In anticipation of a course that I will soon be taking as a part of my first semester as a math Ph.D. student, I will be starting a straightforward series of blog entries containing fundamental results of general (point-set) topology (as well as other courses). Mind you, they will contain nothing extraordinary (I might introduce something interesting every once in a while, though); this is purely to keep track of the minutiae that I might easily forget down the road.
A topology on a set $X$ is a collection $\mathscr T$ of subsets of $X$ having the following properties:
- $\varnothing$ and $X$ are in $\mathscr T$.
- The union of the elements of any subcollection of $\mathscr T$ is in $\mathscr T$.
- The intersection of the elements of any finite subcollection of $\mathscr T$ is in $\mathscr T$.
A set $X$ for which a topology $\mathscr T$ has been specified is called a topological space.
Properly speaking, a topological space is an ordered pair $(X,\mathscr T)$ consisting of a set $X$ and a topology $\mathscr T$ on $X$.
If $X$ is a topological space with topology $\mathscr T$, we say that a subset $U$ of $X$ is an open set of $X$ if $U$ belongs to the collection $\mathscr T$.
Using this terminology, one can say that a topological space is a set $X$ together with a collection of subsets of $X$, called open sets, such that $\varnothing$ and $X$ are both open, and such that arbitrary unions and finite intersections of open sets are open.
If $X$ is any set, the collection of all subsets of $X$ is a topology on $X$; it is called the discrete topology. The collection consisting of $X$ and $\varnothing$ only is also a topology on $X$; we shall call it the indiscrete topology, or the trivial topology.
Let $X$ be a set; let $\mathscr T_f$ be the collection of all subsets $U$ of $X$ such that $X-U$ either is finite or is all of $X$. Then $\mathscr T_f$ is a topology on $X$, called the finite complement topology.
Suppose that $\mathscr T$ and $\mathscr T'$ are two topologies on a given set $X$. If $\mathscr T'\supset\mathscr T$, we say that $\mathscr T'$ is finer than $\mathscr T$; if $\mathscr T'$ properly contains $\mathscr T$, we say that $\mathscr T'$ is strictly finer than $\mathscr T$. We also say that $\mathscr T$ is coarser than $\mathscr T'$, or strictly coarser, in these two respective situations. We say that $\mathscr T$ is comparable with $\mathscr T'$ if either $\mathscr T'\supset\mathscr T$ or $\mathscr T\supset\mathscr T'$.
If $X$ is a set, a basis for a topology on $X$ is a collection $\mathscr B$ of subsets of $X$ (called basis elements) such that
- For each $x\in X$, there is at least one basis element $B$ containing $x$.
- If $x$ belongs to the intersection of two basis elements $B_1$ and $B_2$, then there is a basis element $B_3$ containing $x$ such that $B_3\subset B_1\cap B_2$.
If $\mathscr B$ satisfies these two conditions, then we define the topology $\mathscr T$ generated by $\mathscr B$ as follows: A subset $U$ of $X$ is said to be open in $X$ (that is, to be an element of $\mathscr T$) if for each $x\in U$, there is a basis element $B\in\mathscr B$ such that $x\in B$ and $B\subset U$. Note that each basis element is itself an element of $\mathscr T$.
Let us prove that the collection $\mathscr T$ generated by the basis $\mathscr B$ is, in fact, a topology on $X$. If $U$ is the empty set, it satisfies the defining condition of openness vacuously. Likewise, $X$ is in $\mathscr T$, since for each $x\in X$ there is some basis element $B$ containing $x$ and contained in $X$. Now let us take an indexed family $\left\{U_\alpha\right\}_{\alpha\in J}$ of elements of $\mathscr T$ and show that$$U=\bigcup_{\alpha\in J}U_\alpha$$belongs to $\mathscr T$. Given $x\in U$, there is an index $\alpha$ such that $x\in U_\alpha$. Since $U_\alpha$ is open, there is a basis element $B$ such that $x\in B\subset U_\alpha$. Then $x\in B$ and $B\subset U$, so that $U$ is open, by definition.
