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}$.

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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.

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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 }$.

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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:

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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.

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Conclusion:

Prove that every open interval is a Borel set.

Answer:

Statement:

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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})$.

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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:

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

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

1. Closure under Addition:

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2. Closure under Scalar Multiplication:

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3. Zero Function:

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4. Additive Inverses:

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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:

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

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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.

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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}.$${({\int }_X|f(x)+g(x){|}^{p}d\mu )}^{1/p}\le {({\int }_X|f(x){|}^{p}d\mu )}^{1/p}+{({\int }_X|g(x){|}^{p}d\mu )}^{1/p}.$$
   Thus:
   $$\Vert f+g{\Vert }_p\le \Vert f{\Vert }_p+\Vert g{\Vert }_p.$$$$\Vert f+g{\Vert }_p\le \Vert f{\Vert }_p+\Vert g{\Vert }_p.$$||f+g||_(p) <= ||f||_(p)+||g||_(p).\|f + g\|_p \leq \|f\|_p + \|g\|_p.$$\Vert f+g{\Vert }_p\le \Vert f{\Vert }_p+\Vert g{\Vert }_p.$$
4. Homogeneity and Definiteness:
   The properties of scalar multiplication and non-negativity ensure that $\Vert f{\Vert }_p$$\Vert f{\Vert }_p$||f||_(p)\|f\|_p$\Vert f{\Vert }_p$ satisfies homogeneity and definiteness.

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2. $L^p$$L^p$L^(p)L^p$L^p$-Space is a Metric Space

1. Non-negativity:
   $$d(f,g)=\Vert f-g{\Vert }_p\ge 0,$$$$d(f,g)=\Vert f-g{\Vert }_p\ge 0,$$d(f,g)=||f-g||_(p) >= 0,d(f, g) = \|f – g\|_p \geq 0,$$d(f,g)=\Vert f-g{\Vert }_p\ge 0,$$
   and $d(f,g)=0$$d(f,g)=0$d(f,g)=0d(f, g) = 0$d(f,g)=0$ if and only if $f=g$$f=g$f=gf = g$f=g$ almost everywhere.
2. Symmetry:
   $$d(f,g)=\Vert f-g{\Vert }_p=\Vert g-f{\Vert }_p=d(g,f).$$$$d(f,g)=\Vert f-g{\Vert }_p=\Vert g-f{\Vert }_p=d(g,f).$$d(f,g)=||f-g||_(p)=||g-f||_(p)=d(g,f).d(f, g) = \|f – g\|_p = \|g – f\|_p = d(g, f).$$d(f,g)=\Vert f-g{\Vert }_p=\Vert g-f{\Vert }_p=d(g,f).$$
3. Triangle Inequality:
   For $f,g,h\in L^p$$f,g,h\in L^p$f,g,h inL^(p)f, g, h \in L^p$f,g,h\in L^p$:
   $$d(f,h)=\Vert f-h{\Vert }_p=\Vert f-g+g-h{\Vert }_p\le \Vert f-g{\Vert }_p+\Vert g-h{\Vert }_p=d(f,g)+d(g,h).$$$$d(f,h)=\Vert f-h{\Vert }_p=\Vert f-g+g-h{\Vert }_p\le \Vert f-g{\Vert }_p+\Vert g-h{\Vert }_p=d(f,g)+d(g,h).$$d(f,h)=||f-h||_(p)=||f-g+g-h||_(p) <= ||f-g||_(p)+||g-h||_(p)=d(f,g)+d(g,h).d(f, h) = \|f – h\|_p = \|f – g + g – h\|_p \leq \|f – g\|_p + \|g – h\|_p = d(f, g) + d(g, h).$$d(f,h)=\Vert f-h{\Vert }_p=\Vert f-g+g-h{\Vert }_p\le \Vert f-g{\Vert }_p+\Vert g-h{\Vert }_p=d(f,g)+d(g,h).$$

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Conclusion:

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Section-C

Long Answer Questions

Show that second countable space is always first countable, but the converse is not true.

Answer:

Statement:

1. Every second countable space is first countable.
2. The converse is not true; i.e., a first countable space may not be second countable.

