📂Topology

Subbasis in Topology

Definition ¹

For a topological space $\left( X , \mathscr{T} \right)$, let $\mathscr{S} \subset \mathscr{T}$.

When $\displaystyle \mathscr{B} = \left\{ \left. B = \bigcap_{ i = 1}^{n} S_{i} \ \right| \ S_{i} \in \mathscr{S} \right\}$ becomes a basis of $\mathscr{T}$, $\mathscr{S}$ is called a subbasis of $\mathscr{T}$.

Explanation

The reason a subbasis is hard to accept is that, in mathematics, the prefix ‘sub-’ usually indicates that something is a subset while still retaining the original properties. For example, a subgroup is when a subset satisfies the conditions of a group, and a subspace is when a subset satisfies the conditions of a space. In this sense, one could say that a subbasis is harder because of its terminology than its concept.

To begin with, by definition, if $\mathscr{S}$ has become a subbasis, then it being a subset of the basis $\mathscr{B}$, that is $\mathscr{S} \subset \mathscr{B}$, is trivial. However, since $\mathscr{S}$ must construct $\mathscr{B}$ as the set of all finite intersections in order to become a basis, one could describe it as still being immature as a basis.

Rather, considering the mathematics we have seen so far, it is more natural to say that a subbasis is the basis of a basis. The problem is that, even after one has accepted the ‘sub-’ part, the definition of a subbasis itself is still complicated and bizarre. Regarding this aspect, it is more comfortable to simply accept it as something you have to learn later for the sake of products of topological spaces. Once you have studied that to some extent, you will also come to understand why finite intersections in particular are considered.

Now let us grasp the concept a little through an example.

Example

Show that $\mathscr{S} = \left\{ (- \infty , b ), ( a , \infty ) \ | \ a,b \in \mathbb{R} \right\}$ is a subbasis of the metric space $\mathbb{R}$.

Solution

Since the set of all open intervals $(a,b)$ can be formed as the intersection of the two open intervals $( - \infty , b )$ and $( a , \infty )$, $\mathscr{S}$ is a subbasis.

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