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Source: Topology Without Tears by Sidney A. Morris. Topology book and Videos on Pure Mathematics, Topology and Writing Proofs supplementing the book

The previous chapter: Morris-Topology Without Tears-CH01

2024-08-08 Book Notes

Notes

Introduction

• the Euclidean topology on the set of real numbers is one of the central characters
• just like every vector space has a basis and every vector is a linear combination of members of basis, a topological space every open set can be expressed as a union of members of the basis.

2.1 The Euclidean Topology on $\mathbb{R}$

Notation. Whenever we refer to the topological space $\mathbb{R}$ without specifying the topology, we mean $\mathbb{R}$ with the euclidean topology.

Remarks

(i) The "euclidean topology" $\mathcal{T}$ is a topology

• the given information is property ($\star$)

• we have to prove that $\mathcal{T}$ satisfies conditions (i), (ii) and (iii) of Definitions 1.1.1 in Morris-Topology Without Tears-CH01

  Proof. First, to prove condition (i) in definition 1.1.1, let $x\in \mathbb{R}$, as already wrote in the definition 2.1.1: "a subset $S$ of $\mathbb{R}$", subset $S$ could just be $\mathbb{R}$. And we could also find $a,b$ in $\mathbb{R}$ satisfy the property that $a<b$ let $x\in (a,b)\subseteq \mathbb{R}$. Thus, by default setting, $\mathbb{R}$ is in $\mathcal{T}$ (be open in the euclidean topology).

  The next step, we should prove $\mathcal{T}$ satisfies condition (ii), which means that the union of any members of subset of $\mathbb{R}$ is in $\mathcal{T}$, in other word, the union should hold the property ($\star$). Let $\{A_j:j\in J\}$, for some index set $J$, be a family of members of $\mathcal{T}$. And let $x\in \bigcup _{j\in J}{A}_j$. For some $k\in J$, then $x\in A_k$ (which means that at least there is one $A_k$ where $x$ belongs to). As $A_k\in \mathcal{T}$, there should exist $a$ and $b$ in $\mathbb{R}$ with $a<b$ such that $x\in (a,b)\subseteq A_k$. As $k\in J$, $A_k\subseteq \bigcup _{j\in J}{A}_j$ and so $x\in (a,b)\subseteq \bigcup _{j\in J}{A}_j$. Thus, we know that the union of any members of subset of $\mathbb{R}$ is in $\mathcal{T}$.

  Finally, we should prove $\mathcal{T}$ satisfies condition (iii), which means that the intersection of any two subset of $\mathbb{R}$ is in $\mathcal{T}$. As the same logics above, the intersection should hold the property ($\star$). Let $A_1$ and $A_2$ be in $\mathcal{T}$, and $y\in A_1\cap A_2$. As the property of the open subset in the euclidean topology on $\mathbb{R}$, we could know that $y\in (a,b)\subseteq A_1$ and also $y\in (c,d)\subseteq A_2$. Let $e$ be greater than $a$ and $c$, and $f$ the smaller of $b$ and $d$. Easily to know that $e<y<f$, thus, $y\in (e,f)$. We could deduce that $y\in (e,f)\subseteq A_1\cap A_2$, for the intersection of any two subset in $\mathcal{T}$ hold the property.

(ii) Let $r,s\in \mathbb{R}$ with $r<s$. In the euclidean topology $\mathcal{T}$ on $\mathbb{R}$, the open interval $(r,s)$ does indeed belong to $\mathcal{T}$ and so is an open set.

• Proof.
  Let $x\in (r,s)$, we could let $a=r$ and $b=s$, leading that $x\in (a,b)\subseteq (r,s)$, which means that $(r,s)$ is an open set in $\mathbb{R}$.

(iii) The open intervals $(r,\mathrm{\infty })$ and $(-\mathrm{\infty },r)$ are open sets in $\mathbb{R}$, for every real number $r$.

