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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 next chapter: Morris-Topology Without Tears-CH02

2024-06-06 Book Notes

Notes

Introduction

• Playing with Mathematics is just like playing sports
• Writing mathematical proofs is a kind of skill
• A Mathematical proof is a watertight argument
  • Beginning: Setups information
  • Processing: Logical argument
  • End: Proofs required
• One should participate

Topology

Let's give some examples,

Let $X=\{a,b,c,d,e,f\}$ and ${\mathcal{T}}_1=\{\{a\},\{a,b\},\{a,c\},\{a,b,c\},\mathrm{\varnothing },X\}$.

Then, ${\mathcal{T}}_1$ is a topology on $X$ as it satisfies axiom(i), (ii) and (iii) of Definitions 1.1.1.

• for (i), $X$ and $\mathrm{\varnothing }$ do belong to ${\mathcal{T}}_1$;
• for (ii), the union of any number of sets (for $\{a\}$, count up the combinations one by one, $\{a\},\{a,b\},\{a,c\},\{a,b,c\},\{a\},\{X\}$) belong to ${\mathcal{T}}_1$,
• for (iii), the intersection of any two sets, let's just list up every combination of $\{a,b\}$ here, as $\{a\},\{a,b\},\{a\},\{a,b\},\{a,b\},\mathrm{\varnothing }$, skip the rest of them, all the intersections belong to ${\mathcal{T}}_1$.

Let ${\mathcal{T}}_2=\{\{a\},\{a,b\},\{a,c\},\mathrm{\varnothing },X\}$.

Then, ${\mathcal{T}}_2$ is not a topology on $X$ as it does not satisfy axiom(ii) of Definition 1.1.1.

• The union of $\{a,b\}$ and $\{a,c\}$ is $\{a,b,c\}$ is not included in ${\mathcal{T}}_2$

Let ${\mathcal{T}}_3=\{\{a\},\{a,b\},\{a,c\},\{b,c\},\{a,b,c\},\mathrm{\varnothing },X\}$.

Then, ${\mathcal{T}}_3$ is not a topology on $X$ as it does not satisfy axiom(iii) of Definition 1.1.1.

• The intersection of $\{a,b\}$ and $\{b,c\}$, $\{b\}$ does not belong to ${\mathcal{T}}_3$.

Interestingly, if let $\mathbb{N}$ be the set of all natural numbers, and let ${\mathcal{T}}_{\mathcal{4}}$ consist of $\mathbb{N}$, $\mathrm{\varnothing }$, and all finite subsets of $\mathbb{N}$. Then ${\mathcal{T}}_4$ is not a topology on $\mathbb{N}$. Since, we could combine an infinite union like $\{2,3,...,n,...\}$, and this infinite union does not belong to ${\mathcal{T}}_4$.

$\mathcal{T}$ in Definition 1.1.6 does satisfy the conditions of Definitions 1.1.1 and so is indeed a topology.

We could check out that $\mathcal{T}$ satisfies the conditions of Definition 1.1.1 and so is indeed a topology.

The 1st example of Proof.

1.1.8 Example. If $X=\{a,b,c\}$ and $\mathcal{T}$ is a topology on $X$ with $\{a\}\in \mathcal{T}$, $\{b\}\in \mathcal{T}$, $\{c\}\in \mathcal{T}$, prove that $\mathcal{T}$ is the discrete topology.

Proof.
Let's focus on the given information first. The set $X$ has 3 elements, thus it has 8 distinct subsets:
$S_1=\{\mathrm{\varnothing }\}$,
$S_2=\{a\}$,
$S_3=\{b\}$,
$S_4=\{c\}$,
$S_5=\{a,b\}$,
$S_6=\{a,c\}$,
$S_7=\{b,c\}$,
and $S_8=\{a,b,c\}$
We are required to prove that each of these subsets is in $\mathcal{T}$. There are 2 lines involved the requirement which $\mathcal{T}$ is a topology on $X$, and $\{a\}$, $\{b\}$, $\{c\}$ is already in $\mathcal{T}$ partially.
For given $\mathcal{T}$ is a topology in $X$, following the definition, we know that $\mathrm{\varnothing }$ and $X$ should be in $\mathcal{T}$. Thus, we could suggest that $S_1$ and $S_8$ is in $\mathcal{T}$. Besides, the unions of each element should also be in $\mathcal{T}$, which infers that $S_5$ ~ $S_7$ are also in $\mathcal{T}$. And for the given information, $S_2$ ~ $S_4$ is also in $\mathcal{T}$, finally we could prove that $\mathcal{T}$ is the discrete topology which means that all the subsets of $X$ are in $\mathcal{T}$.

