Definition of basis of topological space

Question

I'm currently studying topology, and at present, I'm learning the concept of the basis of a topological space. The textbook I'm using is "Topology" by Munkres, and the definition of the basis in the book is as follows:

Definition 1 If $X$ is a set, a basis for a topology on $X$ is a collection $\mathcal{B}$ of subsets of $X$ (called basis elements) such that

1. For each $x \in X$, there is at least one basis element $B$ containing $x$.
2. 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 $B_3\subset B_1\cap B_2$.

However, I've also learned from other textbooks that a basis can also be defined in the following way.

Definition 2 Let $(X, \tau)$ be a topological space and $\mathcal{B}$ be a collection of open sets in $\tau$. If any open set in $\tau$ can be expressed as the union of a sub - collection of sets from $\mathcal{B}$, then $\mathcal{B}$ is called a basis.

My Confusion

• In Definition 1, the basis is a collection of subsets of $X$ that satisfies two properties. There is no pre-specified topology. It only states that it is the basis for some topology. However, in Definition 2, a specific topology $\tau$ is pre-specified, and a collection of subsets is selected from $\tau$, and this collection satisfies those properties. So, for a basis, the question is whether a topology should be pre-specified or there is no need for pre-specifying a topology.
• Are these two definitions in conflict? What is the connection between them? Are they the same? Or is the first definition more general?

Answer 1

Let us see what is happening here.

Suppose that $X$ is a non-empty set. A collection $\mathcal{B}$ of subsets of $X$ is called a basis for a topology on $X$ if (i) For each $x\in X$ there exists a $B\in\mathcal{B}$ such that, $x\in B$ and (ii) For every $B_1,B_2\in\mathcal{B}$ for which $B_1\cap B_2\neq\phi$ whenever, $y\in B_1\cap B_2$ there exists a $B_3\in\mathcal{B}$ such that $y\in B_3$ and $B_3\subseteq B_1\cap B_2$.

Note that in this definition a priori there is no topology on $X$. But in fact, there exists a smallest topology on $X$ containing the sets in $\mathcal{B}$. To see this define, $\tau_\mathcal{B}:=\{U\subseteq X\,:\,\forall\,x\in U\, \exists B\in\mathcal{B}\,\text{such that}\,x\in B\,\text{and} \,B\subseteq U\}$. It is easy to check that $\tau_\mathcal{B}$ is the smallest topology on $X$ containing $\mathcal{B}$ as a subset. This is called the topology on $X$ generated by $\mathcal{B}$. Now why this definition is important ? Let us illustrate it with an easy example. Consider the set $X=C[0,1]$. It is a vector space over $\mathbb{R}$. For each $p\in [0,1]$ define, $ev_p:X\to\mathbb{R}$ by, $ev_p(f)=f(p)$. Then with respect to norm topology on $X$ these maps are continuous. But if anyone asks what's the smallest topology on $X$ for which these maps are continuous ? It's hard to specify all the open sets in a concrete way. But this can be done by using the concept for basis for a topology. Let us see how can we do it ?

We want all the $g_p$ to be continuous. So for any open set $V$ in $\mathbb{R}$ we want $g_p^{-1}(V)$ to be open in $X$. So, for every finite set $F$ of $[0,1]$ we want, $\bigcap_{i€F} g_i^{-1}(V_i)$ to be open in $[0,1]$. So this topology must have these open sets. One can also see that subsets of these form form a basis for a topology on $X$. So the smallest topology containing these subsets is precisely the smallest topology on $X$ wrt which these maps are continuous. Now you ask why are we interested in smallest such topologies. One common intention to reduce the number of open sets so that some non-compact subsets becomes compact in new topology without hampering the continuity of the maps in question. This allows us to use compactness in different important fields of mathematics. You can see Banach-Alaoglu theorem to better understand what I am saying.

