Points in Product Spaces are Near Open Sets

Theorem
Let $\family {X_i}_{i \mathop \in I}$ be an indexed family of topological spaces where $I$ is an arbitrary index set.

Let $X = \displaystyle \prod_{i \mathop \in I} X_i$ be the product space of $\family {X_i}_{i \mathop \in I}$.

Let $U$ be nonempty open subset of $X$.

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

For each point $y$ in $X$, let $\map K y = \set{i \in I : y_i \ne x_i}$.

Then there exists a point $u$ in $U$ such that $\map K u$ is finite.

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

The topology on the product space $X$ is the Tychonoff topology which has the natural basis as a (synthetic) basis.

Then:
 * $\exists Q \in \BB : q \in Q \subseteq U$

where $\BB$ is the natural basis for Tychonoff topology on $X$.

From Natural Basis of Tychonoff Topology:
 * $\displaystyle Q = \prod_{i \mathop \in I} Q_i$ where:
 * for all $i \in I : Q_i$ is open in $X_i$
 * $J = \set {i \in I : Q_i \ne X_i}$ is finite

Let $u$ be the point defined by:
 * $u_i = \begin{cases} q_i & i \in J \\ x_i & i \notin J \end{cases}$.

By definition of $u$:
 * $\forall i \in I : u_i \in Q_i$

By definition of the Cartesian product:
 * $u \in Q \subseteq U$

By definition of $u$:
 * $\forall i \in I : i \notin J \implies u_i = x_i$

So by Rule of Transposition:
 * $\forall i \in I : u_i \ne x_i \implies i \in J$:

Thus:
 * $\map K u \subseteq J$

From Subset of Finite Set is Finite and since $J$ is finite, so is $\map K u$.