The Proof above lead 1.1.9 Proposition which could be regarded as a more generalized proposition.

We have to prove it without list up a few special cases, but the general case of an arbitrary non-empty set $X$. According to the proof in 1.1.8, we know the core of this question is to prove that every subset of $X$ is a union of singleton subsets of $X$, and we already know that all the singleton subsets are in $\mathcal{T}$. Thus, we could record the fact in the following way:

Exercises 1.1

1. Let $X=\{a,b,c,d,e,f\}$. Determine whether or not each of the following collections of subsets of $X$ is a topology on $X$:
   (a) $\mathcal{T}=\{X,\mathrm{\varnothing },\{a\},\{a,f\},\{b,f\},\{a,b,f\}\}$;
   (b) $\mathcal{T}=\{X,\mathrm{\varnothing },\{a,b,f\},\{a,b,d\},\{a,b,d,f\}\}$;
   (c) $\mathcal{T}=\{X,\mathrm{\varnothing },\{f\},\{e,f\},\{a,f\}\}$.
   $$
   Answer:
   Reminding of the definition of topology, if all the unions of each element also be in $\mathcal{T}$, we could say that $\mathcal{T}$ is a topology on $X$.
   Following this line, $\mathcal{T}$ in (a) and (b) are topology on $X$, but $\mathcal{T}$ in (c) is not. Because, the union of $\{e,f\}$ and $\{a,f\}$, namely $\{a,e,f\}$ is not in $\mathcal{T}$.
   $$

2. Let $X=\{a,b,c,d,e,f\}$. Which of the following collections of subsets of $X$ is a topology on $X$? (Justify your answers.)
   (a) $\mathcal{T}=\{X,\mathrm{\varnothing },\{c\},\{b,d,e\},\{b,c,d,e\},\{b\}\}$;
   (b) $\mathcal{T}=\{X,\mathrm{\varnothing },\{a\},\{b,d,e\},\{a,b,d\},\{a,b,d,e\}\}$;
   (c) $\mathcal{T}=\{X,\mathrm{\varnothing },\{b\},\{a,b,c\},\{d,e,f\},\{b,d,e,f\}\}$;
   $$
   Answer:
   According to the definition of topology, $\mathcal{T}$ in (a) is not a topology on $X$, because the union of $\{b\}$ and $\{c\}$ is not in $\mathcal{T}$. Similarly, $\mathcal{T}$ in (c) is not a topology on $X$, because the union of $\{a,b,c\}$ and $\{d,e,f\}$ is not in $\mathcal{T}$. However, $\mathcal{T}$ in (b) is a topology on $X$, for every combination of the unions of elements is in $\mathcal{T}$.
   $$

3. If $X=\{a,b,c,d,e,f\}$ and $\mathcal{T}$ is the discrete topology on $X$, which of the following statements are true?
   (a) $X\in \mathcal{T}$; (b) $\{X\}\in \mathcal{T}$; (c) $\{\mathrm{\varnothing }\}\in \mathcal{T}$; (d) $\mathrm{\varnothing }\in \mathcal{T}$;
   (e) $\mathrm{\varnothing }\in X$; (f) $\{\mathrm{\varnothing }\}\in X$; (g) $\{a\}\in \mathcal{T}$; (h) $a\in \mathcal{T}$;
   (i) $\mathrm{\varnothing }\subseteq X$; (j) $\{a\}\in X$; (k) $\{\mathrm{\varnothing }\}\subseteq X$; (l) $a\in X$;
   (m) $X\subseteq \mathcal{T}$; (n) $\{a\}\subseteq \mathcal{T}$; (o) $\{X\}\subseteq \mathcal{T}$; (p) $a\subseteq \mathcal{T}$.
   $$
   Answer:
   According to the given information: definition of topology, and proposition 1.1.9 which generated the definition of discrete topology ($\mathcal{T}=\{X,\mathrm{\varnothing },\{a\},\{b\},\{c\},\{d\},\{e\},\{f\},\{a,b\},...\}$), (a), (d), (g) are true for the definition of topology; (e) and (l) are true for the definition of a set; (o) is true for every $x\in \{X\}$, namely $\mathrm{\varnothing }$ and $\{a,b,c,d,e,f\}$ are also in $\mathcal{T}$. → which could echo to the notion of topology as it involves "sets of sets".
   $$

