Catesian Product of Open Real Intervals is Open in Real Number Plane

Theorem

Let $\struct {\R^2, d}$ denote the real number plane with the Euclidean metric.

Let $\openint a b$ and $\openint c d$ be open real intervals.

Then their Cartesian product:

$S = \openint a b \times \openint c d$

is an open set of $\struct {\R^2, d}$.

Proof

Let $P = \tuple {x, y} \in S$.

Then by definition:

$a < x < b$

and:

$c < y < d$

Let $\epsilon = \min \set {x - a, b - x, y - c, d - y}$

By definition, $\epsilon > 0$.

Consider the open ball $\map {B_\epsilon} P$.

Let $Q = \tuple {x_0, y_0} \in \map {B_\epsilon} P$ be an arbitrary point in $\map {B_\epsilon} P$.

By definition of $\map {B_\epsilon} P$:

$\size {x - x_0} < \epsilon$

and:

$\size {y - y_0} < \epsilon$

Hence we have:

$\size {x - x_0} < x - a$

$\size {x - x_0} < b - x$

$\size {y - y_0} < y - c$

$\size {y - y_0} < d - y$

and so:

$Q \in S$

Thus we have that:

$\map {B_\epsilon} P \subseteq S$

Hence the result by definition of open set.

$\blacksquare$

Sources

• 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $2$: Continuity generalized: metric spaces: Exercise $2.6: 11$