Real Analysis MSQs and MSQs : Point Set Topology - I

1. Let \(G\) be the set of all irrational numbers. The interior and the closure of \(G\) are denoted by \(G^{\circ}\) and \(\bar{G}\), respectively. Then

(a.) \(G^{0}=\emptyset\), \(\bar{G}=G\)
(b.) \(G^{0}=\mathbb{R}\), \(\bar{G}=\mathbb{R}\)
(c.) \(G^{0}=\emptyset\), \(\bar{G}=\mathbb{R}\)
(d.) \(G^{0}=G\), \(\bar{G}=\mathbb{R}\)

Explanation for Question 1:
For any \(x \in (\mathbb{R} \setminus \mathbb{Q})^\circ \), it implies that there exists an \(r > 0\) such that \(B(x, r) \subseteq \mathbb{R} \setminus \mathbb{Q}\). This means that \(B(x, r) \cap \mathbb{Q} = \varnothing\), which is a contradiction. Therefore, \((\mathbb{R} \setminus \mathbb{Q})^\circ = \varnothing\).
If \(x \in \mathbb{Q}\), then the set \(\{y \mid y = x + \frac{1}{\sqrt{n}}, n \in \mathbb{N}\}\) has a limit point at \(x\). This implies that \(\mathbb{Q} \subseteq \overline{\mathbb{R} \setminus \mathbb{Q}}\), and since \(\mathbb{R} \setminus \mathbb{Q} \subseteq \overline{\mathbb{R} \setminus \mathbb{Q}}\), we conclude that \(\overline{\mathbb{R} \setminus \mathbb{Q}} = \mathbb{R}\).

2. The set \(\left\{x \in \mathbb{R}: x^{2}>x\right\}\) is the same as

(a.) the interval \((0,0)\)
(b.) the complement of the interval \((0,1)\)
(c.) the complement of the interval \([0,1]\)
(d.) the interval \([0,1]\)

Explanation for Question 2:
Let \(A = \{x \in \mathbb{R} \mid x^2 > x\}\), and \(a \in A\). This means \(a^2 - a > 0\), and we can simplify to \(a > 0\) and \(a - 1 > 0\), or \(a < 0\) and \(a - 1 < 0\). In other words, \(a\) belongs to either \((1, \infty)\) or \((-\infty, 0)\). So, \(A = \mathbb{R} - [0, 1]\).

3. The set of all boundary points of \(\mathbb{Q}\) in \(\mathbb{R}\) is

(a.) \(\mathbb{R}\)
(b.) \(\mathbb{R} \setminus \mathbb{Q}\)
(c.) \(\mathbb{Q}\)
(d.) Empty set

Explanation for Question 3:
A point \(a\) is considered a boundary point of \(\mathbb{Q}\) if for every \(r > 0\), both \(B(a, r) \cap \mathbb{Q} \neq \varnothing\) and \(B(a, r) \cap \mathbb{Q}^c \neq \varnothing\). This condition holds true for all \(a \in \mathbb{Q\). Therefore, \(\mathbb{Q}\) is the set of all boundary points of \(\mathbb{Q}\).

4. Lim superior \(\left\{\frac{(-1)^{n}}{2^{n}}: n=1,2, \ldots\right\}\) is

(a.) 0
(b.) \(\frac{1}{4}\)
(c.) 1
(d.) -1

Explanation for Question 4:
Let \(A = \left\{-\frac{1}{2}, \frac{1}{4}, -\frac{1}{8}, \frac{1}{16}, \cdots\right\}\). If we calculate the supremum of each subset, it becomes evident that \(\limsup A = 0\). This is due to the fact that if a sequence has a limit, the \(\limsup A\) is equal to the limit.

5. If \(Y=\left\{\frac{x}{1+|x|} \mid x \in \mathbb{R}\right\}\), then the set of all limit points of \(Y\) is

(a.) \((-1,1)\)
(b.) \((-1,1]\)
(c.) \([0,1]\)
(d.) \([-1,1]\)

Explanation for Question 5:
The function \(\frac{x}{1 + |x|}\) is continuous on \(\mathbb{R}\) with limits as \(x \rightarrow \infty\) and \(x \rightarrow -\infty\). These limits are \(1\) and \(-1\), respectively. Therefore, \(-1 < \frac{x}{1 + |x|} < 1\). The set \(Y = (1, -1)\), and the set of limit points of \(Y\) is \([-1, 1]\).

