Subspace of Product Space is Homeomorphic to Factor Space/Proof 2/Continuous Mapping
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 $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$.
Let $z \in X$.
Let $i \in I$.
Let $Y_i = \set {x \in X: \forall j \in I \setminus \set i: x_j = z_j}$.
Let $\upsilon_i$ be the subspace topology of $Y_i$ relative to $\tau$.
Let $p_i = \pr_i {\restriction_{Y_i}}$, where $\pr_i$ is the projection from $X$ to $X_i$.
Then:
- $p_i$ is continuous.
Proof
Let $V \in \tau_i$.
Let $\ds U = \prod_{i \mathop \in I} U_i$ where:
- $U_j = \begin{cases} X_j & j \ne i \\ V & j = i \end{cases}$
From Natural Basis of Product Topology, $U$ is an element of the the natural basis.
By definition of the product topology $\tau$ on the product space $\struct {X, \tau}$ the natural basis is a basis for the product topology.
It follows that:
- $U$ is open in $\struct {X, \tau}$
Let $x \in Y_i$.
Now:
\(\ds x\) | \(\in\) | \(\ds \map {p_i^\gets} V\) | ||||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \map {p_i} x\) | \(\in\) | \(\ds V\) | Definition of Inverse Image Mapping of Mapping: $p_i^\gets$ | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \map {\pr_i} x\) | \(\in\) | \(\ds V\) | Definition of Restriction of Mapping $p_i$ | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds x\) | \(\in\) | \(\ds U\) | Definition of $U$ | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds x\) | \(\in\) | \(\ds U \cap Y_i\) | as $x \in Y_i$ |
By set equality:
- $\map {p_i^\gets} V = U \cap Y_i$
By definition of the subspace topology on $Y_i$:
- $\map {p_i^\gets} V \in \upsilon_i$
It follows that $p_i$ is continuous by definition.
$\blacksquare$