Subspace of Product Space is Homeomorphic to Factor Space/Proof 2/Lemma 1
Theorem
Let $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$ be a family of topological spaces where $I$ is an arbitrary index set.
Let $\ds \struct {X, \tau} = \prod_{i \mathop \in I} \struct {X_i, \tau_i}$ be the product space of $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$.
For all $k \in I$, let $\pr_k$ denote the projection from $X$ to $X_k$.
Let $z \in X$.
Let $i, k \in I$.
Let $Y_i = \set {x \in X: \forall j \in I \setminus \set i: x_j = z_j}$.
Let $p_i = \pr_i {\restriction_{Y_i} }$ be the restriction of $\pr_i$ to $Y_i$.
Let $V_k \in \tau_k$.
Let $\map {\pr_k^\gets } {V_k} \cap Y_i \ne \O$.
Then:
- $\map {p_i^\to} {\map {\pr_k^\gets} {V_k} \cap Y_i}$ is open in $\struct{X_i, \tau_i}$
Proof
We have that $p_i$ is a bijection from the lemmas:
Let $x \in X_i$.
Then:
| \(\ds x\) | \(\in\) | \(\ds \map {p_i^\to} {\map {\pr_k^\gets} {V_k} \cap Y_i}\) | ||||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds \map {p_i^{-1} } x\) | \(\in\) | \(\ds \map {\pr_k^\gets} {V_k} \cap Y_i\) | Definition of Direct Image Mapping of Mapping | ||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds \map {p_i^{-1} } x\) | \(\in\) | \(\ds \map {\pr_k^\gets} {V_k}\) | as $\map {p_i^{-1} } x \in Y_i$ | ||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds \map {\pr_k} {\map {p_i^{-1} } x}\) | \(\in\) | \(\ds V_k\) | Definition of Direct Image Mapping of Mapping |
By definition of $p_i$:
- $\map {p_i^{-1} } x = y$
where:
- $\forall j \in I : y_j = \begin {cases} z_j & j \ne i \\ x & j = i \end {cases}$
Case 1: $k = i$
Let $k = i$.
Then:
| \(\ds \map {\pr_i} {\map {p_i^{-1} } x}\) | \(=\) | \(\ds \map {\pr_i} {y}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds y_i\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds x\) |
So:
- $x \in \map {p_i} {\map {\pr_i^\gets} {V_i} \cap Y_i}$ if and only if $x \in V_i$
That is:
- $\map {p_i} {\map {\pr_i^\gets} {V_i} \cap Y_i} = V_i \in \tau_i$
$\Box$
Case 2: $k \ne i$
Let $k \ne i$.
Then:
| \(\ds \map {\pr_k} {\map {p_i^{-1} } x}\) | \(=\) | \(\ds \map {\pr_k} y\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds y_k\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds z_k\) |
So:
- $x \in \map {p_i} {\map {\pr_i^\gets} {V_i} \cap Y_i}$ if and only if $z_k \in V_k$
Since $x \in X_i$ was arbitrary, then:
- $\map {p_i} {\map {\pr_i^\gets} {V_i} \cap Y_i} = X_i$ if and only if $z_k \in V_k$
Let $w \in \map {\pr_k^\gets} {V_k} \cap Y_i$ which is guaranteed since $\map {\pr_k^\gets} {V_k} \cap Y_i \ne \O$
By the definition of inverse image mapping then:
- $w_k = \map {\pr_k} w \in V_k$
By the definition of $Y_i$ then:
- $w_k = z_k$
So $z_k \in V_k$.
It follows that:
- $\map {p_i} {\map {\pr_i^\gets} {V_i} \cap Y_i} = X_i \in \tau_i$
$\Box$
In either case:
- $\map {p_i^\to} {\map {\pr_k^\gets} {V_k} \cap Y_i}$ is open in $\struct {X_i, \tau_i}$.
$\blacksquare$