Now, let us take two elements $U_1$ and $U_2$ of $\mathscr T$ and show that $U_1\cap U_2$ belongs to $\mathscr T$. Given $x\in U_1\cap U_2$, choose a basis element $B_1$ containing $x$ such that $B_1\subset U_1$; choose also a basis element $B_2$ containing $x$ such that $B_2\subset U_2$. The second condition for a basis enables us to choose a basis element $B_3$ containing $x$ such that $B_3\subset B_1\cap B_2$. Then $x\in B_3$ and $B_3\subset U_1\cap U_2$, so $U_1\cap U_2$ belongs to $\mathscr T$, by definition.
Finally, we show by induction that any finite intersection $U_1\cap\cdots\cap U_n$ of elements of $\mathscr T$ is in $\mathscr T$. This fact is trivial for $n=1$; we suppose it true for $n-1$ and prove it for $n$. Now$$(U_1\cap\cdots\cap U_n)=(U_1\cap\cdots\cap U_{n-1})\cap U_n.$$By hypothesis, $U_1\cap\cdots\cap U_{n-1}$ belongs to $\mathscr T$; by the result just proved, the intersection of $U_1\cap\cdots\cap U_{n-1}$ and $U_n$ also belongs to $\mathscr T$.
Another way of describing the topology generated by a basis is given in the following lemma:
Lemma. Let $X$ be a set; let $\mathscr B$ be a basis for a topology $\mathscr T$ on $X$. Then $\mathscr T$ equals the collection of all unions of elements of $\mathscr B$.
Proof. Given a collection of elements of $\mathscr B$, they are also elements of $\mathscr T$. Because $\mathscr T$ is a topology, their union is in $\mathscr T$. Conversely, given $U\in\mathscr T$, choose for each $x\in U$ an element $B_x$ of $\mathscr B$ such that $x\in B_x\subset U$. Then $U=\bigcup_{x\in U}B_x$, so $U$ equals a union of elements of $\mathscr B$. $\blacksquare$
This lemma states that every open set $U$ in $X$ can be expressed as a union of basis elements. This expression for $U$ is not, however, unique.
Here is one way of obtaining a basis for a given topology:
Lemma. Let $X$ be a topological space. Suppose that $\mathscr C$ is a collection of open sets of $X$ such that for each open set $U$ of $X$ and each $x$ in $U$, there is an element $C$ of $\mathscr C$ such that $x\in C\subset U$. Then $\mathscr C$ is a basis for the topology of $X$.
Proof. We must show that $\mathscr C$ is a basis. The first condition for a basis is easy: Given $x\in X$, since $X$ is itself an open set, there is by hypothesis an element $C$ of $\mathscr C$ such that $x\in C\subset X$. To check the second condition, let $x$ belong to $C_1\cap C_2$, where $C_1$ and $C_2$ are elements of $\mathscr C$. Since $C_1$ and $C_2$ are open, so is $C_1\cap C_2$. Therefore, there exists by hypothesis an element $C_3$ in $\mathscr C$ such that $x\in C_3\subset C_1\cap C_2$.
Let $\mathscr T$ be the collection of open sets of $X$; we must show that the topology $\mathscr T'$ generated by $\mathscr C$ equals the topology $\mathscr T$. First, note that if $U$ belongs to $\mathscr T$ and if $x\in U$, then there is by hypothesis an element $C$ of $\mathscr C$ such that $x\in C\subset U$. It follows that $U$ belongs to the topology $\mathscr T'$, by definition. Conversely, if $W$ belongs to the topology $\mathscr T'$, then $W$ equals a union of elements of $\mathscr C$, by the preceding lemma. Since each element of $\mathscr C$ belongs to $\mathscr T$ and $\mathscr T$ is a topology, $W$ also belongs to $\mathscr T$. $\blacksquare$
When topologies are given by bases, it is useful to have a criterion in terms of the bases for determining whether one topology is finer than another. One such criterion is the following:
Lemma. Let $\mathscr B$ and $\mathscr B'$ be bases for the topologies $\mathscr T$ and $\mathscr T'$, respectively, on $X$. Then the following are equivalent:
Another way of describing the topology generated by a basis is given in the following lemma:
Lemma. Let $X$ be a set; let $\mathscr B$ be a basis for a topology $\mathscr T$ on $X$. Then $\mathscr T$ equals the collection of all unions of elements of $\mathscr B$.