4. Let $(X,\mathcal{T})$ be any topological space. Verify that the intersection of any finite number of members of $\mathcal{T}$ is a member of $\mathcal{T}$.
   $$
   Answer:
   Given $(X,\mathcal{T})$ is any topological space, $\mathcal{T}$ is a topology on $X$. According to (iii) in definition of a topology, intersection of any two sets in $\mathcal{T}$ is in $\mathcal{T}$. Thus, it means that, for example, given $\{a,b,c\}$, $\{a,b\}$, $\{a\}$, there are three pair-wise intersections, namely, $\{a,b\}$, $\{a\}$, $\{a\}$. The intersection of all of them is $\{a\}$, which, should be one of those pair-wise intersections. Besides, because $\mathrm{\varnothing }$ is also in $\mathcal{T}$, even the intersection of any two sets is empty, the empty set is also in $\mathcal{T}$. Thus, we could easily prove that any finite number of members of $\mathcal{T}$ is in $\mathcal{T}$.
   $$

5. Let $\mathbb{R}$ be the set of all real numbers. Prove that each of the following collections of subsets of $\mathbb{R}$ is a topology.
   (i) ${\mathcal{T}}_1$ consists of $\mathbb{R}$, $\mathrm{\varnothing }$, and every interval $(-n,n)$, for $n$ any positive integer, where $(-n,n)$ denotes the set $\{x\in \mathbb{R}:-n<x<n\}$;
   (ii) ${\mathcal{T}}_2$ consists of $\mathbb{R}$, $\mathrm{\varnothing }$, and every interval $[-n,n]$, for $n$ any positive integer, where $[-n,n]$ denotes the set $\{x\in \mathbb{R}:-n\le x\le n\}$;
   (iii) ${\mathcal{T}}_3$ consists of $\mathbb{R}$, $\mathrm{\varnothing }$, and every interval $[n,\mathrm{\infty })$, for $n$ any positive integer, where $(-n,n)$ denotes the set $\{x\in \mathbb{R}:n\le x\}$;
   $$
   Answer:
   (i) Considering the definition of a topology on $X$, ${\mathcal{T}}_1$ should consist of $\mathbb{R}$ and $\mathrm{\varnothing }$ first, it is already given. Then we should check if union of any members of sets of ${\mathcal{T}}_1$ and intersection of any two sets of ${\mathcal{T}}_1$ is also belong to ${\mathcal{T}}_1$. Easy to image that union of any members of sets of ${\mathcal{T}}_1$ should be the interval $(-n,n)$ which $n$ is a big enough positive integer. And the intersection of any two sets of ${\mathcal{T}}_1$ must be one set of those two. Thus, we could say that ${\mathcal{T}}_1$ is a topology on $\mathbb{R}$.
   And by using basically same logics, we know it is also true for interval $[-n,n]$ in (ii), but we should also take care of that every interval contains positive integer $n$.
   Finally, back to the definition of topology, the union of any number of sets, whether finite or infinite, in $\mathcal{T}$ belongs to $\mathcal{T}$. According to the given information, we could easily know that the union of any number of sets should be the largest interval, and this specified interval also belongs to "every interval $[n,\mathrm{\infty })$". And for $\{x\in \mathbb{R}:n\le x\}$, given any positive integer $n$, intersection of any two sets of "every interval $[n,\mathrm{\infty })$" should also belong to $[n,\mathrm{\infty })$. Thus, we could prove that ${\mathcal{T}}_3$ is a topology on $\mathbb{R}$.
   $$

