Projection from Product Topology is Open

Theorem
Let $T_1 = \left({S_1, \tau_1}\right)$ and $T_2 = \left({S_2, \tau_2}\right)$ be topological spaces.

Let $T = \left({T_1 \times T_2, \tau}\right)$ be the topological product of $T_1$ and $T_2$, where $\tau$ is the product topology on $S$..

Let $\operatorname{pr}_1: T \to T_1$ and $\operatorname{pr}_2: T \to T_2$ be the first and second projections from $T$ onto its factors.

Then both $\operatorname{pr}_1$ and $\operatorname{pr}_2$ are open.

Proof
First, we prove that $\operatorname{pr}_1$ is open.

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


 * $\displaystyle U = \bigcup_{j \mathop \in J} \bigcap_{k \mathop = 1}^{n_j} \operatorname{pr}_{i_{k,j} }^{-1} \left({ U_{k,j} }\right)$

where $J$ is an arbitrary index set, $n_j \in \N$, $i_{k,j} \in \left\{ {1,2}\right\}$, and $U_{k,j} \in \tau_{i_{k,j} }$.

For all $i \in \left\{ {1, 2}\right\}$, 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 $i \ne i_{k,j}$.

Then, it follows by definition of projection that $\displaystyle \operatorname{pr}_{i_{k,j} }^{-1} \left({ U_{k,j} }\right) = \prod_{i \mathop = 1}^2 V_{i, k, j}$.

Then:

As $\displaystyle \bigcup_{j \mathop \in J} \bigcap_{k \mathop = 1}^{n_j} V_{1,k,j} \in \tau_1$, it follows that $\operatorname{pr}_1$ is open.

The proof for $\operatorname{pr}_2$ is symmetrical.