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Q3E

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Found in: Page 437

### Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

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

1. Open
2. Closed
3. Neither open nor closed
4. Closed
5. Closed
See the step by step solution

## Step 1: To check every point is an interior point

(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.

## Step 2: To check every point is a boundary point

(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.

## Step 3: Use the proper definition of open and 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.

## Step 4: To check every point is a boundary point

(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.

## Step 5: To check every point is a boundary point

(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.

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