I've translated the following from my German textbook, so please correct me if there is something wrong or strange.


Let $X$ be a set and $\mathfrak{T} \subseteq \mathcal{P}(X)$ with fits the following restrictions:

(i) $\emptyset, X \in \mathfrak{T}$
(ii) $\forall U_1, U_2 \in \mathfrak{T}: U_1 \cap U_2 \in \mathfrak{T}$
(iii) Let $I$ be an index set with $\forall i \in I: U_i \in \mathfrak{T} $. Then: $\bigcup_{i \in I} U_i \in \mathfrak{T}$

Then $\mathfrak{T}$ is called a topology and $(X, \mathfrak{T})$ is called a topological space

Let $(X, \mathfrak{T})$ be a topological space. $\mathcal{S} \subseteq \mathfrak{T}$ is called a subbasis of $\mathfrak{T} : \Leftrightarrow \forall U \in \mathfrak{T}: U$ is a union of finite intersections from Elements in $\mathcal{S}$.


Can a finite subbasis generate an infinite topology? Or, more formally, is the following implication true:

$$|\mathcal{S}| \in \mathbb{N} \Rightarrow |\mathfrak{T}| \in \mathbb{N}$$

Can a finite subbasis generate any topology for an infinite space $X$? So, formally:

$$|\mathcal{S}| \in \mathbb{N} \Rightarrow |X| \in \mathbb{N}$$


No. If $\mathcal{S}$ is a finite, set, then there are only finitely many (nonempty) finite subcollections of that set, and so the family $\mathcal{I}$ of intersections of (nonempty) finite subcollections of $\mathcal{S}$ is also finite (saying "finite" here is somewhat redundant). As there are only finitely many subcollections of $\mathcal{I}$, there are only finitely many distinct unions of subcollections of $\mathcal{I}$. Thus any topology generated by a finite subbasis is itself finite.

In particular, the usual metric topology on $\mathcal{R}$ has no finite subbase.

(Even more, given an arbitrary subbasis $\mathcal{S}$ on a set $X$, either $\mathcal{S}$ and $\mathcal{I}$ (as described above) above are both finite, or are both infinite and of the same cardinality. It follows that the topology generated by $\mathcal{S}$ has cardinality bounded above by some cardinal related to $| \mathcal{S} |$. In the finite case, it would be $2^{2^{|\mathcal{S}|}}$, and in the infinite case it is $2^{|\mathcal{S}|}$.)

  • $\begingroup$ Is a subcollection the same as a subset of a set of sets? So $A \subseteq \mathfrak{T}$ is always a subcollection? $\endgroup$ – Martin Thoma Feb 21 '14 at 19:38
  • $\begingroup$ @moose: Yes, they are almost always used as synonyms (as well as "subfamily" and some other similar terms). $\endgroup$ – user642796 Feb 21 '14 at 19:43
  • $\begingroup$ Thanks for your answer. It took me a while and I needed to talk about it with other students to really understand it. Seems quite simple now, just like in this comic) $\endgroup$ – Martin Thoma Feb 24 '14 at 20:33

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