• Proof.
  Let $x\in (r,\mathrm{\infty })$, we could find $a=r$ and $b=a+1$. Thus, $x\in (a,b)\subseteq (r,\mathrm{\infty })$, which means that $(r,\mathrm{\infty })$ is an open set in $\mathbb{R}$. With the same logics, let $x\in (-\mathrm{\infty },r)$, we could find $b=r$ and $a=b-1$. Thus, $x\in (a,b)\subseteq (-\mathrm{\infty },r)$, which means that $(-\mathrm{\infty },r)$ is also an open set in $\mathbb{R}$.

(iv) While every open interval is an open set in $\mathbb{R}$, the converse is false. Not all open sets in $\mathbb{R}$ are intervals.

• For example, the $(1,3)\cup (41,42)$ is open set in $\mathbb{R}$, but it is not an open interval.

(v) For each $c$ and $d$ in $\mathbb{R}$ with $c<d$, the closed interval $[c,d]$ is not an open set in $\mathbb{R}$.

• Proof.
  Suppose there exist $a$ and $b$ in $\mathbb{R}$ with $a<b$ such that $c\in (a,b)\subseteq [c,d]$. As $c\in (a,b)$, we know that $a<c<b$, and $a<\frac{a+c}{2}<c<b$. $\frac{a+c}{2}\in (a,b)$, but obviously not in $[c,d]$, which means that $(a,b)⊊[c,d]$ and this result is contradiction. Thus, we could know that interval $[c,d]$ does not hold the property and is not an open set on $\mathbb{R}$.

(vi) For each $a$ and $b$ in $\mathbb{R}$ with $a<b$, the closed interval $[a,b]$ is a closed set in the euclidean topology on $\mathbb{R}$.

• Proof.
  We could know that the complement of interval $[a,b]$ is $(-\mathrm{\infty },a)\cup (b,\mathrm{\infty })$, which are open sets in $\mathbb{R}$ as we have already proved in the (iii). As the property of a topology, or we could say that as the definition of a closed set, we know that $[a,b]$ is a closed set in the euclidean topology on $\mathbb{R}$.

(vii) Each singleton set $\{a\}$ is closed in $\mathbb{R}$.

• Proof.
  The logics of this proof is just as same as the previous one, as we all know that $(-\mathrm{\infty },a)$ and $(a,\mathrm{\infty })$ are open sets in the euclidean topology, thus, the complement of the union of them, namely the singleton $\{a\}$ is closed in $\mathbb{R}$.

(viii) Note that we could have included (vii) in (vi) simply by replacing $a<b$ by $a\le b$. The singleton set $\{a\}$ is just the degenerate case of the closed interval $[a,b]$.

(ix) The set $\mathbb{Z}$ of all integers is a closed subset of $\mathbb{R}$.

• Proof.
  As same as (vii) and (vii), we know that the complement of $\mathbb{Z}$ is the union $\bigcup _{n=-\mathrm{\infty }}^{\mathrm{\infty }}(n,n+1)$ of $\mathbb{R}$ and it is open in $\mathbb{R}$. Thus, the set $\mathbb{Z}$ of all integers is closed in $\mathbb{R}$.

(x) The set $\mathbb{Q}$ of all rational numbers is neither a closed subset of $\mathbb{R}$ nor an open subset of $\mathbb{R}$.

• Proof.
  Suppose that $(a,b)\subseteq \mathbb{Q}$, where $a,b\in \mathbb{R}$ with $a<b$. And we know that between any two distinct real numbers there is an irrational number (see 20240808Th1623). Thus, there exists $c\in (a,b)$ such that $c\notin \mathbb{Q}$, which contradicts $(a,b)\subseteq \mathbb{Q}$. Thus, $\mathbb{Q}$ is not an open set on $\mathbb{R}$.

  Moreover, suppose that $(a,b)\subseteq \mathbb{R}\setminus \mathbb{Q}$, where $a,b\in \mathbb{R}$ with $a<b$. For the fact that between any two distinct real numbers there is a rational number. We could know that, the complement of $\mathbb{Q}$, namely, $\mathbb{R}\setminus \mathbb{Q}$ is also not open. Therefore, $\mathbb{Q}$ is neither closed nor open on $\mathbb{R}$.

