Projection from Product Topology is Continuous

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Theorem

Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.

Let $T = \struct {T_1 \times T_2, \tau}$ be the product space of $T_1$ and $T_2$, where $\tau$ is the Tychonoff topology on $S$.

Let $\pr_1: T \to T_1$ and $\pr_2: T \to T_2$ be the first and second projections from $T$ onto its factors.


Then both $\pr_1$ and $\pr_2$ are are continuous.


General Result

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 continuous for all $i \in I$.


Proof

From Natural Basis of Tychonoff Topology:Finite Product, a basis for $\tau$ is:

$\BB = \set {U \times V: U \in \tau_1, V \in \tau_2}$

et $U$ be open in $T_1$.

Then $\map {\pr_1^{-1} } U = U \times T_2$ is one of the open sets in the basis in the definition of product topology.

Thus $\pr_1$ is continuous.


The same argument can be applied to $\pr_2$.

$\blacksquare$


Also see


Sources