6. Let $\mathbb{N}$ be the set of all positive integers. Prove that each of the following collections of subsets of $\mathbb{N}$ is a topology.
   (i) ${\mathcal{T}}_1$ consists of $\mathbb{N}$, $\mathrm{\varnothing }$, and every set $\{1,2,...,n\}$, for $n$ any positive integer. (This is called the initial segment topology.)
   (ii) ${\mathcal{T}}_2$ consists of $\mathbb{N}$, $\mathrm{\varnothing }$, and every set $\{n,n+1,...\}$, for $n$ any positive integer. (This is called the final segment topology.)
   $$
   Answer:
   Let's repeat the definition of a topology again, and we could check out that the union of any number of sets $\{1,2,...,n\}$ belong to $\{1,2,...,n\}$. And the intersection of any two sets is also in $\{1,2,...,n\}$. We could say that ${\mathcal{T}}_1$ is a topology on $\mathbb{N}$. For the same reason, it is easy to image that ${\mathcal{T}}_2$ is also a topology on $\mathbb{N}$.
   $$

7. List all possible topologies on the following sets:
   (a) $X=\{a,b\}$;
   (b) $Y=\{a,b,c\}$.
   $$
   Answer:
   (a) ${\mathcal{T}}_{a1}=\{X,\mathrm{\varnothing },\{a\}\}$, ${\mathcal{T}}_{a2}=\{X,\mathrm{\varnothing },\{b\}\}$,
   ${\mathcal{T}}_{a3}=\{X,\mathrm{\varnothing },\{a\},\{b\}\}$ are all possible topologies on $X$.
   (b) ${\mathcal{T}}_{b1}=\{X,\mathrm{\varnothing },\{a\}\}$, ${\mathcal{T}}_{b2}=\{X,\mathrm{\varnothing },\{b\}\}$, ${\mathcal{T}}_{b3}=\{X,\mathrm{\varnothing },\{c\}\}$,
   ${\mathcal{T}}_{b4}=\{X,\mathrm{\varnothing },\{a,b\}\}$, ${\mathcal{T}}_{b5}=\{X,\mathrm{\varnothing },\{a,c\}\}$, ${\mathcal{T}}_{b6}=\{X,\mathrm{\varnothing },\{b,c\}\}$,
   ${\mathcal{T}}_{b7}=\{X,\mathrm{\varnothing },\{a\},\{a,b\}\}$, ${\mathcal{T}}_{b8}=\{X,\mathrm{\varnothing },\{b\},\{a,b\}\}$, ${\mathcal{T}}_{b9}=\{X,\mathrm{\varnothing },\{a\},\{b\},\{a,b\}\}$,
   ${\mathcal{T}}_{b10}=\{X,\mathrm{\varnothing },\{a\},\{a,c\}\}$, ${\mathcal{T}}_{b11}=\{X,\mathrm{\varnothing },\{c\},\{a,c\}\}$, ${\mathcal{T}}_{b12}=\{X,\mathrm{\varnothing },\{a\},\{c\},\{a,c\}\}$,
   ${\mathcal{T}}_{b13}=\{X,\mathrm{\varnothing },\{b\},\{b,c\}\}$, ${\mathcal{T}}_{b14}=\{X,\mathrm{\varnothing },\{c\},\{b,c\}\}$, ${\mathcal{T}}_{b15}=\{X,\mathrm{\varnothing },\{b\},\{c\},\{b,c\}\}$,
   ${\mathcal{T}}_{b16}=\{X,\mathrm{\varnothing },\{a\},\{a,b\},\{a,c\}\}$, ${\mathcal{T}}_{b17}=\{X,\mathrm{\varnothing },\{b\},\{a,b\},\{b,c\}\}$, ${\mathcal{T}}_{b18}=\{X,\mathrm{\varnothing },\{c\},\{a,c\},\{b,c\}\}$,
   ${\mathcal{T}}_{b19}=\{X,\mathrm{\varnothing },\{a\},\{b\},\{c\},\{a,b\},\{a,c\},\{b,c\}\}$are topologies on $Y$.
   $$

