each $i \in I$ such that:
 * $Q_i$ is open in $X_i$

and:
 * $J = \left\{ {i \in I : Q_i \ne X_i} \right\}$ is finite such that:
 * $\displaystyle q \in Q = \prod_{i \mathop \in I} Q_i \subseteq U$

Let $u$ be the point for which $u_i = q_i$ when $i \in J$ and $u_i = x_i$ when $i \notin J$.

Because $u_i \in Q_i$ for each $i \in I$:
 * $u \in U$

Because $u_i \ne x_i \implies i \notin J$:
 * $K \left({u}\right) \subseteq J$

Since $J$ is finite, so is $K \left({u}\right)$.