Projection from Product Topology is Continuous

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $T_1 = \left({A_1, \tau_1}\right)$ and $T_2 = \left({A_2, \tau_2}\right)$ be topological spaces.

Let $T = T_1 \times T_2$ be the topological product of $T_1$ and $T_2$.

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 continuous.


General Result

Let $\family {T_i}_{i \mathop \in I}$ be a family of topological spaces where $I$ is an arbitrary index set.

Let $\displaystyle T = \prod_{i \mathop \in I} T_i$ be the corresponding product space.

Let $\pr_i : T \to T_i$ be the corresponding projection from $T$ onto $T_i$.


Then $\pr_i$ is continuous for all $i \in I$.


Proof

If $U$ is open in $T_1$ then $\operatorname{pr}_1^{-1} \left({U}\right) = U \times T_2$ is one of the open sets in the basis in the definition of product topology.

Thus $\operatorname{pr}_1$ is continuous.

The same argument applies to $\operatorname{pr}_2$.

$\blacksquare$


Sources