Exercises 2.1

1. Prove that if $a,b\in \mathbb{R}$ with $a<b$ then neither $[a,b)$ nor $(a,b]$ is an open subset of $\mathbb{R}$. Also show that neither is a closed subset of $\mathbb{R}$.
   Proof.
   Firstly, suppose $[a,b)$ is an open subset of $\mathbb{R}$, there exists $p$ and $q$ in $\mathbb{R}$ with $p<q$ such that $a\in (p,q)\subseteq [a,b)$. For $a\in (p,q)$, we know $p<a<q$ and also $p<\frac{p+a}{2}<a<q<b$. Thus, $\frac{p+a}{2}\in (p,q)$ but $\frac{p+a}{2}\notin [a,b)$, which means that $(p,q)⊊[a,b)$ as a contradiction. Hence, $[a,b)$ is not an open subset of $\mathbb{R}$. Besides, the complement of $[a,b)$ is $(-\mathrm{\infty },a)\cup \{b\}\cup (b,\mathrm{\infty })$, for $\{b\}$ is closed subset, thus, $\mathbb{R}\setminus [a,b)$ is closed. Then we know $[a,b)$ is neither a closed subset of $\mathbb{R}$.
   For the same logics, we could also prove $(a,b]$ is not an open subset and not a closed subset of $\mathbb{R}$.
   $$

2. Prove that the sets $[a,\mathrm{\infty })$ and $(-\mathrm{\infty },a]$ are closed subsets of $\mathbb{R}$.
   Proof.
   We could prove $\mathbb{R}\setminus [a,\mathrm{\infty })$, namely, $(-\mathrm{\infty },a)$ is open. As same as remark (iii), let $x\in (-\mathrm{\infty },a)$, we could find an interval $(a-ϵ,a)$, where $ϵ>0$, there exists $x\in (a-ϵ,a)\subseteq (-\mathrm{\infty },a)$, thus $(-\mathrm{\infty },a)$ is open, and its complement on $\mathbb{R}$ is closed. I would like to skip the proof of $(-\mathrm{\infty },a]$ as the totally same logics.
   $$

3. Show, by example, that the union of an infinite number of closed subsets of $\mathbb{R}$ is not necessarily a closed subset of $\mathbb{R}$.
   Proof.
   For example, consider a closed subset $(-\mathrm{\infty },n]$, let $ϵ>0$, and the union of an infinite number of $(-\mathrm{\infty },n]\cup (-\mathrm{\infty },n+ϵ]\cup ...$, it could be $(-\mathrm{\infty },\mathrm{\infty })$ which should be open on $\mathbb{R}$.
   $$

4. Prove each of the following statements.
   (i) The set $\mathbb{Z}$ of all integers is not an open subset of $\mathbb{R}$.
   Proof.
   As we know, any interval between two integers is an open interval, and as we have proved previously, open intervals are indeed open. Also, the union of any numbers of open subsets is also open. Which means $\mathbb{R}\setminus \mathbb{Z}$ is open, thus, $\mathbb{Z}$ is not an open subset of $\mathbb{R}$.

   (ii) The set $\mathbb{P}$ of all prime numbers is a closed subset of $\mathbb{R}$ but not an open subset of $\mathbb{R}$.
   Proof.
   As same as the proof on (ii) above, any interval between two primes should also be an open interval. Thus, $\mathbb{P}$ is also a closed subset of $\mathbb{R}$.
   $$
   (iii) The set $\mathbb{I}$ of all irrational numbers is neither a closed subset nor an open subset of $\mathbb{R}$.
   Proof.
   According to remark (x), the set $\mathbb{Q}$ of all rational numbers is neither a closed nor an open subset of $\mathbb{R}$. The set of all irrational numbers $\mathbb{I}=\mathbb{R}\setminus \mathbb{Q}$, should also be neither a closed subset nor an open subset of $\mathbb{R}$. Or, suppose $(a,b)\subseteq \mathbb{I}$, where $a,b\in \mathbb{R}$. However, as we know, between any two distinct real numbers exists a rational number. Thus, we know that $\mathbb{I}$ is not open in $\mathbb{R}$. As the same logics, it could be easily proved that $\mathbb{I}$ is also not closed in $\mathbb{R}$.
   $$

