I am new to Topology, and I was reading up the definition of subbasis, and a point that says that "a Topology generated by the subbasis S is defined to be the collection T of all unions of finite intersections of elements of the subbasis S."

I do not understand this. so for example if we have X={a,b,c,d} and a subbasis S={(a,b},{a,c},{a,b,d},{a}}, how do we find T(S)?


Just follow the definition. Since $S$ is finite, the family of all finite intersections of elements of $S$ is just the family of all intersections of elements of $S$. Intersecting a single element of $S$, we just get $S$ back, so we get the sets $\{a,b\},\{a,c\},\{a,b,d\}$, and $\{a\}$. There are six pairs of sets in $S$ that we can intersect:

$$\begin{align*} &\{a,b\}\cap\{a,c\}=\{a\}\;,\\ &\{a,b\}\cap\{a,b,d\}=\{a,b\}\;,\\ &\{a,b\}\cap\{a\}=\{a\}\;,\\ &\{a,c\}\cap\{a,b,d\}=\{a\}\;,\\ &\{a,c\}\cap\{a\}=\{a\}\;,\text{ and}\\ &\{a,b,d\}\cap\{a\}=\{a\}\;. \end{align*}$$

As you can see, this did not give us anything new: $\{a\}$ and $\{a,b\}$ were already in the subbase $S$.

There are four triples of sets that we can intersect:

$$\begin{align*} &\{a,b\}\cap\{a,c\}\cap\{a,b,d\}=\{a\}\;,\\ &\{a,b\}\cap\{a,c\}\cap\{a\}=\{a\}\;,\\ &\{a,b\}\cap\{a,b,d\}\cap\{a\}=\{a\}\;,\text{ and}\\ &\{a,c\}\cap\{a,b,d\}\cap\{a\}=\{a\}\;. \end{align*}$$

Once again we got nothing new. Finally, we can intersect all four members of $S$, but that also just gives us $\{a\}$ back. Thus, $S$ is already closed under taking finite intersections and is therefore a base for a topology on $X$. That topology is the family of all unions of members of the base, i.e., in this case the family of all unions of members of $S$.

  • The union of no members of $S$ is $\varnothing$.
  • The unions of one-member families from $S$ are the members of $S$: $\{a,b\},\{a,c\},\{a,b,d\}$, and $\{a\}$.
  • The unions of two-member families from $S$ are: $$\begin{align*}&\{a,b\}\cup\{a,c\}=\{a,b,c\}\;,\\&\{a,b\}\cup\{a,b,d\}=\{a,b,d\}\;,\\&\{a,b\}\cup\{a\}=\{a,b\}\;,\\&\{a,c\}\cup\{a,b,d\}=X\,\\&\{a,c\}\cup\{a\}=\{a,c\}\;,\text{ and}\\&\{a,b,d\}\cup\{a\}=\{a,b,d\}\;.\end{align*}$$

I’ll leave it to you to check that the unions of the four $3$-member families from $S$ are $X,\{a,b,c\}$, and $\{a,b,d\}$, and that the union of all four members of $S$ is $X$. The topology generated by the subbase $S$ is therefore


  • $\begingroup$ Thank you so much! I was very confused, and you cleared all my doubts! $\endgroup$ – aswa09 Feb 10 '15 at 18:11
  • $\begingroup$ @ajpk: Glad it helped; you’re very welcome! $\endgroup$ – Brian M. Scott Feb 10 '15 at 18:28

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