Question: In Exercise 3, determine whether each set is open or closed or neither open nor closed.

3. a. \(\left\{ {\left( {x,y} \right):y > {\bf{0}}} \right\}\)

b. \(\left\{ {\left( {x,y} \right):x = {\bf{2}}\,\,\,and\,\,{\bf{1}} \le y \le {\bf{3}}} \right\}\)

c. \(\left\{ {\left( {x,y} \right):x = {\bf{2}}\,\,\,and\,\,{\bf{1}} < y < {\bf{3}}} \right\}\)

d. \(\left\{ {\left( {x,y} \right):xy = {\bf{1}}\,\,\,and\,\,x > {\bf{0}}} \right\}\)

e. \(\left\{ {\left( {x,y} \right):xy \ge {\bf{1}}\,\,\,and\,\,x > {\bf{0}}} \right\}\)

(a)

Consider the set \(\left( {x,y} \right) \in \left\{ {\left( {x,y} \right):y > 0} \right\}\). Choose \(\delta = y\). To check that \(B\left( {\left( {x,y} \right),\delta } \right)\) is wholly contained in \(\left\{ {\left( {x,y} \right):y > 0} \right\}\) or not.

Let \((a,b) \in B\left( {\left( {x,y} \right),\delta } \right)\). Then

\(\begin{array}{l}\sqrt {{{\left( {a - x} \right)}^2} + {{\left( {b - y} \right)}^2}} < \delta \\\sqrt {{{\left( {a - x} \right)}^2} + {{\left( {b - y} \right)}^2}} < y\\\,\,\,\,{\left( {a - x} \right)^2} + {\left( {b - y} \right)^2} < {y^2}\\ \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,0 < {\left( {b - y} \right)^2} < {y^2}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 < b - y < y\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 < b < 2y\end{array}\)

This implies \(\left( {a,b} \right) \in \left\{ {\left( {x,y} \right):y > 0} \right\}\). Hence, \(B\left( {\left( {x,y} \right),\delta } \right)\) is wholly contained in \(\left\{ {\left( {x,y} \right):y > 0} \right\}\).

Thus \(\left\{ {\left( {x,y} \right):y > 0} \right\}\) is open.

(b)

Consider \(\left( {2,3} \right) \in \left\{ {\left( {x,y} \right):x = 2\,\,{\rm{and}}\,\,1 \le y \le 3} \right\}\). Choose \(\delta = \frac{1}{n}\), \(n \in \mathbb{N}\).

Let \((2 - \varepsilon ,3 - \varepsilon ) \notin \left\{ {\left( {x,y} \right):x = 2\,\,{\rm{and}}\,\,1 \le y \le 3} \right\}\) where \(\varepsilon = \frac{1}{k}\) when \(k \to \infty \). Note that

\(\begin{array}{c}\sqrt {{{\left( {2 - \varepsilon - 2} \right)}^2} + {{\left( {3 - \varepsilon - 3} \right)}^2}} < \sqrt {{\varepsilon ^2} + {\varepsilon ^2}} \\ < \sqrt {2{\varepsilon ^2}} \\ < \sqrt 2 \varepsilon \\ < \delta \end{array}\)

This implies \((2 - \varepsilon ,3 - \varepsilon ) \in B\left( {\left( {2,3} \right),\delta } \right)\). Thus \(\left( {2,3} \right)\) is a boundary point of \(\left\{ {\left( {x,y} \right):x = 2\,\,{\rm{and}}\,\,1 \le y \le 3} \right\}\).

Hence \(\left\{ {\left( {x,y} \right):x = 2\,\,{\rm{and}}\,\,1 \le y \le 3} \right\}\) is closed.

(c)

Consider \(\left( {2,1 + \varepsilon } \right) \in \left\{ {\left( {x,y} \right):x = 2\,\,{\rm{and}}\,\,1 < y < 3} \right\}\) where \(\varepsilon = \frac{1}{k}\) when \(k \to \infty \). Choose \(\delta = \frac{1}{n}\), \(n \in \mathbb{N}\). To check that \(B\left( {\left( {2,1 + \varepsilon } \right),\delta } \right)\) is completely contained in \(\left\{ {\left( {x,y} \right):x = 2\,\,{\rm{and}}\,\,1 < y < 3} \right\}\) or not.

Let \(\left( {2 + \varepsilon ,1 + \varepsilon } \right) \in {\mathbb{R}^2}\). Then

\(\begin{array}{c}\left\| {\left( {2 + \varepsilon ,1 + \varepsilon } \right) - \left( {2,1 + \varepsilon } \right)} \right\| = \sqrt {{{\left( {2 + \varepsilon - 2} \right)}^2} + {{\left( {1 + \varepsilon - \left( {1 + \varepsilon } \right)} \right)}^2}} \\ < \sqrt {{\varepsilon ^2} + 0} \\ < \varepsilon \\ < \delta \end{array}\)

This implies \(\left( {2 + \varepsilon ,1 + \varepsilon } \right) \in B\left( {\left( {2,1 + \varepsilon } \right),\delta } \right)\). But \(\left( {2 + \varepsilon ,1 + \varepsilon } \right) \notin \left\{ {\left( {x,y} \right):x = 2\,\,{\rm{and}}\,\,1 < y < 3} \right\}\).