5. If $F$ is a non-empty finite subset of $\mathbb{R}$, show that $F$ is closed in $\mathbb{R}$ but that $F$ is not open in $\mathbb{R}$.
   Proof.
   Suppose $F$ is open in $\mathbb{R}$, $(a,b)\subseteq F$, where $a,b\in \mathbb{R}$ with $a<b$. As the given information, $F$ is a non-empty finite subset of $\mathbb{R}$, we could find $x\in (a,b)$ but $x\notin F$, however, it is contractive. Thus, $F$ is not open in $\mathbb{R}$.
   $$

6. If $F$ is a non-empty countable subset of $\mathbb{R}$, prove that $F$ is not an open set, but that $F$ may or may not be a closed set depending on the choice of $F$.
   Proof.
   Suppose $F$ is open in $\mathbb{R}$, $(a,b)\subseteq F$, where $a,b\in \mathbb{R}$ with $a<b$. As the given information, $F$ is a non-empty countable subset of $\mathbb{R}$, we could always find $x\in (a,b)$ but $x$ is not countable. As this contradiction, $F$ is not open in $\mathbb{R}$.
   If $F$ contain limit points, $\mathbb{R}\setminus F$ could be open, thus $F$ will be regarded as closed.
   $$

7. (i) Let $S=\{0,1,1/2,1/3,..,1/n,...\}$. Prove that the set $S$ is closed in the euclidean topology on $\mathbb{R}$.
   Proof.
   We could prove $\mathbb{R}\setminus S$ is open. If $x\in \mathbb{R}\setminus S$, $x$ should locate in the intervals divided by the isolated elements of $S$. In those each open interval, we could always find $x\in (a,b)$, where $(a,b)$ is small enough be the subset of $\mathbb{R}\setminus S$. Thus, we know $\mathbb{R}\setminus S$ is open on $\mathbb{R}$, which means $S$ is closed in the euclidean topology on $\mathbb{R}$.
   $$
   (ii) Is the set $T=\{1,1/2,1/3,...,1/n,...\}$ closed in $\mathbb{R}$?
   Proof.
   What we should do is to check out whether $\mathbb{R}\setminus T$ is open in the euclidean topology on $\mathbb{R}$. We could say $\mathbb{R}\setminus T=(-\mathrm{\infty },0)\cup \{0\}\cup (0,1)\cup (1,\mathrm{\infty })\cup ...\cup (\frac{1}{n+1},\frac{1}{n})$. As any singleton is closed in the euclidean topology on $\mathbb{R}$, we could easily know that the $\mathbb{R}\setminus T$ is closed, which means that $T$ is open.
   $$
   (iii) Is the set $\{\sqrt{2},2\sqrt{2},3\sqrt{2},...n\sqrt{2},...\}$ closed in $\mathbb{R}$?
   Proof.
   We could easily know that the set $\{\sqrt{2},2\sqrt{2},3\sqrt{2},...n\sqrt{2},...\}$ contains infinite points with certain interval between them. The complement of the set should be infinite open intervals and should be open. Thus, $\{\sqrt{2},2\sqrt{2},3\sqrt{2},...n\sqrt{2},...\}$ is closed in $\mathbb{R}$.
   $$

