Subspace of Product Space is Homeomorphic to Factor Space/Proof 2/Continuous Mapping

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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.

$\blacksquare$

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