VMOU MT-02 SOLVED ASSIGNMENT | MA/M.SC. MT- 02(Real Analysis and Topology) | July-2024 & January-2025

1. (i). Define base for a topology.
   (ii). Define a $\sigma$$\sigma$sigma\sigma$\sigma$-ring.
   (iii). What is Embedding?
   (iv). What is Bolazano -Weirstrass property?

2. Give an example of a locally connected space which is not connected.
3. Prove that every open interval is a Borel set.
4. Show that the space $L_2$$L_2$L_(2)L_2$L_2$ of a square summable function is a linear space.
5. Show that $L^p$$L^p$L^(p)L^p$L^p$-space is a normed metric space.

6. (i). Show that second countable space is always first countable but converse is not true.
   (ii). Prove that the sequence of functions in $L^P$$L^P$L^(P)L^P$L^P$ space has at most one limit.
7. (i). Show that one point compactification of set of rational numbers $Q$$Q$QQ$Q$ is not Hausdorff.
   (ii) Prove that every second countable regular space is normal space.

Section-A

Very Short Answer Type Questions

Define base for a topology.

Answer:

Definition of Base for a Topology

1. Covering Property: Every open set in the topology can be written as a union of elements of $\mathcal{B}$$\mathcal{B}$B\mathcal{B}$\mathcal{B}$.
2. Intersection Property: If $B_1,B_2\in \mathcal{B}$$B_1,B_2\in \mathcal{B}$B_(1),B_(2)inBB_1, B_2 \in \mathcal{B}$B_1,B_2\in \mathcal{B}$ and $x\in B_1\cap B_2$$x\in B_1\cap B_2$x inB_(1)nnB_(2)x \in B_1 \cap B_2$x\in B_1\cap B_2$, then there exists a $B_3\in \mathcal{B}$$B_3\in \mathcal{B}$B_(3)inBB_3 \in \mathcal{B}$B_3\in \mathcal{B}$ such that $x\in B_3\subseteq B_1\cap B_2$$x\in B_3\subseteq B_1\cap B_2$x inB_(3)subeB_(1)nnB_(2)x \in B_3 \subseteq B_1 \cap B_2$x\in B_3\subseteq B_1\cap B_2$.

Define a $\sigma$$\sigma$sigma\sigma$\sigma$-ring.

Answer:

Definition of a $\sigma$$\sigma$sigma\sigma$\sigma$-Ring

1. Closure under set difference: If $A,B\in \mathcal{R}$$A,B\in \mathcal{R}$A,B inRA, B \in \mathcal{R}$A,B\in \mathcal{R}$, then $A\setminus B\in \mathcal{R}$$A\setminus B\in \mathcal{R}$A\\B inRA \setminus B \in \mathcal{R}$A\setminus B\in \mathcal{R}$.
   $$A\setminus B=\{x\in A:x\notin B\}.$$$$A\setminus B=\{x\in A:x\notin B\}.$$A\\B={x in A:x!in B}.A \setminus B = \{ x \in A : x \notin B \}.$$A\setminus B=\{x\in A:x\notin B\}.$$
2. Closure under countable unions: If $\{A_n{\}}_{n=1}^{\mathrm{\infty }}$$\{A_n{\}}_{n=1}^{\mathrm{\infty }}${A_(n)}_(n=1)^(oo)\{ A_n \}_{n=1}^\infty$\{A_n{\}}_{n=1}^{\mathrm{\infty }}$ is a countable collection of sets in $\mathcal{R}$$\mathcal{R}$R\mathcal{R}$\mathcal{R}$, then their union is also in $\mathcal{R}$$\mathcal{R}$R\mathcal{R}$\mathcal{R}$:
   $$\bigcup _{n=1}^{\mathrm{\infty }}{A}_n\in \mathcal{R}.$$$$\bigcup _{n=1}^{\mathrm{\infty }} A_n\in \mathcal{R}.$$uuu_(n=1)^(oo)A_(n)inR.\bigcup_{n=1}^\infty A_n \in \mathcal{R}.$$\bigcup _{n=1}^{\mathrm{\infty }}{A}_n\in \mathcal{R}.$$
3. Contains the empty set: The empty set $\mathrm{\varnothing }$$\mathrm{\varnothing }$O/\emptyset$\mathrm{\varnothing }$ belongs to $\mathcal{R}$$\mathcal{R}$R\mathcal{R}$\mathcal{R}$.