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Definitions:

1. Second Countable Space:

• $\mathcal{B}$$\mathcal{B}$B\mathcal{B}$\mathcal{B}$ is a collection of open sets,
• For every open set $U\subseteq X$$U\subseteq X$U sube XU \subseteq X$U\subseteq X$, there exists a subcollection $\{B_{\alpha }\in \mathcal{B}\}$$\{B_{\alpha }\in \mathcal{B}\}${B_( alpha)inB}\{B_\alpha \in \mathcal{B}\}$\{B_{\alpha }\in \mathcal{B}\}$ such that $U=\bigcup _{\alpha }{B}_{\alpha }$$U=\bigcup _{\alpha } B_{\alpha }$U=uuu_(alpha)B_( alpha)U = \bigcup_{\alpha} B_\alpha$U=\bigcup _{\alpha }{B}_{\alpha }$,
• $\mathcal{B}$$\mathcal{B}$B\mathcal{B}$\mathcal{B}$ is countable.

2. First Countable Space:

• For each $x\in X$$x\in X$x in Xx \in X$x\in X$, there exists a countable collection $\{U_n{\}}_{n=1}^{\mathrm{\infty }}$$\{U_n{\}}_{n=1}^{\mathrm{\infty }}${U_(n)}_(n=1)^(oo)\{U_n\}_{n=1}^\infty$\{U_n{\}}_{n=1}^{\mathrm{\infty }}$ of open sets such that:$$\text{For any open set}U\ni x,\text{there exists}{U}_n\text{such that}x\in U_n\subseteq U.$$$$\text{For any open set}U\ni x,\text{there exists}{U}_n\text{such that}x\in U_n\subseteq U.$$“For any open set “U∋x,” there exists “U_(n)” such that “x inU_(n)sube U.\text{For any open set } U \ni x, \text{ there exists } U_n \text{ such that } x \in U_n \subseteq U.$$\text{For any open set}U\ni x,\text{there exists}{U}_n\text{such that}x\in U_n\subseteq U.$$

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Part 1: A Second Countable Space is Always First Countable

Proof:

1. Neighborhood Basis Construction:
   • Fix $x\in X$$x\in X$x in Xx \in X$x\in X$.
   • Consider all basis elements $B_i\in \mathcal{B}$$B_i\in \mathcal{B}$B_(i)inBB_i \in \mathcal{B}$B_i\in \mathcal{B}$ such that $x\in B_i$$x\in B_i$x inB_(i)x \in B_i$x\in B_i$.
   • Let $\{B_{i_k}{\}}_{k=1}^{\mathrm{\infty }}$$\{B_{i_k}{\}}_{k=1}^{\mathrm{\infty }}${B_(i_(k))}_(k=1)^(oo)\{B_{i_k}\}_{k=1}^\infty$\{B_{i_k}{\}}_{k=1}^{\mathrm{\infty }}$ be the collection of such basis elements. This is a countable set because $\mathcal{B}$$\mathcal{B}$B\mathcal{B}$\mathcal{B}$ is countable.
2. Verification of Neighborhood Basis:
   • For any open set $U\ni x$$U\ni x$U∋xU \ni x$U\ni x$, there exists a subcollection of basis elements $\{B_{\alpha }\in \mathcal{B}\}$$\{B_{\alpha }\in \mathcal{B}\}${B_( alpha)inB}\{B_\alpha \in \mathcal{B}\}$\{B_{\alpha }\in \mathcal{B}\}$ such that $U=\bigcup _{\alpha }{B}_{\alpha }$$U=\bigcup _{\alpha } B_{\alpha }$U=uuu_(alpha)B_( alpha)U = \bigcup_{\alpha} B_\alpha$U=\bigcup _{\alpha }{B}_{\alpha }$.
   • In particular, since $x\in U$$x\in U$x in Ux \in U$x\in U$, there exists $B_{i_k}\subseteq U$$B_{i_k}\subseteq U$B_(i_(k))sube UB_{i_k} \subseteq U$B_{i_k}\subseteq U$ such that $x\in B_{i_k}$$x\in B_{i_k}$x inB_(i_(k))x \in B_{i_k}$x\in B_{i_k}$.
   • Thus, $\{B_{i_k}\}$$\{B_{i_k}\}${B_(i_(k))}\{B_{i_k}\}$\{B_{i_k}\}$ forms a countable neighborhood basis for $x$$x$xx$x$.
3. Conclusion:
   Since every point $x\in X$$x\in X$x in Xx \in X$x\in X$ has a countable neighborhood basis, $X$$X$XX$X$ is first countable.