6. The set \(\left\{\frac{1}{n} \sin \frac{1}{n}: n \in \mathbb{N}\right\}\) has

(a.) One limit point and it is 0
(b.) One limit point and it is 1
(c.) One limit point and it is -1
(d.) Three limit points and these are -1, 0, and 1

Explanation for Question 6:
For the inequality \(-\frac{1}{n} \leq \frac{1}{n} \sin \frac{1}{n} \leq \frac{1}{n}\), as \(n\) approaches infinity, \(\frac{1}{n} \sin \frac{1}{n}\) approaches \(0\).

7. The set \(U=\left\{x \in \mathbb{R} \mid \sin x=\frac{1}{2}\right\}\) is

(a.) open
(b.) closed
(c.) both open and closed
(d.) neither open nor closed

Explanation for Question 7:
The set \(U = \{x \in \mathbb{R} \mid \sin x = \frac{1}{2}\}\) can be described as \(U = \{x \in \mathbb{R} \mid x = \sin^{-1}\left(\frac{1}{2}\right) + 2n\pi, n \in \mathbb{N}\}\), where \(a = \sin^{-1}\left(\frac{1}{2}\right)\). Thus, \(U\) is a set of discrete points and is closed.

8. Let \(S\) be the set of all numbers of the form \(\sum_{k=0}^{\infty} \frac{a_{k}}{3^{k}} ; a_{k}=0\) or 2. Which of the following is true?

(a.) \(S\) is countable
(b.) \(S\) is dense in \([0,1]\)
(c.) \(S\) is closed in \([0,1]\)
(d.) \(S=[0,1]\)

Explanation for Question 8:
The set \(S = \left\{x = \sum_{k=0}^{\infty} \frac{a_{k}}{3^{k}} \mid a_{k} = 0 \text{ or } 2\}\) is the Cantor set.

9. Consider the following subsets of \(\mathbb{R}\): \(E=\left\{\frac{n}{n+1}: n \in \mathbb{N}\right\}\), \(F=\left\{\frac{1}{1-x}: 0 \leq x<1\right\}\) Then

(a.) Both \(E\) and \(F\) are closed
(b.) \(E\) is closed and \(F\) is NOT closed
(c.) \(E\) is NOT closed and \(F\) is closed
(d.) Neither \(E\) nor \(F\) is closed

Explanation for Question 9:
- \(1 \in E^{\prime}\), but \(1 \notin E\), which means \(E\) is not closed. - The function \(\frac{1}{1-x}\) is continuous, and at \(x = 0\), \(\frac{1}{1-x} = 1\). - As \(x\) approaches \(1\), \(\frac{1}{1-x}\) approaches \(\infty\). - Thus, \(F = [1, \infty)\), which is a closed set in \(\mathbb{R}\).

10. How many subsets \(A\) of \(\{1,2,3,4\}\) are there such that \((A \cup\{1,2\})-(A \cap\{1,2\})\) has exactly one point?

(a.) 1
(b.) 2
(c.) 3
(d.) 4

Explanation for Question 10:
There are four possibilities for \(A\): \(\{1\}, \{2\}, \{1, 2, 3\}, \{1, 2, 4\}\). This implies that the number of possible sets \(A\) is greater than or equal to \(4\).

11. Let \(A=\{m+n \sqrt{2}: m, n \in \mathbb{Z}\}\), where \(\mathbb{Z}\) stands for the set of all integers. Then

(a.) \(A\) is dense in \(\mathbb{R}\)
(b.) \(A\) has only countably many limit points in \(\mathbb{R}\)
(c.) \(A\) has no limit point in \(\mathbb{R}\)
(d.) Only irrational numbers can be limit points of \(A\)

Explanation for Question 11:
For \(A_{\alpha} = \{m + n\alpha \mid m, n \in \mathbb{Z}\}\), \(A_{\alpha}\) is dense for any irrational \(\alpha\) .

12. Let \(X=\left\{\frac{1}{n}: n \in \mathbb{Z}, n \geq 1\right\}\) and let \(\bar{X}\) be its closure. Then

(a.) \(\bar{X} \backslash X\) is a single point
(b.) \(\bar{X} \backslash X\) is open in \(\mathbb{R}\)
(c.) \(\bar{X} \backslash X\) is infinite but not open in \(\mathbb{R}\)
(d.) \(\bar{X} \backslash X=\phi\)

Explanation for Question 12:
The closure of \(X\), denoted \(\bar{X}\), is equal to the union of \(X\) and the singleton set \(\{0\}\).