# Why Is The Concept Of Subbasis For A Topology Essential?

I know that there are topologies that have to be defined in terms of a basis, for example, the standard topology on $$\mathbb{R}$$. I'm wondering if there is an examples of a topology that necessarily has to be defined in terms of a sub-basis?

Here are the relevant definitions from Munkres:

Basis:

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 $$B_3 \subset B_1 \cap B_2$$.

If $$\mathcal{B}$$ satisfies these two conditions, then we define the topology $$\mathcal{T}$$ generated by $$\mathcal{B}$$ as follows: A subset $$U$$ of $$X$$ is said to be open in $$X$$ (that is, to be an element of $$\mathcal{T}$$) if for each $$x \in U$$, there is a basis element $$B \in \mathcal{B}$$ such that $$x \in B$$ and $$B \subset U$$. Note that each basis element is itself an element of $$\mathcal{T}$$.

Subbasis:

A subbasis $$\mathcal{S}$$ for a topology on $$X$$ is a collection of subsets of $$X$$ whose union equals $$X$$. The topology generated by the subbasis $$\mathcal{S}$$ is defined to be the collection $$\mathcal{T}$$ of all unions of finite intersections of elements of $$\mathcal{S}$$.

Now is there an example of a topology that necessarily has to be defined in terms of a subbasis?

The product topology on an infinite Cartesian product of topological space is a candidate, but even there one can simply characterize the topology in terms of a basis. Am I right?

• It's not essential, and neither is the notion of a basis. But both notions are useful. Commented Sep 27, 2017 at 0:34
• Why do you think the standard topology on $\mathbb{R}$ must be defined in terms of a basis? I could just as well say that the topology consists of all sets $U$ with the property that for all $x\in U$, there is an open interval $(a,b)$ with $x\in (a,b)\subseteq U$. Commented Sep 27, 2017 at 0:35
• @AlexKruckman yes, that is true. But here you are using a basis for characterizing open sets. Isn't the collection of all open intervals a basis for the standard topology? Commented Sep 27, 2017 at 5:05
• Or I could say the topology contains all sets whose complements are closed under taking limits of Cauchy sequences. Is a basis for the topology hiding in this description? Commented Sep 27, 2017 at 13:10
• Or I could say it's the subspace topology inherited from the standard topology on $\mathbb{C}$, which I then describe in some other way. Commented Sep 27, 2017 at 13:12

If $\mathcal{S}$ is a subbasis for $\mathcal{T}$, then the set of all finite intersections of elements of $\mathcal{S}$ is a basis for $\mathcal{T}$. So there's never going to be a context where it's much harder to describe a basis than a subbasis (although describing a subbasis might be slightly simpler).

• yes, it is so. But what is not clear to me is any situation where we have to start with a sub-basis in order to specify a topology. Any example of a situation where we cannot give a topology in terms of simply a sub-basis, just as we cannot give the standard topology on $\mathbb{R}$ without giving a basis for it. Infering from Munkres' presentation, I assume that the concept of sub-basis is to follow, and not precede, that a basis. Commented Sep 27, 2017 at 4:58

There are situations (for example, weak and weak* topologies in functional analysis) where we have a set $X$ and some functions $f_i:X\to Y_i$ from $X$ into some topological spaces $Y_i$, and we want to topologize $X$ so that these functions $f_i$ become continuous. Of course, we could just give $X$ the discrete topology; then all functions from $X$ to any topological space will be continuous. But that's usually making far more subsets of $X$ open than we actually need. Suppose we want the smallest topology $T$ on $X$ making the $f_i$"s continuous. So, for each $i$ and each open $U\subseteq Y_i$, we need $f_i^{-1}(U)\in T$. Of course, to be a topology, $T$ must also include finite intersections and arbitrary unions of whatever sets are in $T$. An efficient way to define the desired $T$ is to say that the sets $f_i^{-1}(U)\in T$ (for $U$ open in $Y_i$) constitute a subbase.

More generally, given any family $\mathcal F$ of subsets of a set $X$, there is a smallest topology on $X$ such that all the sets in $\mathcal F$ are open. That topology has $\mathcal F$ as a subbase.

A subbase can be a handy tool. The main examples of spaces most easily described by subbases are ordered topological spaces $(X,<)$, where the subbase is given by all sets of the form $L(a) = \{x: x < a\}$, $a \in X$ together with all sets of the form $U(a) = \{x: x > a\}$, $a \in X$. (lower and upper sets are open). The base generated by this subbase also includes the open intervals, as $(a,b) = U(a) \cap L(b)$.

Another example is the standard subbase for $\prod_{i \in I} X_i$, the Cartesian product of topological spaces $X_i$, is given by $\{\pi_i^{-1}[O]: i \in I, O \subseteq X_i \text{ open }\}$. The corresponding base has open sets that depend on finitely many coordinates, as is well-known.

It's easy to show that if we have a subbase $\mathcal{S}$ for the topology of $X$, that

a function $f: Y \to X$ is continuous iff $f^{-1}[S]$ is open in $Y$ for all $S \in \mathcal{S}$.

An immediate consequence is that a function $Y \to\prod_{i \in I} X_i$ is continuous iff $\pi_i \circ f$ is continuous for all $i \in I$.

A deeper fact is the Alexander subbase theorem:

$X$ is compact iff every open cover of $X$ by members of $\mathcal{S}$ has a finite subcover.

We quite easily get that a product of compact spaces is compact: suppose all $X_i$ are compact and take an open cover $\{O_j: j \in J\}$ of $X = \prod_{i \in I} X_i$ by subbase elements, so that we write each $O_j = \pi_{i(j)}^{-1}[U_j]$, with $U_j$ open in $X_{i(j)}$. Claim: there is some $i_0 \in I$ such that $\mathcal{O_i} = \{U_j : i(j) = i_0\}$ is an open cover of $X_{i_0}$, because if this were not the case we could pick $p_i \in X_i\setminus \bigcup \mathcal{O}_i$ for each $i$, and then $p := (p_i)_{i \in I}$ would not be covered by any member of $\mathcal{O}$, so this cannot be. For such an $i_0$ we find a finite subcover of $\mathcal{O}_i$, say $U_{j_1}, \ldots U_{j_n}$, where all $i(j_k) = i_0$, and then note that the corresponding $O_{j_1} = \pi_{i_0}^{-1}[U_{j_1}], \ldots,O_{j_n} = \pi_{i_0}^{-1}[U_{j_n}]$ are a finite subcover of $\mathcal{O}$. By Alexander's subbase theorem the product is compact.

As an exercise for the interested reader we also can prove easily from Alexander's subbase theorem that an ordered topological space $(X,<)$ is compact iff every subset of $X$ has a supremum.

I think applications as these show that subbases can be handy tools for compactness or continuity proofs.