Subspace of Product Space is Homeomorphic to Factor Space/Proof 1/Lemma 1
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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:
- $Y_i = \prod_{j \mathop \in I} Z_j$
Proof
\(\ds \) | \(\) | \(\ds x \in Y_i\) | ||||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \forall j \in I: \, \) | \(\ds x_j\) | \(=\) | \(\ds \begin {cases} z_j & j \ne i \\ x_i \in X_i & i = j \end {cases}\) | Definition of $Y_i$ | |||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \forall j \in I: \, \) | \(\ds x_j\) | \(\in\) | \(\ds Z_j\) | Definition of $Z_j$ for all $j \in I$ | |||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds x\) | \(\in\) | \(\ds \prod_{j \mathop \in I} Z_j\) | Definition of Cartesian Product |
The result follows by definition of set equality.
$\blacksquare$