8. Let $X$ be an infinite set and $\mathcal{T}$ is a topology on $X$. If every infinite subset of $X$ is in $\mathcal{T}$, prove that $\mathcal{T}$ is the discrete topology.
   $$
   Answer:
   According to the definition of a discrete topology, if $\mathcal{T}$ is a collection of all subsets of $X$, then we should call $\mathcal{T}$ is a discrete topology on $X$. We also know the given information that every infinite subset of $X$ is in $\mathcal{T}$, and $X$ is a non-empty set, precisely, an infinite set. Since every infinite subsets of $X$ is in $\mathcal{T}$. Thus, exactly what should we do here is, to prove every finite subset of $X$ is in $\mathcal{T}$.

   Proof:
   Consider $S=X\setminus \{x\}$, for some $x\in X$. Since the $\{x\}$ is any singleton in $X$ and $X$ is an infinite set, $S$ is also infinite and thus in $\mathcal{T}$. As the given information, every infinite subset of $X$ is in $\mathcal{T}$.
   Since, $X=S\cup \{x\}=(X\setminus \{x\})\cup \{x\}$, and according to the proposition which suggests that the union of any members of open sets is also open, $\{x\}$ should be both "in $\mathcal{T}$" and "out of $\mathcal{T}$" (see following section). $\{x\}$ should be in $\mathcal{T}$ for if it is not, every infinite set in $\mathcal{T}$ which is the union of it and its complement will not be in $\mathcal{T}$ anymore.
   Then, for $\{x\}$ could be any singleton in $X$, we could say every singleton and any finite members union of singletons are also in $\mathcal{T}$. Thus, $\mathcal{T}$ is a discrete topology for its all subsets are in $\mathcal{T}$. $◻$

   Comments: I edited this proof at 09/22/2024, for the original one was wrong.

1.2 Open Sets, Closed Sets, and Clopen Sets

We call "members of $\mathcal{T}$" as "open sets" for convenience.

And we could get following proposition:

Why infinite intersections of open sets are not always open? Just give one counterexample could explain it: Let $\mathbb{N}$ be the set of all positive integers and let $\mathcal{T}$ consist of $\mathrm{\varnothing }$ and each subset $S$ of $\mathbb{N}$ such that the complement of $S$ in $\mathbb{N}$, namely $\mathbb{N}\setminus S$ is a finite set (Comments: it is confusing, due to the complex expression. The core definition is let $\mathbb{N}\setminus S$ be finite, thus, each $S$ is infinite open set on $\mathcal{T}$). $\mathcal{T}$ is a topology on $\mathbb{N}$, for each natural number $n$, define the set $S_n$ as :

Example of $S_n$ for better understanding:
$S_1=\{1,2,3,...\}$;
$S_2=\{1,3,4,...\}$;
$S_3=\{1,4,5,...\}$;
$S_4=\{1,5,6,...\}$...

Each $S_n$ is an open set in the topology $\mathcal{T}$, its complement is finite (such as, $\mathrm{\varnothing }$, $\{2\}$, $\{3\}$...), however, $\bigcap _{n=1}^{\mathrm{\infty }}{S}_n=\{1\}$ is not open, which is the intersection of the open sets $S_n$.

It is interesting that, the definition of "open set" and "closed set" is quite dialectical. For example, let $(X,\mathcal{T})$ is an indiscrete space, then we know that $X$, and $\mathrm{\varnothing }$ belongs to $\mathcal{T}$, they are open sets. Obviously, their complements in $X$, namely, $\mathrm{\varnothing }$ and $X$ are closed sets.

Proof. I would like to skip (i) as I have mentioned previously. Let's prove (iii) first, let finite number sets $S_1$, $S_2$,..$S_n$ be closed sets. Then we should prove that $S_1\cup S_2\cup ...\cup S_n$ is a closed set. It is equal to prove that $X\setminus (S_1\cup S_2\cup ...\cup S_n)$ is an open set. We know that,

And, as we defined previously, $S_1$, $S_2$,...$S_n$ are closed sets, $X\setminus S_1$, $X\setminus S_2$,...$X\setminus S_n$ are open sets. According to Proposition 1.2.2, the intersection of finite number of open sets is also an intersection, thus, the right side of formula above is an open set, meaning that (iii) is true.

