# Projection from Product Topology is Open/General Result

## 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 $\displaystyle S = \prod_{i \mathop \in I} S_i$ be the corresponding product space.

Let $\tau$ denote the Tychonoff 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

If $U \in \tau$, it follows from the definition of Tychonoff topology that $U$ can be expressed as:

$\displaystyle U = \bigcup_{j \mathop \in J} \bigcap_{k \mathop = 1}^{n_j} \map {\pr_{i_{k,j} }^{-1}} { U_{k,j} }$

where $J$ is an arbitrary index set, $n_j \in \N$, $i_{k,j} \in I$, and $U_{k,j} \in \tau_{i_{k,j} }$.

For all $i' \in I$, define $V_{i', k, j} \in \tau_{i'}$ by $V_{i', k, j} = U_{k,j}$ if $i' = i_{k,j}$, and $V_{i', k, j} = S_{i'}$ if $i' \ne i_{k,j}$.

For all $i \in I$, we have:

 $\displaystyle \map {\pr_i} U$ $=$ $\displaystyle \bigcup_{j \mathop \in J} \map {\pr_i} { \bigcap_{k \mathop = 1}^{n_j} \map {\pr_{i_{k,j} }^{-1} } { U_{k,j} } }$ Image of Union under Relation: Family of Sets $\displaystyle$ $=$ $\displaystyle \bigcup_{j \mathop \in J} \map {\pr_i} { \bigcap_{k \mathop = 1}^{n_j} \prod_{i' \mathop \in I} V_{i', k, j} }$ Definition of Projection $\displaystyle$ $=$ $\displaystyle \bigcup_{j \mathop \in J} \map {\pr_i} { \prod_{i' \mathop \in I} \bigcap_{k \mathop = 1}^{n_j} V_{i', k, j} }$ Cartesian Product of Intersections: Family of Sets $\displaystyle$ $=$ $\displaystyle \bigcup_{j \mathop \in J} \bigcap_{k \mathop = 1}^{n_j} V_{i,k,j}$ Definition of Projection

As:

$\displaystyle \bigcup_{j \mathop \in J} \bigcap_{k \mathop = 1}^{n_j} V_{i,k,j} \in \tau_i$

it follows that $\pr_i$ is open.

$\blacksquare$