8. (i) Let $(X,\mathcal{T})$ be a topological space. A subset $S$ of $X$ is said to be an $F_{\sigma }$-set if it is the union of a countable number of closed sets. Prove that all open intervals $(a,b)$ and all closed intervals $[a,b]$, are $F_{\sigma }$-sets in $\mathbb{R}$.
   Proof.
   According to the given definition, we are asked to prove that all open intervals and all closed intervals are the subsets of $\mathbb{R}$ and are the union of two countable number of closed sets in the euclidean topology on $\mathbb{R}$.
   First, for the all open intervals $(a,b)$, we could consider it as the union of a sequence of closed intervals $[a+\frac{1}{n},b-\frac{1}{n}]$ for $n\in \mathbb{N}$. As $n$ grows up to $\mathrm{\infty }$, $(a,b)=\bigcup _{n=1}^{\mathrm{\infty }}[a+\frac{1}{n},b-\frac{1}{n}]$. Every interval $[a+\frac{1}{n},b-\frac{1}{n}]$ is closed for containing the limit points. Thus, we could say that all open intervals $(a,b)$ are $F_{\sigma }$-sets in $\mathbb{R}$.
   Then, for the all closed intervals $[a,b]$, we could regard $[a,b]$ as $(a,b)\cup \{a\}\cup \{b\}$. We have already proved that all open intervals $(a,b)$ are $F_{\sigma }$-sets in $\mathbb{R}$. Adding two singleton $\{a\}$ and $\{b\}$ to this union does not change the property of it.
   $$
   (ii) Let $(X,\mathcal{T})$ be a topological space. A subset $S$ of $X$ is said to be an $G_{\delta }$-set if it is the intersection of a countable number of open sets. Prove that all open intervals $(a,b)$ and all closed intervals $[a,b]$ are $G_{\delta }$-sets in $\mathbb{R}$.
   Proof.
   As same as (ii), we have to prove that all open intervals and all closed intervals are the subsets of $\mathbb{R}$ and are the intersections of countable number of open sets on the euclidean topology on $\mathbb{R}$.
   Open intervals $(a,b)$ and closed intervals $[a,b]$ could be expressed similarly as $\bigcap _{n=1}^{\mathrm{\infty }}(a-\frac{1}{n},b+\frac{1}{n})$.
   $$
   (iii) Prove that the set $\mathbb{Q}$ of rationals is an $F_{\sigma }$-set in $\mathbb{R}$.
   Proof.
   Set $\mathbb{Q}$ of rationals could be regarded as a union of each singleton rational number. As we have already proved that each singleton is closed in the euclidean topology on $\mathbb{R}$. Thus, $\mathbb{Q}$ is an $F_{\sigma }$-set in $\mathbb{R}$.
   $$
   (iv) Verify that the complement of an $F_{\sigma }$-set is a $G_{\delta }$-set and the complement of a $G_{\delta }$-set is an $F_{\sigma }$-set.
   Proof.
   For the given definitions, let $(X,\mathcal{T})$ be a topological space. An $F_{\sigma }$-set is the union of a countable number of closed sets in $\mathcal{T}$ on $X$, and a $G_{\delta }$-set is the intersection of a countable number of open sets in $\mathcal{T}$ on $X$.
   Let $C_j$ is any closed sets in $\mathcal{T}$ on $X$, where $j\in J$, $J$ is the index set of the closed sets. Then we could express an $F_{\sigma }$-set as $\bigcup C_j$, and the complement of $F_{\sigma }$-set could be expressed as $X\setminus (\bigcup C_j)$. We could also write it into $\bigcap (X\setminus C_j)$. For $X\setminus C_j$ are open sets, $\bigcap (X\setminus C_j)$ is exactly in the form of a $G_{\delta }$-set.
   Following the same logics, we could also prove that a $G_{\delta }$-set is an $F_{\sigma }$-set.

2.2 Basis for a Topology

Proof.
What we are required to prove:

• (i) If $S$ is a union of open intervals, then it is an open set
• (ii) If $S$ is an open set, then it is a union of open intervals

For (i), assume that $S$ is a union of open intervals, according to [Remarks 2.1.2](#Remarks 2.1.2)(ii), every open interval is an open set. Thus, $S$ is a union of open sets and so $S$ is an open set.

For (ii), assume that $S$ is open in $\mathbb{R}$. Then for each $x\in S$, there exists an interval $I_x=(a,b)$ such that $x\in I_x\subseteq S$. Now we have to prove that $S=\bigcup _{x\in S}{I}_x$. Or, we could express it in the way of proving: the statement (a) $S\subseteq \bigcup _{x\in S}{I}_x$, and (b) $\bigcup _{x\in S}{I}_x\subseteq S$. For (a), let $y\in S$, then $y\in I_y$. So $y\in \bigcup _{x\in S}{I}_x$ as $y\in I_y\subseteq S$ and $I_y$ should be a subset of the union of any interval $I_x$. For (b), let $z\in \bigcup _{x\in S}{I}_x$, then $z\in I_t$, for some $t\in S$. As each $I_x\subset S$, $I_t$ should also be a subset of $S$, thus, $z\in S$. All as required, we could say that $S=\bigcup _{x\in S}{I}_x$, (ii) is proved also, and we could claim that a subset $S$ of $\mathbb{R}$ is open if and only if it is a union of open intervals.