With the same logics, let any number of closed sets $S_1$, $S_2$,..,$S_n$ be closed set. Then we should prove that $S_1\cap S_2\cap ...\cap S_n$ is a closed set. We might to prove that $X\setminus (S_1\cap S_2\cap ...\cap S_n$) is an open set.

According to Proposition 1.2.2, the union of any number of open sets is open set. So, we know that the complement of $S_1\cap S_2\cap ...\cap S_n$ is also open set, it means that $S_1\cap S_2\cap ...\cap S_n$ is closed set.

We could easily figure out the followings:

• In every topological space $(X,\mathcal{T})$ both $X$ and $\mathrm{\varnothing }$ are clopen
• In a discrete space all subsets of $X$ are clopen
• In an indiscrete space the only clopen subsets are $X$ and $\mathrm{\varnothing }$

Exercises 1.2

1. List all subsets of set $X$ in Example 1.1.2. Write down, next to each set, whether it is (i) clopen; (ii) neither open nor closed; (iii) open but not closed; (iv) closed but not open.
   $$
   Answer:
   I won't list up all of them, writing down 64 subsets and detecting whether they are open, closed, or clopen are too heavy. Just some of them..., $X$, $\mathrm{\varnothing }$, and $\{a\}$ is clopen, $\{b\},\{c\}$ is neither open nor closed, ..., $\{a,b,c,d,e\}$ is neither open nor closed, and $\{b,c,d,e,f\}$ is clopen.
   $$

2. Let $(X,\mathcal{T})$ be a topological space with the property that every subset is closed. Prove that it is a discrete space.
   $$
   Answer:
   The given information is in a topological space $(X,\mathcal{T})$, any subset $S$ of $X$ is closed, to prove all the subsets of $X$ belongs to $\mathcal{T}$.
   Proof. In a topological space $(X,\mathcal{T})$, let $S$ be a subset of $X$, according to the definitions of open/closed set, the complement of $S$ should be open, thus we got: $\mathrm{\forall }(X\setminus S)\in \mathcal{T}$. According to Proposition 1.2.2, the union of any number of open sets should be an open set, $\bigcup (X\setminus S_n)\in \mathcal{T}$. It is easy to image that $\bigcup (X\setminus S_n)=\bigcup S_n$, which means that the collection of all subsets of $X$ belongs to $\mathcal{T}$, exactly the definition of discrete space.
   $$

3. Observe that if $X,\mathcal{T}$ is a discrete space or an indiscrete space, then every open set is a clopen set. Find a topology $\mathcal{T}$ on the set $X=\{a,b,c,d\}$ which is not discrete and is not indiscrete but has the property that every open set is clopen.
   $$
   Answer:
   Consider a topology $\mathcal{T}=\{,X,\mathrm{\varnothing },\{a,b\},\{c,d\}\}$ on $X$, it might satisfy the lines.
   $$

4. Let $X$ be an infinite set. If $\mathcal{T}$ is a topology on $X$ such that every infinite subset of $X$ is closed, prove that $\mathcal{T}$ is the discrete topology.
   $$
   Answer:
   If $\mathcal{T}$ is a topology on $X$, and the property that, every infinite subset of $X$ is closed. Let any infinite subset $S$ of $X$ is closed, thus, any complement of infinite subset $X\setminus S$ is open. If $A=X\setminus S$ is infinite, the union of any number of them should also be open. And, any singleton $\{x\}=X\setminus S$ is also open, leading any finite union is also open. Since, we have known that $\mathcal{T}$ is a topology on $X$, and any subset (finite and infinite) is open, $\mathcal{T}$ is a discrete topology.
   $$

5. Let $X$ be an infinite set and $\mathcal{T}$ a topology on $X$ with the property that the only infinite subset of $X$ which is open is $X$ itself. Is $(X,\mathcal{T})$ necessarily an indiscrete space?
   $$
   Answer:
   Let $A=\{a\}$ a finite subset of $X$, then we could easily know $X\setminus A$ is a infinite subset of $X$. According to the given condition, "the only infinite subset of $X$ which is open is $X$ itself", thus, $X\setminus A$ is close, and singleton $\{a\}$ is open. However, once there is more element than $\mathrm{\varnothing }$ and $X$ in $\mathcal{T}$, $\mathcal{T}$ should not be an indiscrete space, then we could say that in such case, $(X,\mathcal{T})$ is necessarily an indiscrete space.