Thus \(\left\{ {\left( {x,y} \right):x = 2\,\,{\rm{and}}\,\,1 < y < 3} \right\}\) is not open.

Consider \(\left( {2,1} \right) \notin \left\{ {\left( {x,y} \right):x = 2\,\,{\rm{and}}\,\,1 < y < 3} \right\}\). Choose \(\delta = \frac{1}{n}\), \(n \in \mathbb{N}\). To check that \(B\left( {\left( {2,1} \right),\delta } \right)\) contains a point of \(\left\{ {\left( {x,y} \right):x = 2\,\,{\rm{and}}\,\,1 < y < 3} \right\}\) or not.

Let \(\left( {2,1 + \varepsilon } \right) \in {\mathbb{R}^2}\) where \(\varepsilon = \frac{1}{k}\) when \(k \to \infty \). Then

\(\begin{array}{c}\left\| {\left( {2,1 + \varepsilon } \right) - \left( {2,1} \right)} \right\| = \sqrt {{{\left( {2 - 2} \right)}^2} + {{\left( {1 + \varepsilon - 1} \right)}^2}} \\ < \sqrt {0 + {\varepsilon ^2}} \\ < \varepsilon \\ < \delta \end{array}\)

This implies \(\left( {2,1 + \varepsilon } \right) \in B\left( {\left( {2,1} \right),\delta } \right)\) and also in \(\left\{ {\left( {x,y} \right):x = 2\,\,{\rm{and}}\,\,1 < y < 3} \right\}\).Hence \(\left( {2,1} \right)\) is a boundary point of \(\left\{ {\left( {x,y} \right):x = 2\,\,{\rm{and}}\,\,1 < y < 3} \right\}\). But \(\left( {2,1} \right) \notin \left\{ {\left( {x,y} \right):x = 2\,\,{\rm{and}}\,\,1 < y < 3} \right\}\).

Thus \(\left\{ {\left( {x,y} \right):x = 2\,\,{\rm{and}}\,\,1 < y < 3} \right\}\) is not closed.

Therefore \(\left\{ {\left( {x,y} \right):x = 2\,\,{\rm{and}}\,\,1 < y < 3} \right\}\) is neither open nor closed.

(d)

The given set \(\left\{ {\left( {x,y} \right):xy = 1\,\,{\rm{and}}\,\,x > 0} \right\}\) can be written as \(\left\{ {\left( {x,y} \right):\,\,x > 0\,{\rm{and}}\,y > 0\,{\rm{such}}\,{\rm{that }}xy = 1} \right\} = \left\{ {\left( {\frac{p}{q},\frac{q}{p}} \right):\,\,p,q \in \mathbb{N}} \right\}\).

Let \(\left( {\frac{p}{q},\frac{q}{p}} \right) \in \left\{ {\left( {\frac{p}{q},\frac{q}{p}} \right):\,\,p,q \in \mathbb{N}} \right\}\). Choose \(\delta = \frac{1}{n}\), \(n \in \mathbb{N}\). We know that every\(B\left( {\left( {\frac{p}{q},\frac{q}{p}} \right),\delta } \right)\) contains at least an irrational point.

This implies every point in \(\left( {\frac{p}{q},\frac{q}{p}} \right) \in \left\{ {\left( {\frac{p}{q},\frac{q}{p}} \right):\,\,p,q \in \mathbb{N}} \right\}\) is the boundary point. Hence \(\left( {\frac{p}{q},\frac{q}{p}} \right) \in \left\{ {\left( {\frac{p}{q},\frac{q}{p}} \right):\,\,p,q \in \mathbb{N}} \right\}\) is closed.

That is \(\left\{ {\left( {x,y} \right):xy = 1\,\,{\rm{and}}\,\,x > 0} \right\}\) closed.

(e)

Let \(\left( {x,\frac{1}{x}} \right) \in \left\{ {\left( {x,y} \right):xy \ge 1\,\,{\rm{and}}\,\,x > 0} \right\}\). Choose \(\delta = \frac{1}{n}\), \(n \in \mathbb{N}\).

Let \((x,\frac{1}{x} - \varepsilon ) \notin \left\{ {\left( {x,y} \right):xy \ge 1\,\,{\rm{and}}\,\,x > 0} \right\}\) where \(\varepsilon = \frac{1}{k}\) when \(k \to \infty \). Note that:

\(\begin{array}{c}\sqrt {{{\left( {x - x} \right)}^2} + {{\left( {\frac{1}{x} - \varepsilon - \frac{1}{x}} \right)}^2}} < \sqrt {{\varepsilon ^2}} \\ < \varepsilon \\ < \delta \end{array}\)

This implies \((x,\frac{1}{x} - \varepsilon ) \in B\left( {\left( {x,\frac{1}{x}} \right),\delta } \right)\). Thus \(\left( {x,\frac{1}{x}} \right)\) is a boundary point of\(\left\{ {\left( {x,y} \right):xy \ge 1\,\,{\rm{and}}\,\,x > 0} \right\}\).

Hence, \(\left\{ {\left( {x,y} \right):xy \ge 1\,\,{\rm{and}}\,\,x > 0} \right\}\) is closed.