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Part 2: A First Countable Space May Not Be Second Countable

Counterexample:

1. First Countable:
   • In the discrete topology, the singleton set $\{x\}$$\{x\}${x}\{x\}$\{x\}$ is open for every $x\in \mathbb{R}$$x\in \mathbb{R}$x inRx \in \mathbb{R}$x\in \mathbb{R}$.
   • For any point $x\in \mathbb{R}$$x\in \mathbb{R}$x inRx \in \mathbb{R}$x\in \mathbb{R}$, a countable neighborhood basis is $\{\{x\}\}$$\{\{x\}\}${{x}}\{\{x\}\}$\{\{x\}\}$, which is trivially countable.
   • Thus, $\mathbb{R}$$\mathbb{R}$R\mathbb{R}$\mathbb{R}$ with the discrete topology is first countable.
2. Not Second Countable:
   • In the discrete topology, every subset of $\mathbb{R}$$\mathbb{R}$R\mathbb{R}$\mathbb{R}$ is open.
   • A basis for this topology is the collection of all singleton sets $\{\{x\}:x\in \mathbb{R}\}$$\{\{x\}:x\in \mathbb{R}\}${{x}:x inR}\{\{x\} : x \in \mathbb{R}\}$\{\{x\}:x\in \mathbb{R}\}$.
   • Since $\mathbb{R}$$\mathbb{R}$R\mathbb{R}$\mathbb{R}$ is uncountable, the basis is uncountable.
   • Therefore, $\mathbb{R}$$\mathbb{R}$R\mathbb{R}$\mathbb{R}$ with the discrete topology is not second countable.

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Conclusion:

1. Every second countable space is first countable because a countable basis ensures the existence of a countable neighborhood basis for each point.
2. The converse is false; for example, $\mathbb{R}$$\mathbb{R}$R\mathbb{R}$\mathbb{R}$ with the discrete topology is first countable but not second countable.

Prove that the sequence of functions in $L^p$$L^p$L^(p)L^p$L^p$ space has at most one limit.

Answer:

Statement:

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Definitions:

1. $L^p$$L^p$L^(p)L^p$L^p$-Space:

2. Convergence in $L^p$$L^p$L^(p)L^p$L^p$-Norm:

3. Uniqueness of Limits:

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Proof:

Step 1: Assumption of Two Limits

Step 2: Triangle Inequality

Step 3: Taking the Limit

Step 4: Norm Zero Implies Equality Almost Everywhere

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Conclusion:

Show that one point compactification of the set of rational numbers $Q$$Q$QQ$Q$ is not Hausdorff.

Answer:

Statement:

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Definitions:

1. One-Point Compactification:

• The open sets of $X^{\ast }$$X^{\ast }$X^(**)X^*$X^{\ast }$ include:
  • All open sets of $X$$X$XX$X$,
  • Sets of the form $U_{\mathrm{\infty }}=X^{\ast }\setminus K$$U_{\mathrm{\infty }}=X^{\ast }\setminus K$U_( oo)=X^(**)\\KU_\infty = X^* \setminus K$U_{\mathrm{\infty }}=X^{\ast }\setminus K$, where $K\subseteq X$$K\subseteq X$K sube XK \subseteq X$K\subseteq X$ is compact in $X$$X$XX$X$.