6. "I would like to skip this question".
   $$

7. "I would like to skip this question".

1.3 The Finite-Closed Topology

Sometimes, it is nature to describe the topology by saying which sets are closed.

1.3.3 Example

Let $\mathcal{T}$ be the finite-closed topology on a set $X$. If $X$ has at least 3 distinct clopen subsets, prove that $X$ is a finite set.

Proof. For given information that $\mathcal{T}$ is a finite-closed topology, which means that $X$ and all finite subsets of $X$ are closed. Then, there are at least 3 distinct clopen subsets, for every topology, $\mathrm{\varnothing }$ and $X$ are clopen. Thus, we could know that there should be a subset $S$ of $X$ is clopen. For $S$ is open, its complement $X\setminus S$ is closed, since, "the closed subsets of $X$ are $X$ and all finite subsets of $X$". Further,
because of the distinctness, $S\ne X$ and $S\ne \mathrm{\varnothing }$, closed $S$ and closed $X\setminus S$ should be finite. Additionally, $X=S\cup X\setminus S$, the union of finite subsets should also be finite.

Let's review some basics of function:

1.3.9 Examples.

Let $(Y,\mathcal{T})$ be a topological space and $X$ a non-empty set. Further, let $f$ be a function from $X$ into $Y$. Put ${\mathcal{T}}_1=\{f^{-1}(S):S\in \mathcal{T}\}$. Prove that ${\mathcal{T}}_1$ is a topology on $X$.

Proof.
For the given information that $f$ is a function(projection) from $X$ into $Y$. And as we know the definition of a topology, $Y$ should belongs to $\mathcal{T}$, and as we know $X=f^{-1}(Y)$, which could lead $X\in {\mathcal{T}}_1$. Follow the same logics, $\mathrm{\varnothing }$ is also in ${\mathcal{T}}_1$. It satisfies the condition (i) of topology.

Then we have to verify condition (ii) and condition (iii), first let $\{A_j:j\in J\}$ be a collection of members of ${\mathcal{T}}_1$, and $J$ as some index set. As $A_j=f^{-1}{B}_j$, where $B_j\in \mathcal{T}$. We could easily know that $A_j\in {\mathcal{T}}_1$. As $\bigcup _{j\in J}{A}_j=\bigcup _{j\in J}{f}^{-1}(B_j)=f^{-1}(\bigcup _{j\in J}{B}_j)$, and $\mathcal{T}$ is a topology which leading $\bigcup _{j\in J}{B}_j\in \mathcal{T}$. Thus, $f^{-1}(\bigcup _{j\in J}{B}_j)\in {\mathcal{T}}_1$, which means that $\bigcup _{j\in J}{A}_j\in {\mathcal{T}}_1$, and satisfies the condition (ii). For the same logics, let $A_1$ and $A_2$ be in ${\mathcal{T}}_1$, and $B_1$, $B_2$ are their inverse images in $\mathcal{T}$. Since $A_1\cap A_2=f^{-1}(B_1)\cap f^{-1}(B_2)=f^{-1}(B_1\cap B_2)$, and we have already known that $f^{-1}(B_1\cap B_2)\in {\mathcal{T}}_1$. Hence $A_1\cap A_2\in {\mathcal{T}}_1$. All conditions are verified, ${\mathcal{T}}_1$ is a topology on $X$.

Exercises 1.3

I would like to do exercises in next week.

1.4 Comments

It takes long time to understand the core of topology and practice, but it is interesting and I might overestimate the hardness of self-learning topology. Just in my institution, it just like to learn a language which is similar to your mother-tongue, because some fields of knowledge from mathematics is already installed, such of how one could give out definitions and prove. The rest of learning might be imaging and practicing how to express the relationships and contexts with beautiful formats.

Indeed, expressing and explaining the definition and playing it with your own viewpoints is also important. For me, it just like some kind of process to liberate yourself.

References

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