Now coming to definition 2. In this case you already have a topology. It is analogous to vector space (in this case topological space) and it's basis (basis for the existing topology). It's easy (?) to study properties of vector spaces in terms of its basic elements. These elements carry all the algebraic information of a vector space ie. they determine your vector space uniquely up to isomorphisms. Same thing happens with basis for a existing topology. Continuity, limits can be defined in terms of basis for the existing topology. Sometimes it is convenient because there are usually concrete definitions of basic open sets but no such concrete descriptions of a general open set. So studying basis for existing topology makes our life easier.

Two definitions has two different use cases. Definition 1 helps us to define complicated topologies using simple starting "open sets" and the definition 2 helps us to manipulate continuity and limits in a very efficient manner.

But these two definitions don't conflict each other. Since, in $\tau_\mathcal{B}$, $\mathcal{B}$ then becomes a basis for the existing topology. Also you can try to show that if $\mathcal{C}$ is already a basis for some existing topology then the topology generated by $\mathcal{C}$ is exactly the pre-existing topology.

I think it will be now clear to you.

Answer 2

So, for a basis, the question is whether a topology should be pre-specified or there is no need for pre-specifying a topology.

That is really a philosophical question.

Suppose you have a space $(X, \mathcal{T})$ and some collection of subsets $\mathcal{B} \subseteq \mathcal{P} (X)$.

It is possible for $\mathcal{B}$ to satisfy Munkres' definition without satisfying the second definition. In this case, $\mathcal{B}$ is still a basis, just not one that "corresponds to" $\mathcal{T}$. In other words, the collection $\mathcal{B}$ does not carry much meaning in the context of $\mathcal{T}$. For example, a set $U \in \mathcal{T}$ may be not be expressible as the union of some subset of $\mathcal{B}$, or there may be some union of elements of $\mathcal{B}$ that is not contained in $\mathcal{T}$. One reason why bases are nice is that they "simplify" thinking about the topology in some sense - sets are open if and only if they equal to some arbitrary union of basis elements.

But, a $\mathcal{B}$ that satisfies your second definition automatically satisfies Munkres' definition. First, note that $X \in \mathcal{T}$ by definition of a topology. So, there must be some union of elements of $\mathcal{B}$ that equals $X$. This means there is some element of $\mathcal{B}$ that contains $x$, for any $x \in X$. Now suppose $x \in B_1 \cap B_2$, where $B_1, B_2 \in \mathcal{B}$. $B_1$ and $B_2$ are themselves unions of elements of $\mathcal{T}$ (note that $B_1 = \bigcup \{ B_1 \}$ and $B_2 = \bigcup \{ B_2 \}$). Since $\mathcal{T}$ is closed to finite intersection by the definition of a topology, then $B_1 \cap B_2 \in \mathcal{T}$. By hypothesis, there is some union of elements of $\mathcal{B}$ that equals $B_1 \cap B_2$. This means there is some element of $B_3 \in \mathcal{B}$ such that $x \in B_3$ and $B_3 \subseteq B_1 \cap B_2$, as desired. Therefore, Munkres' definition can be viewed as a property that is satisfied by any basis.

Munkres' definition is useful because given $\mathcal{B}$ satisfying its preconditions, there exists a topology for which $\mathcal{B}$ is a basis (where "basis" here is per your second definition). Specifically, consider the collection $\mathcal{S} \subseteq \mathcal{P} (X)$ such that $U \in \mathcal{S}$ if and only if for each $x \in U$, there is some $B \in \mathcal{B}$ such that $x \in B$ and $B \subseteq U$. $\mathcal{S}$ is a topology and has $\mathcal{B}$ as a basis, facts which you can easily check.

Answer 3

Definition 1 defines when $\mathcal{B}$ is a basis.

Definition 2 defines when $\mathcal{B}$ is a basis-for-$\tau$.

Here are two facts:

• If $\tau$ is some topology and $\mathcal{B}$ is a basis-for-$\tau$, then $\mathcal{B}$ is a basis.
• If $\mathcal{B}$ is a basis, then it is a basis-for-$\tau_\mathcal{B}$, where $\tau_\mathcal{B}$ is a topology whose open sets are exactly the unions of $\mathcal{B}$ elements.