2. Hausdorff Space:

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Proof:

Step 1: The Rational Numbers $\mathbb{Q}$$\mathbb{Q}$Q\mathbb{Q}$\mathbb{Q}$

• Locally compact: Every point $q\in \mathbb{Q}$$q\in \mathbb{Q}$q inQq \in \mathbb{Q}$q\in \mathbb{Q}$ has a compact neighborhood (e.g., $[q-ϵ,q+ϵ]\cap \mathbb{Q}$$[q-ϵ,q+ϵ]\cap \mathbb{Q}$[q-epsilon,q+epsilon]nnQ[q – \epsilon, q + \epsilon] \cap \mathbb{Q}$[q-ϵ,q+ϵ]\cap \mathbb{Q}$).
• Not compact: The space $\mathbb{Q}$$\mathbb{Q}$Q\mathbb{Q}$\mathbb{Q}$ is not compact because it is not closed in $\mathbb{R}$$\mathbb{R}$R\mathbb{R}$\mathbb{R}$, and sequences like $q_n=1+1/n$$q_n=1+1/n$q_(n)=1+1//nq_n = 1 + 1/n$q_n=1+1/n$ do not have convergent subsequences in $\mathbb{Q}$$\mathbb{Q}$Q\mathbb{Q}$\mathbb{Q}$.

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Step 2: Neighborhoods in ${\mathbb{Q}}^{\ast }$${\mathbb{Q}}^{\ast }$Q^(**)\mathbb{Q}^*${\mathbb{Q}}^{\ast }$

1. In $\mathbb{Q}$$\mathbb{Q}$Q\mathbb{Q}$\mathbb{Q}$, the open sets are the usual open sets inherited from $\mathbb{R}$$\mathbb{R}$R\mathbb{R}$\mathbb{R}$.
2. For the point $\mathrm{\infty }$$\mathrm{\infty }$oo\infty$\mathrm{\infty }$, open neighborhoods are of the form:$$U_{\mathrm{\infty }}={\mathbb{Q}}^{\ast }\setminus K,$$$$U_{\mathrm{\infty }}={\mathbb{Q}}^{\ast }\setminus K,$$U_( oo)=Q^(**)\\K,U_\infty = \mathbb{Q}^* \setminus K,$$U_{\mathrm{\infty }}={\mathbb{Q}}^{\ast }\setminus K,$$where $K\subseteq \mathbb{Q}$$K\subseteq \mathbb{Q}$K subeQK \subseteq \mathbb{Q}$K\subseteq \mathbb{Q}$ is compact.

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Step 3: Hausdorff Property and Failure

1. Let $x\in \mathbb{Q}$$x\in \mathbb{Q}$x inQx \in \mathbb{Q}$x\in \mathbb{Q}$ and $\mathrm{\infty }$$\mathrm{\infty }$oo\infty$\mathrm{\infty }$. Suppose there exist disjoint open sets $U$$U$UU$U$ and $V$$V$VV$V$ such that:
   • $x\in U$$x\in U$x in Ux \in U$x\in U$,
   • $\mathrm{\infty }\in V$$\mathrm{\infty }\in V$oo in V\infty \in V$\mathrm{\infty }\in V$.
2. By definition of neighborhoods of $\mathrm{\infty }$$\mathrm{\infty }$oo\infty$\mathrm{\infty }$, $V=U_{\mathrm{\infty }}={\mathbb{Q}}^{\ast }\setminus K$$V=U_{\mathrm{\infty }}={\mathbb{Q}}^{\ast }\setminus K$V=U_( oo)=Q^(**)\\KV = U_\infty = \mathbb{Q}^* \setminus K$V=U_{\mathrm{\infty }}={\mathbb{Q}}^{\ast }\setminus K$ for some compact $K\subseteq \mathbb{Q}$$K\subseteq \mathbb{Q}$K subeQK \subseteq \mathbb{Q}$K\subseteq \mathbb{Q}$. Thus, $K$$K$KK$K$ is closed and bounded in $\mathbb{Q}$$\mathbb{Q}$Q\mathbb{Q}$\mathbb{Q}$.
3. Since $K$$K$KK$K$ is compact, it contains only finitely many points of $\mathbb{Q}$$\mathbb{Q}$Q\mathbb{Q}$\mathbb{Q}$. However, $\mathbb{Q}$$\mathbb{Q}$Q\mathbb{Q}$\mathbb{Q}$ is dense in $\mathbb{R}$$\mathbb{R}$R\mathbb{R}$\mathbb{R}$, so any open set $U\subseteq \mathbb{Q}$$U\subseteq \mathbb{Q}$U subeQU \subseteq \mathbb{Q}$U\subseteq \mathbb{Q}$ containing $x$$x$xx$x$ will intersect $K$$K$KK$K$. This implies $U\cap V\ne \mathrm{\varnothing }$$U\cap V\ne \mathrm{\varnothing }$U nn V!=O/U \cap V \neq \emptyset$U\cap V\ne \mathrm{\varnothing }$, contradicting the assumption that $U$$U$UU$U$ and $V$$V$VV$V$ are disjoint.