Consider 2.2.7 Example. Let $X=\{a,b,c\}$ and $\mathcal{B}=\{\{a\},\{c\},\{a,b\},\{b,c\}\}$. Then $\mathcal{B}$ is not a basis for any topology on $X$. Suppose that $\mathcal{B}$ is a basis for a $\mathcal{T}=\{X,\mathrm{\varnothing },\{a\},\{c\},\{a,c\},\{a,b\},\{b,c\}\}$. However, $\mathcal{T}$ is not a topology since the set $\{b\}=\{a,b\}\cap \{b,c\}$ is not in $\mathcal{T}$, which is contradictory.

We are led to ask: if $\mathcal{B}$ is a collection of subsets of $X$, under what conditions is $\mathcal{B}$ a basis for a topology?

Comment: does it mean that $\mathcal{B}$ is some kind of following thing: let a topology $\mathcal{T}$ on some set $X$, $\mathcal{B}=\mathcal{T}\setminus (X\cup \mathrm{\varnothing })$?

Proof.
The proof should be bi-direction, firstly, we have to check that if $\mathcal{B}$ is a basis for a topology $\mathcal{T}$ then $\mathcal{B}$ should have property (a) and (b). As we know that $\mathcal{T}$ is a topology, according to the definition of a basis for a topology, the union of members of $\mathcal{B}$ could generate every open set in this topology. For those open sets are just the unions of members of $\mathcal{B}$.

Then we should prove conversely, assume that $\mathcal{B}$ has properties (a) and (b) and let $\mathcal{T}$ be the collection of all subsets of $X$ which are unions of members of $\mathcal{B}$. We have to prove that $\mathcal{T}$ is a topology on $X$. For property (a), $X=\bigcup _{B\in \mathcal{B}}$ and thus we know $X\in \mathcal{T}$. And $\mathrm{\varnothing }$ is an empty union of members of $\mathcal{B}$ and so $\mathrm{\varnothing }\in \mathcal{B}$. We know that $\mathcal{T}$ does satisfy the Definitions 1.1.1.

Then for property (b), let $\{T_j\}$ be a family of members of $\mathcal{T}$. Then each $T_j$ is a union of members of $\mathcal{B}$. Hence, the union of all the $T_j$ is also a union of members of $\mathcal{B}$, so is in $\mathcal{T}$. Finally let $C$ and $D$ be in $\mathcal{T}$. We have to verify that $C\cap D\in \mathcal{T}$. We could set up that, for some index set $K$ and $J$, let $C=\bigcup _{k\in K}{B}_k$ and $D=\bigcup _{j\in J}{B}_j$, with $B_k\in \mathcal{B}$ and $B_j\in \mathcal{B}$. Thus, we got following:

As the assumption (b), intersection of any $B_1$ and $B_2$ is a union of members of $\mathcal{B}$, thus, $C\cap D$ should also be union of members of $\mathcal{B}$ (a union of union of members of $\mathcal{B}$). Therefore, $C\cap D\in \mathcal{T}$, we know that $\mathcal{T}$ is indeed a topology.

Give another 2.2.9 Example. Let $\mathcal{B}$ be the collection of all "open rectangles", $\{⟨x,y⟩:⟨x,y⟩\in {\mathbb{R}}^2,a<x<b,c<y<d\}$ in the plane which have each side parallel to the X or Y axis. Then $\mathcal{B}$ is a basis for a topology (so called euclidean topology as default) on the plane.

2.2.10 Remark. As a generalization of 2.2.9 Example, let $\mathcal{B}$ be the collection of all subsets $\{⟨x_1,x_2,...,x_n⟩:x_n\in {\mathbb{R}}^n,a_i<x_i<b_i,i=1,2,...,n\}$ of ${\mathbb{R}}^n$ with sides parallel to the axes. This collection is a basis for the euclidean topology on ${\mathbb{R}}^n$.