Remarks:

• A $\sigma$$\sigma$sigma\sigma$\sigma$-ring is not required to contain the universal set $X$$X$XX$X$.
• If a $\sigma$$\sigma$sigma\sigma$\sigma$-ring does contain $X$$X$XX$X$, then it becomes a $\sigma$$\sigma$sigma\sigma$\sigma$-algebra.

What is Embedding?

Answer:

Definition of Embedding

General Definition:

1. $f$$f$ff$f$ is injective (one-to-one), i.e., $f(a_1)=f(a_2)⟹a_1=a_2$$f(a_1)=f(a_2)⟹a_1=a_2$f(a_(1))=f(a_(2))Longrightarrowa_(1)=a_(2)f(a_1) = f(a_2) \implies a_1 = a_2$f(a_1)=f(a_2)⟹a_1=a_2$.
2. $f$$f$ff$f$ preserves the structure of $A$$A$AA$A$ in $B$$B$BB$B$.

What is Bolazano-Weirstrass property?

Answer:

Definition of Bolzano-Weierstrass Property

Every bounded sequence in a Euclidean space (or a metric space) has a convergent subsequence.

Explanation:

1. Bounded Sequence:
   • A sequence $\{x_n\}$$\{x_n\}${x_(n)}\{x_n\}$\{x_n\}$ in a metric space $(X,d)$$(X,d)$(X,d)(X, d)$(X,d)$ is bounded if there exists a real number $M>0$$M>0$M > 0M > 0$M>0$ and a point $p\in X$$p\in X$p in Xp \in X$p\in X$ such that:$$d(x_n,p)\le M\text{for all}n.$$$$d(x_n,p)\le M\text{for all}n.$$d(x_(n),p) <= M quad”for all “n.d(x_n, p) \leq M \quad \text{for all } n.$$d(x_n,p)\le M\text{for all}n.$$
2. Convergent Subsequence:
   • A subsequence $\{x_{n_k}\}$$\{x_{n_k}\}${x_(n_(k))}\{x_{n_k}\}$\{x_{n_k}\}$ of $\{x_n\}$$\{x_n\}${x_(n)}\{x_n\}$\{x_n\}$ is convergent if there exists a point $x\in X$$x\in X$x in Xx \in X$x\in X$ such that:$$\underset{k\to \mathrm{\infty }}{lim}{x}_{n_k}=x,$$$$\underset{k\to \mathrm{\infty }}{lim} x_{n_k}=x,$$lim_(k rarr oo)x_(n_(k))=x,\lim_{k \to \infty} x_{n_k} = x,$$\underset{k\to \mathrm{\infty }}{lim}{x}_{n_k}=x,$$meaning $d(x_{n_k},x)\to 0$$d(x_{n_k},x)\to 0$d(x_(n_(k)),x)rarr0d(x_{n_k}, x) \to 0$d(x_{n_k},x)\to 0$ as $k\to \mathrm{\infty }$$k\to \mathrm{\infty }$k rarr ook \to \infty$k\to \mathrm{\infty }$.

Section-B

Short Answer Questions

Give an example of a locally connected space which is not connected.

