Give an example of topologies $\mathcal{T}$ and $\mathcal{T}'$ on $\{1,2,3\}$ such that $\mathcal{T} \cup \mathcal{T}'$ is not a topology.

By definition: A topology on a set X is a collection $\mathcal{T}$ of subsets of X having the following properties: i) $\emptyset$ and X are in $\mathcal{T}$

ii) The union of the elements of any subcollection of $\mathcal{T}$ is in $\mathcal{T}$

iii) The intersection of the elements of any finite subcollection of $\mathcal{T}$ is in $\mathcal{T}$.

On this example i'm trying to mold it with an example from the book where X=$\{a,b,c\}$ What i'm trying to understand to tackle the question, is what is the difference between the two sets that are not a topology of X, with the other sets that are topologies. Not a topology of X Different topologies in X


Just take $\mathcal{T}=\{\emptyset,\{1,2\},\{1,2,3\}\}$ and $\mathcal{T}'=\{\emptyset,\{1,3\},\{1,2,3\}\}$.

Both are topologies, but $\mathcal{T}\cup\mathcal{T}'$ is not since, for exemple, $\{1,2\}, \{1,3\}\in \mathcal{T}\cup\mathcal{T}'$ but $\{1\}=\{1,2\}\cap\{1,3\}$ is not in $\mathcal{T}\cup\mathcal{T}'$ (so ii) is not verified).

  • $\begingroup$ Can you elaborate, but wouldn't $\{1\}$ be on the set $\{1,2,3\}$? I'm new with topology, and I want to get an understanding. Thank You edit: Okay that gives a clear picture. I'll look into it more though. $\endgroup$ Feb 21 '18 at 20:33
  • $\begingroup$ @Killercamin , no $\{1\}$ is not in $\{1,2,3\}$ : it is $1$ who is in $\{1,2,3\}$! And even if it were, it doesn't change the fact that he wouldn't be in $\mathcal{T}\cup \mathcal{T}'$.... $\endgroup$
    – Netchaiev
    Feb 21 '18 at 20:40
  • $\begingroup$ I see, thank you very much. $\endgroup$ Feb 21 '18 at 21:19

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.