Proof. Given a collection of elements of $\mathscr B$, they are also elements of $\mathscr T$. Because $\mathscr T$ is a topology, their union is in $\mathscr T$. Conversely, given $U\in\mathscr T$, choose for each $x\in U$ an element $B_x$ of $\mathscr B$ such that $x\in B_x\subset U$. Then $U=\bigcup_{x\in U}B_x$, so $U$ equals a union of elements of $\mathscr B$. $\blacksquare$
This lemma states that every open set $U$ in $X$ can be expressed as a union of basis elements. This expression for $U$ is not, however, unique.
Here is one way of obtaining a basis for a given topology:
Lemma. Let $X$ be a topological space. Suppose that $\mathscr C$ is a collection of open sets of $X$ such that for each open set $U$ of $X$ and each $x$ in $U$, there is an element $C$ of $\mathscr C$ such that $x\in C\subset U$. Then $\mathscr C$ is a basis for the topology of $X$.
Proof. We must show that $\mathscr C$ is a basis. The first condition for a basis is easy: Given $x\in X$, since $X$ is itself an open set, there is by hypothesis an element $C$ of $\mathscr C$ such that $x\in C\subset X$. To check the second condition, let $x$ belong to $C_1\cap C_2$, where $C_1$ and $C_2$ are elements of $\mathscr C$. Since $C_1$ and $C_2$ are open, so is $C_1\cap C_2$. Therefore, there exists by hypothesis an element $C_3$ in $\mathscr C$ such that $x\in C_3\subset C_1\cap C_2$.
Let $\mathscr T$ be the collection of open sets of $X$; we must show that the topology $\mathscr T'$ generated by $\mathscr C$ equals the topology $\mathscr T$. First, note that if $U$ belongs to $\mathscr T$ and if $x\in U$, then there is by hypothesis an element $C$ of $\mathscr C$ such that $x\in C\subset U$. It follows that $U$ belongs to the topology $\mathscr T'$, by definition. Conversely, if $W$ belongs to the topology $\mathscr T'$, then $W$ equals a union of elements of $\mathscr C$, by the preceding lemma. Since each element of $\mathscr C$ belongs to $\mathscr T$ and $\mathscr T$ is a topology, $W$ also belongs to $\mathscr T$. $\blacksquare$
When topologies are given by bases, it is useful to have a criterion in terms of the bases for determining whether one topology is finer than another. One such criterion is the following:
Lemma. Let $\mathscr B$ and $\mathscr B'$ be bases for the topologies $\mathscr T$ and $\mathscr T'$, respectively, on $X$. Then the following are equivalent:
- $\mathscr T'$ is finer than $\mathscr T$.
- For each $x\in X$ and each basis element $B\in\mathscr B$ containing $x$, there is a basis element $B'\in\mathscr B'$ such that $x\in B'\subset B$.
Proof. $(2)\implies(1)$. Given an element $U$ of $\mathscr T$, we wish to show that $U\in\mathscr T'$. Let $x\in U$. Since $\mathscr B$ generates $\mathscr T$, there is an element $B\in\mathscr B$ such that $x\in B\subset U$. Condition $(2)$ tells us there exists an element $B'\in\mathscr B'$ such that $x\in B'\subset B$. Then $x\in B'\subset U$, so $U\in\mathscr T'$, by definition.
$(1)\implies(2)$. We are given $x\in X$ and $B\in\mathscr B$, with $x\in B$. Now $B$ belongs to $\mathscr T$ by definition and $\mathscr T\subset\mathscr T'$ by condition $(1)$; therefore, $B\in\mathscr T'$. Since $\mathscr T'$ is generated by $\mathscr B'$, there is an element $B'\in\mathscr B'$ such that $x\in B'\subset B$. $\blacksquare$
We now define three topologies on the real line $\mathbb R$, all of which are of interest.
If $\mathscr B$ is the collection of all open intervals in the real line,$$(a,b)=\{x|a<x<b\},$$the topology generated by $\mathscr B$ is called the standard topology on the real line. Whenever we consider $\mathbb R$, we shall suppose it is given this topology unless we specifically state otherwise.
If $\mathscr B'$ is the collection of all half-open intervals of the form$$[a,b)=\{x|a\leq x<b\},$$where $a<b$, the topology generated by $\mathscr B'$ is called the lower limit topology on $\mathbb R$. When $\mathbb R$ is given the lower limit topology, we denote it by $\mathbb R_l$.