Answer:

Example of a Locally Connected Space that is Not Connected

Example:

Verification:

1. Locally Connected:
   • A space is locally connected if every point has a basis of connected open neighborhoods.
   • In $X$$X$XX$X$:
     • For any point $x\in (0,1)$$x\in (0,1)$x in(0,1)x \in (0, 1)$x\in (0,1)$, a neighborhood of the form $(a,b)\cap (0,1)$$(a,b)\cap (0,1)$(a,b)nn(0,1)(a, b) \cap (0, 1)$(a,b)\cap (0,1)$, where $a<x<b$$a<x<b$a < x < ba < x < b$a<x<b$, is connected.
     • Similarly, for $x\in (2,3)$$x\in (2,3)$x in(2,3)x \in (2, 3)$x\in (2,3)$, a neighborhood of the form $(a,b)\cap (2,3)$$(a,b)\cap (2,3)$(a,b)nn(2,3)(a, b) \cap (2, 3)$(a,b)\cap (2,3)$ is connected.
   • Thus, $X$$X$XX$X$ is locally connected.
2. Not Connected:
   • $X$$X$XX$X$ is not connected because it can be written as the union of two disjoint non-empty open sets:$$X=(0,1)\cup (2,3).$$$$X=(0,1)\cup (2,3).$$X=(0,1)uu(2,3).X = (0, 1) \cup (2, 3).$$X=(0,1)\cup (2,3).$$
   • There is no way to connect points from $(0,1)$$(0,1)$(0,1)(0, 1)$(0,1)$ to $(2,3)$$(2,3)$(2,3)(2, 3)$(2,3)$ using a path or a connected subset.

Conclusion:

Prove that every open interval is a Borel set.

Answer:

Statement:

Proof:

Definition of a Borel Set:

Open Intervals are Open Sets:

Open Sets are Borel Sets:

Construction of Open Intervals:

1. Open intervals $(a,b)$$(a,b)$(a,b)(a, b)$(a,b)$ are basic building blocks of the topology on $\mathbb{R}$$\mathbb{R}$R\mathbb{R}$\mathbb{R}$.
2. Any open set in $\mathbb{R}$$\mathbb{R}$R\mathbb{R}$\mathbb{R}$ is a union of (possibly infinitely many) open intervals.
3. Since $\mathcal{B}(\mathbb{R})$$\mathcal{B}(\mathbb{R})$B(R)\mathcal{B}(\mathbb{R})$\mathcal{B}(\mathbb{R})$ is closed under countable unions, any union of open intervals, including the intervals themselves, must also be in $\mathcal{B}(\mathbb{R})$$\mathcal{B}(\mathbb{R})$B(R)\mathcal{B}(\mathbb{R})$\mathcal{B}(\mathbb{R})$.

Conclusion:

Show that the space $L_2$$L_2$L_(2)L_2$L_2$ of square summable functions is a linear space.

Answer:

Statement:

Definition of $L_2$$L_2$L_(2)L_2$L_2$:

Proof: $L_2$$L_2$L_(2)L_2$L_2$ is a Linear Space

1. Closure under Addition:

2. Closure under Scalar Multiplication:

3. Zero Function:

4. Additive Inverses:

Conclusion:

1. Closure under addition,
2. Closure under scalar multiplication,
3. Contains the zero function,
4. Contains additive inverses.

Show that $L^p$$L^p$L^(p)L^p$L^p$-space is a normed metric space.

Answer:

Statement:

Definition of $L^p$$L^p$L^(p)L^p$L^p$-Space:

To Prove:

1. $L^p$$L^p$L^(p)L^p$L^p$-space is a normed space.
2. The norm induces a metric, making $L^p$$L^p$L^(p)L^p$L^p$ a metric space.

Proof:

