Projection from Product Topology is Open/General Result/Proof

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:

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.