Finally, let $K$ denote the set of all numbers of the form $1/n$, for $n\in\mathbb Z_+$, and let $\mathscr B''$ be the collection of all open intervals $(a,b)$, along with all sets of the form $(a,b)-K$. The topology generated by $\mathscr B''$ will be called the $K$-topology on $\mathbb R$. When $\mathbb R$ is given this topology, we denote it by $\mathbb R_K$.
$(1)\implies(2)$. We are given $x\in X$ and $B\in\mathscr B$, with $x\in B$. Now $B$ belongs to $\mathscr T$ by definition and $\mathscr T\subset\mathscr T'$ by condition $(1)$; therefore, $B\in\mathscr T'$. Since $\mathscr T'$ is generated by $\mathscr B'$, there is an element $B'\in\mathscr B'$ such that $x\in B'\subset B$. $\blacksquare$
We now define three topologies on the real line $\mathbb R$, all of which are of interest.
If $\mathscr B$ is the collection of all open intervals in the real line,$$(a,b)=\{x|a<x<b\},$$the topology generated by $\mathscr B$ is called the standard topology on the real line. Whenever we consider $\mathbb R$, we shall suppose it is given this topology unless we specifically state otherwise.
If $\mathscr B'$ is the collection of all half-open intervals of the form$$[a,b)=\{x|a\leq x<b\},$$where $a<b$, the topology generated by $\mathscr B'$ is called the lower limit topology on $\mathbb R$. When $\mathbb R$ is given the lower limit topology, we denote it by $\mathbb R_l$.
Finally, let $K$ denote the set of all numbers of the form $1/n$, for $n\in\mathbb Z_+$, and let $\mathscr B''$ be the collection of all open intervals $(a,b)$, along with all sets of the form $(a,b)-K$. The topology generated by $\mathscr B''$ will be called the $K$-topology on $\mathbb R$. When $\mathbb R$ is given this topology, we denote it by $\mathbb R_K$.
Lemma. The topologies of $\mathbb R_l$ and $\mathbb R_K$ are strictly finer than the standard topology on $\mathbb R$, but are not comparable with one another.
Proof. Let $\mathscr T$, $\mathscr T'$, and $\mathscr T''$ be the topologies of $\mathbb R$, $\mathbb R_l$, and $\mathbb R_K$, respectively. Given a basis element $(a,b)$ for $\mathscr T$ and a point $x$ of $(a,b)$, the basis element $[x,b)$ for $\mathscr T'$ contains $x$ and lies in $(a,b)$. On the other hand, given the basis element $[x,d)$ for $\mathscr T'$, there is no open interval $(a,b)$ that contains $x$ and lies in $[x,d)$. Thus $\mathscr T'$ is strictly finer than $\mathscr T$.
A similar argument applies to $\mathbb R_K$. Given a basis element $(a,b)$ for $\mathscr T$ and a point $x$ of $(a,b)$, this same interval is a basis element for $\mathscr T''$ that contains $x$. On the other hand, given the basis element $B=(-1,1)-K$ for $\mathscr T''$ and the point $0$ of $B$, there is no open interval $(a,b)$ that contains $0$ and lies in $B$.
To show that $\mathbb R_l$ and $\mathbb R_K$ are not comparable with one another, observe that given a basis element $[x,d)$ for $\mathscr T'$, there is no set for $\mathscr T''$ that contains $x$ and lies in $[x,d)$ because they are all open. On the other hand, given the basis element $B=(-1,1)-K$ for $\mathscr T''$ and the point $0$ of $B$, there is no half-open interval $[a,b)$ for $\mathscr T'$ that contains $0$ and lies in $B$. $\blacksquare$
Since the topology generated by a basis $\mathscr B$ may be described as the collection of arbitrary unions of elements of $\mathscr B$, what happens if you start with a given collection of sets and take finite intersections of them as well as arbitrary unions? This question leads to the notion of a subbasis for a topology.
A subbasis $\mathscr S$ for a topology on $X$ is a collection of subsets of $X$ whose union equals $X$. The topology generated by the subbasis $\mathscr S$ is defined to be the collection $\mathscr T$ of all unions of finite intersections of elements of $\mathscr S$.
I will later check that $\mathscr T$ is indeed a topology...