1
$\begingroup$

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)?

$\endgroup$
2
$\begingroup$

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

$$\big\{\varnothing,\{a\},\{a,b\},\{a,c\},\{a,b,c\},\{a,b,d\},X\big\}\;.$$

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.