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:


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 $\mathcal{T}$.


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?

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    $\begingroup$ It's not essential, and neither is the notion of a basis. But both notions are useful. $\endgroup$ – Alex Kruckman Sep 27 '17 at 0:34
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    $\begingroup$ 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 $x\in (a,b)\subseteq U$. $\endgroup$ – Alex Kruckman Sep 27 '17 at 0:35
  • $\begingroup$ @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? $\endgroup$ – Saaqib Mahmood Sep 27 '17 at 5:05
  • $\begingroup$ 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? $\endgroup$ – Alex Kruckman Sep 27 '17 at 13:10
  • $\begingroup$ 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. $\endgroup$ – Alex Kruckman Sep 27 '17 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).

  • $\begingroup$ 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. $\endgroup$ – Saaqib Mahmood Sep 27 '17 at 4:58

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.


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.


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