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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

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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).

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  • $\begingroup$ 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

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