I was reading Topology from Munkres and got confused by the definition of a subbasis. What is/are the difference between basis and subbasis in a topology?


4 Answers 4


Bases and subbases "generate" a topology in different ways. Every open set is a union of basis elements. Every open set is a union of finite intersections of subbasis elements.

For this reason, we can take a smaller set as our subbasis, and that sometimes makes proving things about the topology easier. We get to use a smaller set for our proof, but we pay for it; with a subbasis we need to worry about finite intersections, whereas we did not have to worry about that in the case of a basis.

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    $\begingroup$ Is it correct to say that if an element $x$ belongs to a sub-basis element $S_{1}$ then it also belongs to the basis element $S_{1}$ because $S_{1}$ is simply a finite intersection of $S_{1}$ with itself? $\endgroup$ Commented Mar 9, 2020 at 7:58
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    $\begingroup$ Yes. One way to say it is that if I have a subbasis S, then there is a basis B associated with that subbasis which is all the finite intersections of elements of S. And $S \subset B$. Remember also that bases/subbases for a topology are not unique: some bases/subbases for a topology $\mathcal{T}$ may include your set $S_1$ while others exclude it. $\endgroup$
    – Eric Auld
    Commented Mar 10, 2020 at 8:33

Consider $S=\{\{0,1\},\{0,2\}\}$. What is the topological space $T(S)$ generated by $S?$ By definition, $S$ will then be a subbasis of $T(S)$.

Well, we want to all requirements to hold true and find that $T(S) = \{\emptyset, \{0\}, \{0,1\}, \{0,2\}, \{0,1,2\}\}$ (check this!).

Is $S$ a basis? No, because you cannot write $\{0\}$ as a union of any elements in $S$.

So you see that subbasis and basis are two different notions, even for a very basic example.

A subbasis can be thought of, and is actually defined to be, the "smallest set that becomes my topological space if I complete it under the property of being a topological space, i.e. fulfiling the axioms of topological space".

The two terms are related nevertheless. Every basis is a subbasis, and in one of the equivalent definitions of subbasis you will find that you already get a basis from your subbasis.

  • $\begingroup$ Shouldn't a subbasis be able to produce every element of the topology by building the union of intersections of elements of the subbasis, where the intersections have to be finite? Something like: Let $(X, T)$ be the topological space and $S$ be a subbasis. Then: $\forall U \in T: U = \cup_{A = M_i \cap M_j \cap ... with M_{i,j} \in S}$ where all $A$ should be finite. How do you make the empty set this way? $\endgroup$ Commented Feb 4, 2014 at 17:04
  • $\begingroup$ @moose You can make the empty set with the empty union of empty intersection. In other words the union of intersection of no sets. Kind of like how the vector space $\{0\}$ is generated by the basis $\varnothing$ since $0$ is an empty sum. $\endgroup$ Commented May 24, 2014 at 23:21
  • $\begingroup$ @Pratyush What do you mean by "the empty union of empty intersection"? Could you clarify this please? $\endgroup$ Commented Feb 5, 2015 at 18:55
  • $\begingroup$ @MadHatter I think I made a typo ... it should be "or" instead of "of". $\endgroup$ Commented Feb 6, 2015 at 8:41
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    $\begingroup$ @Prayutsh I think that it would be better to give a bit deeper explanation. moose question arised because David goes from subbase to topology directly, not mentioning basis. If you first take basis $B=\{\{0\},\{0,1\},\{0,2\}\}$ generated by $S$ (as all finite intersections of elements of $S$), then you can see how to generate emptyset, since in case of $B$ you form unions over its subsets, not elements (like in case of intersections and S). So $\emptyset\subseteq B$ and $\bigcup\emptyset=\emptyset$. By the way, empty intersection doesn't work. $\endgroup$ Commented Feb 6, 2015 at 9:32

The collection of sets $(-\infty,b)$ and $(a,\infty)$ for $a,b\in \mathbb{R}$ constitute a sub-basis for the standard topology on $\mathbb{R}$. The collection of sets $(a,b)$ for $a,b\in \mathbb{R}$ constitute a basis for the standard topology on $\mathbb{R}$. I suggest that you look at the definitions of "basis" and "sub-basis" and convince yourself that the claims that I have made are correct; this is probably the best way to answer your question.

The basic idea is that a basis is the collection of all finite intersections of sub-basis elements. The open sets in a topology are all possible unions of basis elements. So, the open sets in a topology are all possible unions of finite intersections of sub-basis elements.

I hope that answers your question!

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    $\begingroup$ The "all" is very important, which hasn't been provided in other answers. $\endgroup$ Commented Sep 17, 2019 at 19:07

If we ignore, momentarily, the fact that we are trying to generate a topology, a subbasis is any old collection of subsets of the space. Thus, any basis is a subbasis. However, a basis $\mathcal B$ must satisfy the criterion that if $U,V\in\mathcal B$ and $x$ is an arbitrary point in both $U$ and $V$, then there is some $W$ belonging to $\mathcal B$ such that $x\in W\subseteq U\cap V$.

To really understand what that means you have to look at examples. The family $\{X\}$ is a basis for the indiscrete topology on $X$. The family $\mathcal P(X)$, the powerset of $X$, is a basis for the discrete topology on $X$. If $X$ is large enough and you remove some subset $S\subset X$ from $\mathcal P(X)$, you can still have a basis if $S$ is not singleton but $\mathcal P(X)\setminus \{x\}$ is merely a subbasis when $x\in X$.

A more ambitious example is the definition of the topology on the product of two spaces $X$ and $Y$. The family $$\{U\times Y:U\subseteq X, U\text{ is open}\}\cup\{X\times V:V\subseteq Y,V\text{ is open}\}$$ is a subbasis but not necessarily a basis of the box topology on $X\times Y$.

Finally, suppose a family $\mathcal B$ is a basis. You can generate a topology $\tau_0$ using $\mathcal B$ as a subbasis and you can generate a topology $\tau_1$ using $\mathcal B$ as a basis. The topologies are actually the same, as you can see here. You just have to do more work to generate a topology from a subbasis $\mathcal S$, i.e. you have to include all finite intersections of members of $\mathcal S$ because it is not guaranteed that they are unions of members of $\mathcal S$.

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    $\begingroup$ could you explain a bit why in the example of the product of two spaces $X,Y$; $\{U\times Y:U\subseteq X, U\text{ is open}\}\cup\{X\times V:V\subseteq Y,V\text{ is open}\}$ is a subbasis but not basis ? $\endgroup$
    – ElleryL
    Commented Sep 18, 2019 at 4:16
  • $\begingroup$ @ElleryL the basis for the topology on $X \times Y$ is sets of the form $U \times V$ where $U,V$ are open in $X,Y$ respectively. If we have any set where say, $V$ is not all of $Y$ but some proper open subset, then there is no way to construct it using only unions from the above subbasis. Instead we need to intersect $U\times X$ with $X \times V$ to get $U \times V$ $\endgroup$ Commented Feb 22, 2023 at 20:17

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