Answer :
Let's solve each part step-by-step.
### (i) Verification that [tex]\( R_{(a, b)} \subset D \)[/tex]
Let [tex]\((a, b)\)[/tex] be any point in the disc [tex]\(D = \{(x, y): x^2 + y^2 < 1\}\)[/tex]. Define [tex]\(r = \sqrt{a^2 + b^2}\)[/tex].
The open rectangle [tex]\(R_{(a, b)}\)[/tex] has vertices at the points [tex]\((a \pm \frac{1-r}{8}, b \pm \frac{1-r}{8})\)[/tex]. We need to show that every point in this rectangle is within the disc [tex]\(D\)[/tex].
Since [tex]\((a, b)\)[/tex] is in disc [tex]\(D\)[/tex], we know that [tex]\(r = \sqrt{a^2 + b^2} < 1\)[/tex].
Consider the maximum distance any point [tex]\((x, y)\)[/tex] in [tex]\(R_{(a, b)}\)[/tex] could be from [tex]\((a, b)\)[/tex]. The maximum deviation in the [tex]\(x\)[/tex] direction is [tex]\(\frac{1 - r}{8}\)[/tex] and similarly for the [tex]\(y\)[/tex] direction. Hence, the farthest distance from [tex]\((a, b)\)[/tex] to any [tex]\((x, y) \in R_{(a, b)}\)[/tex] is:
[tex]\[ \sqrt{\left(\frac{1-r}{8}\right)^2 + \left(\frac{1-r}{8}\right)^2} = \sqrt{2} \cdot \frac{1-r}{8} = \frac{\sqrt{2} \cdot (1-r)}{8} \][/tex]
Given that [tex]\(r < 1\)[/tex], it is evident that:
[tex]\[ \frac{\sqrt{2} \cdot (1-r)}{8} < \sqrt{2}/8 < \sqrt{2}/4 \][/tex]
[tex]\[ \sqrt{2}/4 < 1 \][/tex]
Thus, any point [tex]\((x, y)\)[/tex] in [tex]\(R_{(a, b)}\)[/tex] will still satisfy [tex]\(x^2 + y^2 < 1\)[/tex].
Therefore, [tex]\(R_{(a, b)} \subset D\)[/tex].
### (ii) Showing that [tex]\( D = \bigcup_{(a, b) \in D} R_{(a, b)} \)[/tex]
From part (i), we verified that for any point [tex]\((a, b) \in D\)[/tex], the rectangle [tex]\(R_{(a, b)} \subset D\)[/tex]. This means that every point in [tex]\(D\)[/tex] lies within some [tex]\(R_{(a, b)}\)[/tex].
Thus, the disc [tex]\(D\)[/tex] can be expressed as the union of all such rectangles:
[tex]\[ D = \bigcup_{(a, b) \in D} R_{(a, b)} \][/tex]
### (iii) Deducing that [tex]\( D \)[/tex] is an open set in [tex]\( R^2 \)[/tex]
An open set in [tex]\(\mathbb{R}^2\)[/tex] is defined as a set where for every point in the set, there exists an [tex]\(\epsilon\)[/tex]-neighborhood contained entirely within that set.
From (ii), for any point [tex]\((a, b)\)[/tex] in [tex]\(D\)[/tex], the corresponding rectangle [tex]\(R_{(a, b)}\)[/tex]—which is an [tex]\(\epsilon\)[/tex]-neighborhood around [tex]\((a, b)\)[/tex]—is entirely contained within [tex]\(D\)[/tex].
Since [tex]\(D\)[/tex] can be written as the union of such open rectangles, each surrounding its point completely within [tex]\(D\)[/tex], [tex]\(D\)[/tex] is itself an open set in [tex]\(R^2\)[/tex].
### (iv) Every open disc [tex]\(\left\{(x, y): (x - a)^2 + (y - b)^2 < c^2, \ a, b, c \in \mathbb{R} \right\}\)[/tex] is open in [tex]\( R^2 \)[/tex].
For an arbitrary disc centered at [tex]\((a, b)\)[/tex] with radius [tex]\(c\)[/tex]:
[tex]\[ D' = \{(x, y) \in \mathbb{R}^2: (x - a)^2 + (y - b)^2 < c^2\} \][/tex]
We can use a similar argument as above for [tex]\(D = \{(x, y): x^2 + y^2 < 1\}\)[/tex].
Given any point [tex]\((x_0, y_0) \in D'\)[/tex], the distance from the center [tex]\((a, b)\)[/tex] to the point [tex]\((x_0, y_0)\)[/tex] is less than [tex]\(c\)[/tex]:
[tex]\[ \sqrt{(x_0 - a)^2 + (y_0 - b)^2} < c \][/tex]
Let [tex]\( r_0 = \sqrt{(x_0 - a)^2 + (y_0 - b)^2} \)[/tex] and construct a rectangle [tex]\( R_{(x_0, y_0)} \)[/tex] with vertices at:
[tex]\[ (x_0 \pm \frac{c - r_0}{8}, y_0 \pm \frac{c - r_0}{8}) \][/tex]
This rectangle [tex]\( R_{(x_0, y_0)} \)[/tex] is entirely within [tex]\( D' \)[/tex] by the same reasoning used in (i).
Since, for every point in [tex]\( D' \)[/tex], there is an open rectangle around it fully contained within [tex]\( D' \)[/tex]:
[tex]\[ D' = \bigcup_{(x_0, y_0) \in D'} R_{(x_0, y_0)} \][/tex]
[tex]\(\Rightarrow D' \)[/tex] is an open set in [tex]\( \mathbb{R}^2\)[/tex].
