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I know that every topology is generated by a basis. Is it true that every topology has a subbasis which generates it? If not, what makes it not possible to "synthesize" a subbasis out of a basis?

Having studied linear algebra, I am intuitively comfortable with the idea of basis, but it is not clear to me why we are interested in subbasis in the first place.

Jennifer
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    Any topology T is a base for itself. Any base for T is a sub-base for T, Various bases and sub-bases are useful for constructing different kinds of topologies. A collection of subsets of $X$ is a sub-base for a topology on $ X iff $\cup Y=X$. – DanielWainfleet Mar 30 '16 at 16:45

2 Answers2

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Of course it is true, just take the topology itself.

When it comes to the concept of a subbasis, it is not important to identify a subbasis of a given topology (whereas for bases it can be important), because this concept appears in the context of constructing a topology out of it (by brute force, namely, taking all unions and finite intersections, the space itself and $\varnothing$).

We are interested in the concept of subbasis because sometimes we want a certain collection of subsets to be open and thus we consider the coarsest topology that contains this collection. With a basis you cannot always do this, because the collection needs to satisfy ceratin properties for it to be a basis for $\textit{some}$ topology.

Marco Flores
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Thm:

Any class $\mathcal{A}$ of subsets of a non-empty set $X$ is the subbase for a unique topology on $X$. That is, finite intersections of members of $\mathcal{A}$ form a base for a topology $\mathcal{T}$ on $X$.

Proof:

We show that the class $\mathcal{B}$ of finite intersections of members of $\mathcal{A}$ satisfies the two conditions for it to be a base for a topology on $X$:

$(i)$ $X = \bigcup\{B: B \in \mathcal{B} \} $

$(ii)$ For any $G, H \in \mathcal{B}$, $\:G \cap H$ is the union of members of $\mathcal{B}$.

Note $X \in \mathcal{B}$, since $X$ by definition is the empty intersection of members of $\mathcal{A}$; so

$X = \bigcup\{B: B \in \mathcal{B} \} $

Furthermore, if $G, H \in \mathcal{B}$, then $G$ and $H$ are finite intersections of members $\mathcal{A}$. Hence $G \cap H$ is also a finite intersection of members of $\mathcal{A}$ and therefore belongs to $\mathcal{B}$.

Accordingly, $\mathcal{B}$ is a base for a topology $\mathcal{T}$ on $X$ for which $\mathcal{A}$ is a subbase.

It is a straight forward exercise to show the uniqueness.

Community
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JKnecht
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