Subspace of Product Space is Homeomorphic to Factor Space/Proof 1/Lemma 1

Theorem

Let $\family {X_i}_{i \mathop \in I}$ be a family of sets where $I$ is an arbitrary index set.

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

Let $z \in X$.

Let $i \in I$.

Let $Y_i = \set {x \in X: \forall j \in I \setminus \set i: x_j = z_j}$

For all for all $j \in I$ let:

$Z_j = \begin{cases} X_i & i = j \\

\set{z_j} & j \ne i \end{cases}$

Then:

$Y_i = \prod_{j \mathop \in I} Z_j$

$\ds $

$\ds x \in Y_i$

$\ds \leadstoandfrom \ \ $

$\ds \forall j \in I: \, $

$\ds x_j$

$=$

$\ds \begin {cases} z_j & j \ne i \\ x_i \in X_i & i = j \end {cases}$

Definition of $Y_i$

$\ds \leadstoandfrom \ \ $

$\ds \forall j \in I: \, $

$\ds x_j$

$\in$

$\ds Z_j$

Definition of $Z_j$ for all $j \in I$

$\ds \leadstoandfrom \ \ $

$\ds x$

$\in$

$\ds \prod_{j \mathop \in I} Z_j$

The result follows by definition of set equality.

$\blacksquare$