Projection from Product Topology is Open/General Result/Proof
Jump to navigation
Jump to search
Theorem
Let $\family {T_i}_{i \mathop \in I} = \family {\struct {S_i, \tau_i} }_{i \mathop \in I}$ be an indexed family of topological spaces where $I$ is an arbitrary index set.
Let $\ds S = \prod_{i \mathop \in I} S_i$ be the corresponding product space.
Let $\tau$ denote the product topology on $S$.
Let $\pr_i: S \to S_i$ be the corresponding projection from $S$ onto $S_i$.
Then $\pr_i$ is open for all $i \in I$.
Proof
Let $U \in \tau$.
It follows from the definition of product topology that $U$ can be expressed as:
- $\ds U = \bigcup_{j \mathop \in J} \bigcap_{k \mathop = 1}^{n_j} \map {\pr_{i_{k, j} }^{-1} } {U_{k, j} }$
where:
- $J$ is an arbitrary index set
- $n_j \in \N$
- $i_{k, j} \in I$
- $U_{k, j} \in \tau_{i_{k, j} }$.
For all $i' \in I$, define $V_{i', k, j} \in \tau_{i'}$ by:
- $V_{i', k, j} = \begin {cases} U_{k, j} & : i' = i_{k, j} \\ S_{i'} & : i' \ne i_{k, j} \end {cases}$
For all $i \in I$, we have:
\(\ds \map {\pr_i} U\) | \(=\) | \(\ds \bigcup_{j \mathop \in J} \map {\pr_i} {\bigcap_{k \mathop = 1}^{n_j} \map {\pr_{i_{k,j} }^{-1} } { U_{k,j} } }\) | Image of Union under Relation: Family of Sets | |||||||||||
\(\ds \) | \(=\) | \(\ds \bigcup_{j \mathop \in J} \map {\pr_i} {\bigcap_{k \mathop = 1}^{n_j} \prod_{i' \mathop \in I} V_{i', k, j} }\) | Definition of Projection | |||||||||||
\(\ds \) | \(=\) | \(\ds \bigcup_{j \mathop \in J} \map {\pr_i} {\prod_{i' \mathop \in I} \bigcap_{k \mathop = 1}^{n_j} V_{i', k, j} }\) | Cartesian Product of Intersections: General Case | |||||||||||
\(\ds \) | \(=\) | \(\ds \bigcup_{j \mathop \in J} \bigcap_{k \mathop = 1}^{n_j} V_{i,k,j}\) | Definition of Projection |
As:
- $\ds \bigcup_{j \mathop \in J} \bigcap_{k \mathop = 1}^{n_j} V_{i, k, j} \in \tau_i$
it follows that $\pr_i$ is open.
$\blacksquare$