Exercises 2.2

1. Prove that disc $\{⟨x,y⟩:x^2+y^2<1\}$ is an open subset of ${\mathbb{R}}^2$, and then that every open disc in the plane is an open set.
   (i) Let $⟨a,b⟩$ be any point in the disc $D=\{⟨x,y⟩:x^2+y^2<1\}$. Put $r=\sqrt{a^2+b^2}$. Let $R_{⟨a,b⟩}$ be the open rectangle with vertices at the points $⟨a\pm \frac{1-r}{8},b\pm \frac{1-r}{8}⟩$. Verify that $R_{⟨a,b⟩}\subset D$.
   Proof.
   To prove that disc $D$ is an open subset of ${\mathbb{R}}^2$, and every open disc in the plane is an open set, in other words, they are the members of euclidean topology in ${\mathbb{R}}^2$.
   First of all, we are required to prove a collection of the open rectangles $R_{⟨a,b⟩}$ with $⟨a\pm \frac{1-r}{8},b\pm \frac{1-r}{8}⟩$ any point $(a,b)$ is in the disc $D$.
   It is obvious that we have to check whether for any point $(a,b)$ where $(a\pm \frac{1-r}{8}{)}^2+(b\pm \frac{1-r}{8}{)}^2$ is smaller than 1. As we could easily image, once the upper bond of $R_{⟨a,b⟩}$ is proved to be in the $D$, we can verify the rectangles are the subsets of $D$. As we already known, $r$ is the distance from $(a,b)$ to $(0,0)$ which is less than 1. Thus, what should be proved is as following: $r+\sqrt{2}\frac{1-r}{8}<1$.
   It is obvious that $\frac{\sqrt{2}}{8}(1-r)<1-r$, which leading that $R_{⟨a,b⟩}\subset D$.

   (ii) Using (i) show that

   $$D=\bigcup _{⟨a,b⟩\in D}{R}_{⟨a,b⟩}.$$

   Proof.
   We have proved in (i), $R_{⟨a,b⟩}\subset D$ as known information. To prove the equation above, we have to prove that $D_{⟨a,b⟩}\subseteq \bigcup _{⟨a,b⟩\in D}R$. This relationship of containment is also given. Consider that for every given point in the disc $D$, where is always some space out of the $a^2+b^2$ but within disc $D$. Thus, by combining those information, we could say $D=\bigcup _{⟨a,b⟩\in D}{R}_{⟨a,b⟩}.$

   (iii) Deduce form (ii) that $D$ is a open set in ${\mathbb{R}}^2$.
   Proof.
   At this time, if we could prove that $D$ is a basis of the topology in ${\mathbb{R}}^2$, we could say that $D$ is open in ${\mathbb{R}}^2$. Let $R_1$ and $R_2$ as any two rectangles in $R$, we should check if $R_1\cap R_2$ is a union of members of $R$. We have already know this conclusion in 2.2.10 Remark. As $R_{⟨a,b⟩},a,b\in \mathbb{R}$ indeed is a basis of euclidean topology. The basis of euclidean topology on ${\mathbb{R}}^2$ generate all open sets. Even giving the limitation that $⟨a\pm \frac{1-r}{8},b\pm \frac{1-r}{8}⟩$ where $r<1$. As $D$ is a collection of all $R_{⟨a,b⟩}$, we could say $D$ is open set in ${\mathbb{R}}^2$, for it is a union of open sets in ${\mathbb{R}}^2$.

   (iv) Show that every disc $\{⟨x,y⟩:(x-a{)}^2+(y-b{)}^2<c^2,a,b,c\in \mathbb{R}\}$ is open in ${\mathbb{R}}^2$.
   Proof.
   We are required to verify the generalization of (iii) which is already proved previously.
   Let point $⟨d,e⟩$ be any point in the disc $D_2=\{⟨x,y⟩:(x-a{)}^2+(y-b{)}^2<c^2,a,b,c\in \mathbb{R}\}$. Put $r=\sqrt{(d-a{)}^2+(e-b{)}^2}$. Let $R_{⟨d,e⟩}$ be the open rectangle with vertices at the points $⟨d\pm \frac{c-r}{8},e\pm \frac{c-r}{8}⟩$. Just like proved in previous steps, generalization could be easily disposed.