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Conclusion:

Prove that every second countable regular space is a normal space.

Answer:

Statement:

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Definitions:

1. Second Countable Space:
   A topological space $X$$X$XX$X$ is second countable if it has a countable basis $\mathcal{B}$$\mathcal{B}$B\mathcal{B}$\mathcal{B}$, i.e., a countable collection of open sets such that every open set in $X$$X$XX$X$ can be expressed as a union of sets in $\mathcal{B}$$\mathcal{B}$B\mathcal{B}$\mathcal{B}$.
2. Regular Space:
   A topological space $X$$X$XX$X$ is regular if:
   • $X$$X$XX$X$ is $T_1$$T_1$T_(1)T_1$T_1$, meaning every singleton set $\{x\}$$\{x\}${x}\{x\}$\{x\}$ is closed, and
   • For every closed set $F\subseteq X$$F\subseteq X$F sube XF \subseteq X$F\subseteq X$ and a point $p\notin F$$p\notin F$p!in Fp \notin F$p\notin F$, there exist disjoint open sets $U$$U$UU$U$ and $V$$V$VV$V$ such that:$$F\subseteq U\text{and}p\in V.$$$$F\subseteq U\text{and}p\in V.$$F sube U quad”and”quad p in V.F \subseteq U \quad \text{and} \quad p \in V.$$F\subseteq U\text{and}p\in V.$$
3. Normal Space:
   A topological space $X$$X$XX$X$ is normal if:
   • $X$$X$XX$X$ is $T_1$$T_1$T_(1)T_1$T_1$, and
   • For any two disjoint closed sets $F_1,F_2\subseteq X$$F_1,F_2\subseteq X$F_(1),F_(2)sube XF_1, F_2 \subseteq X$F_1,F_2\subseteq X$, there exist disjoint open sets $U_1$$U_1$U_(1)U_1$U_1$ and $U_2$$U_2$U_(2)U_2$U_2$ such that:$$F_1\subseteq U_1\text{and}{F}_2\subseteq U_2.$$$$F_1\subseteq U_1\text{and}{F}_2\subseteq U_2.$$F_(1)subeU_(1)quad”and”quadF_(2)subeU_(2).F_1 \subseteq U_1 \quad \text{and} \quad F_2 \subseteq U_2.$$F_1\subseteq U_1\text{and}{F}_2\subseteq U_2.$$

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Proof:

Given:

Step 1: Second Countability and Regularity

• Second countability implies that $X$$X$XX$X$ has a countable basis $\mathcal{B}$$\mathcal{B}$B\mathcal{B}$\mathcal{B}$.
• Regularity means for any closed set $F\subseteq X$$F\subseteq X$F sube XF \subseteq X$F\subseteq X$ and $p\notin F$$p\notin F$p!in Fp \notin F$p\notin F$, there exist disjoint open sets separating $F$$F$FF$F$ and $p$$p$pp$p$.

Step 2: Let $F_1$$F_1$F_(1)F_1$F_1$ and $F_2$$F_2$F_(2)F_2$F_2$ be Two Disjoint Closed Sets

Step 3: Use Regularity to Separate Points

Step 4: Cover $F_1$$F_1$F_(1)F_1$F_1$ Using $\mathcal{B}$$\mathcal{B}$B\mathcal{B}$\mathcal{B}$

Step 5: Ensure $U_1$$U_1$U_(1)U_1$U_1$ and $U_2$$U_2$U_(2)U_2$U_2$ are Disjoint

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Step 6: Conclusion