1. $L^p$$L^p$L^(p)L^p$L^p$-Space is a Normed Space

1. Non-negativity:
   $$\Vert f{\Vert }_p={({\int }_X|f(x){|}^{p}d\mu )}^{1/p}\ge 0,$$$$\Vert f{\Vert }_p={\left({\int }_X |f(x){|}^{p}d\mu \right)}^{1/p}\ge 0,$$||f||_(p)=(int _(X)|f(x)|^(p)d mu)^(1//p) >= 0,\|f\|_p = \left( \int_X |f(x)|^p \, d\mu \right)^{1/p} \geq 0,$$\Vert f{\Vert }_p={({\int }_X|f(x){|}^{p}d\mu )}^{1/p}\ge 0,$$
   and $\Vert f{\Vert }_p=0$$\Vert f{\Vert }_p=0$||f||_(p)=0\|f\|_p = 0$\Vert f{\Vert }_p=0$ if and only if $f(x)=0$$f(x)=0$f(x)=0f(x) = 0$f(x)=0$ almost everywhere (a.e.).
   • This follows from the non-negativity of $|f(x){|}^p$$|f(x){|}^p$|f(x)|^(p)|f(x)|^p$|f(x){|}^p$ and the fact that the integral of a non-negative function is zero only when the function is zero a.e.
2. Scalar Multiplication:
   For $c\in \mathbb{R}$$c\in \mathbb{R}$c inRc \in \mathbb{R}$c\in \mathbb{R}$ (or $\mathbb{C}$$\mathbb{C}$C\mathbb{C}$\mathbb{C}$) and $f\in L^p$$f\in L^p$f inL^(p)f \in L^p$f\in L^p$:
   $$\Vert cf{\Vert }_p={({\int }_X|cf(x){|}^{p}d\mu )}^{1/p}=|c|{({\int }_X|f(x){|}^{p}d\mu )}^{1/p}=|c|\Vert f{\Vert }_p.$$$$\Vert cf{\Vert }_p={\left({\int }_X |cf(x){|}^{p}d\mu \right)}^{1/p}=|c|{\left({\int }_X |f(x){|}^{p}d\mu \right)}^{1/p}=|c|\Vert f{\Vert }_p.$$||cf||_(p)=(int _(X)|cf(x)|^(p)d mu)^(1//p)=|c|(int _(X)|f(x)|^(p)d mu)^(1//p)=|c|||f||_(p).\|cf\|_p = \left( \int_X |cf(x)|^p \, d\mu \right)^{1/p} = |c| \left( \int_X |f(x)|^p \, d\mu \right)^{1/p} = |c| \|f\|_p.$$\Vert cf{\Vert }_p={({\int }_X|cf(x){|}^{p}d\mu )}^{1/p}=|c|{({\int }_X|f(x){|}^{p}d\mu )}^{1/p}=|c|\Vert f{\Vert }_p.$$
3. Triangle Inequality (Minkowski’s Inequality):
   For $f,g\in L^p$$f,g\in L^p$f,g inL^(p)f, g \in L^p$f,g\in L^p$:
   $$\Vert f+g{\Vert }_p={({\int }_X|f(x)+g(x){|}^{p}d\mu )}^{1/p}.$$$$\Vert f+g{\Vert }_p={\left({\int }_X |f(x)+g(x){|}^{p}d\mu \right)}^{1/p}.$$||f+g||_(p)=(int _(X)|f(x)+g(x)|^(p)d mu)^(1//p).\|f + g\|_p = \left( \int_X |f(x) + g(x)|^p \, d\mu \right)^{1/p}.$$\Vert f+g{\Vert }_p={({\int }_X|f(x)+g(x){|}^{p}d\mu )}^{1/p}.$$
   Using Minkowski’s inequality:
   $${({\int }_X|f(x)+g(x){|}^{p}d\mu )}^{1/p}\le {({\int }_X|f(x){|}^{p}d\mu )}^{1/p}+{({\int }_X|g(x){|}^{p}d\mu )}^{1/p}.$$$${\left({\int }_X |f(x)+g(x){|}^{p}d\mu \right)}^{1/p}\le {\left({\int }_X |f(x){|}^{p}d\mu \right)}^{1/p}+{\left({\int }_X |g(x){|}^{p}d\mu \right)}^{1/p}.$$(int _(X)|f(x)+g(x)|^(p)d mu)^(1//p) <= (int _(X)|f(x)|^(p)d mu)^(1//p)+(int _(X)|g(x)|^(p)d mu)^(1//p).\left( \int_X |f(x) + g(x)|^p \, d\mu \right)^{1/p} \leq \left( \int_X |f(x)|^p \, d\mu \right)^{1/p} + \left( \int_X |g(x)|^p \, d\mu \right)^{1/p}.