   Comments: We are required to proved whether all of those discs could be generated by the basis $\mathcal{B}$ of euclidean topology ${\mathbb{R}}^2$. Or, in a more precise way, could those rectangles which are regard as basis of 2D plane cover any disc? If so, we could say any well-shaped generated geometric is a topology of 2D euclidean space. So, what is a topology here? We could simply conceptualize it as a trait within this space, it might have various shapes/looks, however, it must be generated by the basis of this space, for instance, in ${\mathbb{R}}^2$ one of the basis of topology is reflected as open rectangle which is cropped by lines parallel to $X$- or $Y$-axis.

2. Show that the collection of all open discs in ${\mathbb{R}}^2$ is a basis for a topology on ${\mathbb{R}}^2$.
   (i) Let $D_1$ and $D_2$ be any open discs in ${\mathbb{R}}^2$ with $D_1\cap D_2\ne \mathrm{\varnothing }$. If $⟨a,b⟩$ is any point in $D_1\cap D_2$, show that there exists an open disc $D_{⟨a,b⟩}$ with central $⟨a,b⟩$ such that $D_{⟨a,b⟩}\subset D_1\cap D_2$.
   Proof.
   Let $r_1$ and $r_2$ as the radius of $D_1$ and $D_2$, we have to proof that any point in $D_{⟨a,b⟩}$ meet that the sum of the distances to the centrals of $D_1$ and $D_2$ is less than $r_1+r_2$. Taking together, considering the given information that $D_1\cap D_2\ne \mathrm{\varnothing }$, and $D_1$, $D_2$ are open discs in ${\mathbb{R}}^2$,

   (ii) Show that

   $$D_1\cap D_2=\bigcup _{⟨a,b⟩\in D_1\cap D_2}{D}_{⟨a,b⟩}$$

   (iii) Using (ii) and Proposition 2.2.8, prove that the collection of all open discs in ${\mathbb{R}}^2$ is a basis for a topology on ${\mathbb{R}}^2$.

3. Let $\mathcal{B}$ be the collection of all open intervals $(a,b)$ in $\mathbb{R}$ with $a<b$ and $a$ and $b$ rational numbers. Prove that $\mathcal{B}$ is a basis for the euclidean topology on $\mathbb{R}$.

4. A topological space $X,\mathcal{T}$ is said to satisfy the second axiom of countability or to be second countable of there exists a basis $\mathcal{B}$ for $\mathcal{T}$, where $\mathcal{B}$ consists of only a countable number of sets.
   (i)
   (ii)
   (iii)
   (iv)

5. Prove that following statements.
   (i)
   (ii)
   (iii)
   (iv)
   (v)

6. Let ${\mathcal{B}}_1$ be a basis for a topology ${\mathcal{T}}_1$ on a set $X$ and ${\mathcal{B}}_2$ a basis for a topology ${\mathcal{T}}_2$ on a set $Y$. The set $X\times Y$ consists of all ordered pairs $⟨x,y⟩,x\in X$ and $y\in Y$. Let $\mathcal{B}$ be the collection of subsets of $X\times Y$ consisting of all the sets $B_1\times B_2$ where $B_1\in {\mathcal{B}}_1$ and $B_2\in {\mathcal{B}}_2$. Prove that $\mathcal{B}$ is a basis for a topology on $X\times Y$. The topology so defined is called the product topology on $X\times Y$.

7. Using Exercise 3 above and Exercises 2.1 #8, prove that every open subset of $\mathbb{R}$ is an $F_{\sigma }$-set and $G_{\delta }$-set.

2.3 Basis for a Given Topology

We have already known under what conditions a collection $\mathcal{B}$ of subsets of a set $X$ is a basis for some topology on $X$. However, we are sometimes given a topology $\mathcal{T}$ on $X$ and we want to know whether $\mathcal{B}$ is a basis for this specific topology $\mathcal{T}$.

Notation: There is a difference between saying that a collection $\mathcal{B}$ of subsets of $X$ is a basis for some topology, and saying that it is a basis for a given topology.

2.3.1 Example.
Let $\mathcal{B}$ be the collection of a half-open intervals of the form $(a,b],a<b$, where

References

Topology Without Tears by Sidney A. Morris. Topology book and Videos on Pure Mathematics, Topology and Writing Proofs supplementing the book