### (i) Verification that [tex]\( R_{(a, b)} \subset D \)[/tex]
Let [tex]\((a, b)\)[/tex] be any point in the disc [tex]\(D = \{(x, y): x^2 + y^2 < 1\}\)[/tex]. Define [tex]\(r = \sqrt{a^2 + b^2}\)[/tex].
The open rectangle [tex]\(R_{(a, b)}\)[/tex] has vertices at the points [tex]\((a \pm \frac{1-r}{8}, b \pm \frac{1-r}{8})\)[/tex]. We need to show that every point in this rectangle is within the disc [tex]\(D\)[/tex].
Since [tex]\((a, b)\)[/tex] is in disc [tex]\(D\)[/tex], we know that [tex]\(r = \sqrt{a^2 + b^2} < 1\)[/tex].
Consider the maximum distance any point [tex]\((x, y)\)[/tex] in [tex]\(R_{(a, b)}\)[/tex] could be from [tex]\((a, b)\)[/tex]. The maximum deviation in the [tex]\(x\)[/tex] direction is [tex]\(\frac{1 - r}{8}\)[/tex] and similarly for the [tex]\(y\)[/tex] direction. Hence, the farthest distance from [tex]\((a, b)\)[/tex] to any [tex]\((x, y) \in R_{(a, b)}\)[/tex] is:
[tex]\[ \sqrt{\left(\frac{1-r}{8}\right)^2 + \left(\frac{1-r}{8}\right)^2} = \sqrt{2} \cdot \frac{1-r}{8} = \frac{\sqrt{2} \cdot (1-r)}{8} \][/tex]
Given that [tex]\(r < 1\)[/tex], it is evident that:
[tex]\[ \frac{\sqrt{2} \cdot (1-r)}{8} < \sqrt{2}/8 < \sqrt{2}/4 \][/tex]
[tex]\[ \sqrt{2}/4 < 1 \][/tex]
Thus, any point [tex]\((x, y)\)[/tex] in [tex]\(R_{(a, b)}\)[/tex] will still satisfy [tex]\(x^2 + y^2 < 1\)[/tex].
Therefore, [tex]\(R_{(a, b)} \subset D\)[/tex].
### (ii) Showing that [tex]\( D = \bigcup_{(a, b) \in D} R_{(a, b)} \)[/tex]
From part (i), we verified that for any point [tex]\((a, b) \in D\)[/tex], the rectangle [tex]\(R_{(a, b)} \subset D\)[/tex]. This means that every point in [tex]\(D\)[/tex] lies within some [tex]\(R_{(a, b)}\)[/tex].
Thus, the disc [tex]\(D\)[/tex] can be expressed as the union of all such rectangles:
[tex]\[ D = \bigcup_{(a, b) \in D} R_{(a, b)} \][/tex]
### (iii) Deducing that [tex]\( D \)[/tex] is an open set in [tex]\( R^2 \)[/tex]
An open set in [tex]\(\mathbb{R}^2\)[/tex] is defined as a set where for every point in the set, there exists an [tex]\(\epsilon\)[/tex]-neighborhood contained entirely within that set.
From (ii), for any point [tex]\((a, b)\)[/tex] in [tex]\(D\)[/tex], the corresponding rectangle [tex]\(R_{(a, b)}\)[/tex]—which is an [tex]\(\epsilon\)[/tex]-neighborhood around [tex]\((a, b)\)[/tex]—is entirely contained within [tex]\(D\)[/tex].
Since [tex]\(D\)[/tex] can be written as the union of such open rectangles, each surrounding its point completely within [tex]\(D\)[/tex], [tex]\(D\)[/tex] is itself an open set in [tex]\(R^2\)[/tex].
### (iv) Every open disc [tex]\(\left\{(x, y): (x - a)^2 + (y - b)^2 < c^2, \ a, b, c \in \mathbb{R} \right\}\)[/tex] is open in [tex]\( R^2 \)[/tex].
For an arbitrary disc centered at [tex]\((a, b)\)[/tex] with radius [tex]\(c\)[/tex]:
[tex]\[ D' = \{(x, y) \in \mathbb{R}^2: (x - a)^2 + (y - b)^2 < c^2\} \][/tex]
We can use a similar argument as above for [tex]\(D = \{(x, y): x^2 + y^2 < 1\}\)[/tex].
Given any point [tex]\((x_0, y_0) \in D'\)[/tex], the distance from the center [tex]\((a, b)\)[/tex] to the point [tex]\((x_0, y_0)\)[/tex] is less than [tex]\(c\)[/tex]:
[tex]\[ \sqrt{(x_0 - a)^2 + (y_0 - b)^2} < c \][/tex]
Let [tex]\( r_0 = \sqrt{(x_0 - a)^2 + (y_0 - b)^2} \)[/tex] and construct a rectangle [tex]\( R_{(x_0, y_0)} \)[/tex] with vertices at:
[tex]\[ (x_0 \pm \frac{c - r_0}{8}, y_0 \pm \frac{c - r_0}{8}) \][/tex]
This rectangle [tex]\( R_{(x_0, y_0)} \)[/tex] is entirely within [tex]\( D' \)[/tex] by the same reasoning used in (i).
Since, for every point in [tex]\( D' \)[/tex], there is an open rectangle around it fully contained within [tex]\( D' \)[/tex]:
[tex]\[ D' = \bigcup_{(x_0, y_0) \in D'} R_{(x_0, y_0)} \][/tex]
[tex]\(\Rightarrow D' \)[/tex] is an open set in [tex]\( \mathbb{R}^2\)[/tex].
