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

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Theorem

Let $\left({\R^2, \tau_d}\right)$ be the real Euclidean plane.

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

$A := \left\{{\left({x, y}\right) \in \R^2: x y \ge 1}\right\}$


Then $A$ is a closed set in $\left({\R^2, d}\right)$.


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:

$A := \left\{{\left({x, y}\right) \in \R^2: x y < 1}\right\}$

Let $a = \left({x_a, y_a}\right) \in A^2$.

Let $\epsilon = \left\lvert{1 - x_a y_a}\right\rvert$.

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 $\R^2$.

$\blacksquare$


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