Subset of Euclidean Plane whose Product of Coordinates are Greater Than or Equal to 1 is Closed

Theorem

Let $\struct {\R^2, \tau_d}$ be the real number plane with the usual (Euclidean) topology.

Let $A \subseteq R^2$ be the set of all points defined as:

$A := \set {\tuple {x, y} \in \R^2: x y \ge 1}$

Then $A$ is a closed set in $\struct {\R^2, d}$.

Proof

By definition, $\tau_d$ is the topology induced by the Euclidean metric $d$.

Consider the complement of $A$ in $\R^2$:

$A' := \R^2 \setminus A$

Thus:

$A := \set {\tuple {x, y} \in \R^2: x y < 1}$

Let $a = \tuple {x_a, y_a} \in A^2$.

Let $\epsilon = \size {1 - x_a y_a}$.

Then the open $\epsilon$-ball of $a$ in $\R^2$ lies entirely in $A'$.

As $a$ is arbitrary, it follows that any such $a$ has an open $\epsilon$-ball of $a$ in $\R^2$ which lies entirely in $A'$.

Thus, by definition, $A'$ is open in $\R^2$.

So, also by definition, $A$ is closed in $\R^2$.

$\blacksquare$

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $33$. Special Subsets of the Plane: $1$