Projection from Product Topology is Open/General Result

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Let $\left \langle T_i \right \rangle_{i \mathop \in I} = \left \langle \left({S_i, \tau_i}\right) \right \rangle_{i \mathop \in I}$ 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 $\operatorname{pr}_i: S \to S_i$ be the corresponding projection from $S$ onto $S_i$.

Then $\operatorname{pr}_i$ is open for all $i \in I$.


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, $n_j \in \N$, $i_{k,j} \in I$, 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:

\(\displaystyle \operatorname{pr}_i \left({ U }\right)\) \(=\) \(\displaystyle \bigcup_{j \mathop \in J} \operatorname{pr}_i \left({ \bigcap_{k \mathop = 1}^{n_j} \operatorname{pr}_{i_{k,j} }^{-1} \left({ U_{k,j} }\right) }\right)\) Image of Union under Relation: Family of Sets
\(\displaystyle \) \(=\) \(\displaystyle \bigcup_{j \mathop \in J} \operatorname{pr}_i \left({ \bigcap_{k \mathop = 1}^{n_j} \prod_{i' \mathop \in I} V_{i', k, j} }\right)\) Definition of Projection
\(\displaystyle \) \(=\) \(\displaystyle \bigcup_{j \mathop \in J} \operatorname{pr}_i \left({ \prod_{i' \mathop \in I} \bigcap_{k \mathop = 1}^{n_j} V_{i', k, j} }\right)\) 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


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

it follows that $\operatorname{pr}_i$ is open.