More on basis

A topology smaller than other

1. Smaller topology

   If $\tau$ and $\tau'$ are topologies on $X$, where $\tau\subseteq\tau'$, we say that $\tau$ is smaller than $\tau'$. (Not coarser than $\tau'$, as in the book).

2. Condition of smaller topology

   Let $\mathcal{B},\mathcal{B'}$ be bases for topologies on $X$. Then the following are equivalent:

   • $\tau_{\mathcal{B}}$ is smaller than $\tau_{\mathcal{B'}}$.
   • For each $x\in X$ and $B\in \mathcal{B}$ with $x\in B$, there is $B'\in \mathcal{B'}$ with $x\in B'\subseteq B$.

Proof of the lemma

• Suppose $\tau_{\mathcal{B}}\subseteq\tau_{\mathcal{B'}}$, and let $x\in B\in \mathcal{B}$.
• Since $B\in \mathcal{B}\subseteq\tau_{\mathcal{B}}\subseteq\tau_{\mathcal{B'}}$, by definition of the generated topology $\tau_{\mathcal{B'}}$, there is $B'\in \mathcal{B'}$ such that $x\in B'\subseteq B$.
• Now assume the second condition, and let $U\in\tau_{\mathcal{B}}$. We have to prove $U\in\tau_{\mathcal{B'}}$.
• Let $x\in U$. Since $U\in\tau_{\mathcal{B}}$, there is $B\in \mathcal{B}$ with $x\in B\subseteq U$.
• By the assumed condition, there is $B'\in \mathcal{B'}$ such that $x\in B'\subseteq B$.
• Hence $x\in B'\subseteq U$, which proves $U\in\tau_{\mathcal{B'}}$. $\Box$

From a topology to a base

1. Lemma

   Let $(X,\tau)$ be a topological space. Let $\mathcal{C}\subseteq\tau$ be such that for any $U\in\tau$ and $x\in U$, there is $C\in \mathcal{C}$ such that $x\in C\subseteq U$. Then $\mathcal{C}$ is base for a topology on $X$, and $\tau_{\mathcal{C}}=\tau$.

2. Proof

   • The collection $\mathcal{C}$ satisfies (B1) by hypothesis.
   • Let $C_{1},C_{2}\in \mathcal{C}$ and $x\in C_{1}\cap C_{2}$. Since $C_{1}\cap C_{2}$ is open, by hypothesis there is $C_{3}\in \mathcal{C}$ such that $x\in C_{3}\subseteq C_{1}\cap C_{2}$. So $\mathcal{C}$ satisfies (B2) and is base for a topology.
   • Now, to prove $\tau_{\mathcal{C}}=\tau$, let $U\in \tau_{\mathcal{C}}$. Since $U$ is union of elements of $\mathcal{C}$, and these are in $\tau$, then $U\in\tau$.
   • Finally, if $U\in\tau$, then for any $x\in U$, by our hypothesis, there is $C\in\mathcal{C}$ such that $x\in C\subseteq U$. Hence $U\in\tau_{\mathcal{C}}$. $\Box$

Why bases are useful

• We will now give several examples of topological spaces defined by bases. As a matter of fact, most of the topologies we meet in the course are given that way.
• Many theorems on certain topologies are easier to prove using properties of the base that generate them.

Standard topology on $\mathbb{R}$

1. The standard topology

   If $X=\mathbb{R}$, the collection of all open intervals $(a,b)\subseteq \mathbb{R}$ with $a<b$, for $a,b\in\mathbb{R}$ is a base for a topology, called standard topology.

Sorgenfrey line

1. The Sorgenfrey line

   The collection of all intervals of the form

   $$[a,b)=\{x\in \mathbb{R}\mid a\leq x<b\}$$

   for $a,b\in \mathbb{R}$, is also a base for a topology on $\mathbb{R}$. The topological space obtained is called the Sorgenfrey line.

2. Sorgenfrey line is bigger

   The Sorgenfrey topology is strictly bigger than the standard topology.

Subbases

Subbases

1. Subbase

   Let $\mathcal{S}\subseteq P(X)$. We say that $\mathcal{S}$ is subbase of a topology on $X$ if $\cup \mathcal{S}=X$.

2. Topology generated by a subbase

   The collection of all finite intersections of elements of a subbase $\mathcal{S}$ is a base for a topology on $X$.

Proof

• Let $\mathcal{B}$ be the collection of all finite intersections of elements of $\mathcal{S}$. Observe that $\mathcal{S}\subseteq\mathcal{B}$
• Given $x\in X$, by our hypothesis, there is $S\in \mathcal{S}$ such that $x\in S$. Since $S\in \mathcal{B}$, we have (B1).

• Let $B_{1},B_{2}\in \mathcal{B}$, and $x\in B_{1}\cap B_{2}$. Suppose that there are $S_{1},\ldots,S_{n},S'_{1},\ldots,S'_{m}\in \mathcal{S}$ such that:

  $$B_{1}=S_{1}\cap\cdots\cap S_{n},\quad B_{2}=S'_{1}\cap\cdots\cap S'_{m}.$$

• Then, taking

  $$B_{3}=S_{1}\cap\cdots\cap S_{n}\cap S'_{1}\cap\cdots\cap S'_{m},$$

  we obtain $B_{3}\in \mathcal{B}$, and $x\in B_{3}\subseteq B_{1}\cap B_{2}$. Hence (B2) is satisfied. $\Box$

Exercises

1. Let $X=\{1,2,3,4,5,6\}$, and $\mathcal{S}=\{\{1,2,3\},\{3,4,5\},\{4,5,6\}\}$. Enumerate the open sets of $\tau_{\mathcal{S}}$.
2. Show that if $\mathcal{S}$ is a subbase for a topology on $X$, then $\tau_{\mathcal{S}}$ is equal to the intersection of the topologies on $X$ that contain $\mathcal{S}$.
3. Let $\mathcal{S}=\{[a,b]\subseteq \mathbb{R}\mid a<b,a,b\in \mathbb{Q}\}$. Prove that $\mathcal{S}$ is a subbase for a topology. Prove also that if $\mathcal{B}=\mathcal{S}\cup\{\{a\}\mid a\in \mathbb{Q}\}$, then $\mathcal{B}$ is base of a topology, and $\tau_{\mathcal{S}}=\tau_{\mathcal{B}}$.
4. Given a natural number $n$, denote by $[n]=\{kn\in \mathbb{N}\mid k\in \mathbb{N}\}$. Prove that $\mathcal{B}=\{[n]\mid n\in \mathbb{N}\}$ is a base for a topology on $\mathbb{N}$. Show also that $\mathcal{S}=\{[p^{r}]\mid p\text{ is prime},r\in \mathbb{N}\}$ is a subbase for a topology on $\mathbb{N}$, and $\tau_{\mathcal{S}}=\tau_{\mathcal{B}}$.

Links

• Lower limit topology - Wikipedia, the free encyclopedia
• Subbase - Wikipedia, the free encyclopedia