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A first course in algebraic topology by C. Kosniowski asks us to prove that $\mathcal{U}=\{\emptyset\}\cup\{\mathbb{R}\}\cup\{(a,b);a,b\in \mathbb{R},a<b\}$ is NOT a topology. I've been trying to set a family of sub sets from this like $\{(a_{j},b_{j});a_{j},b_{j}\in \mathbb{R}, j\in J\}$.

Is it ok to say the following?:

$\cup_{j\in J}(a_{j},b_{j})=(min\{a_{j};j\in J\},max\{b_{j};j\in J\})$ which the largest interval and is contained on $\mathbb{R}$?

Can I use that to prove that $\mathcal{U}$ is NOT a topology?

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Hint: Before jumping to the general case of trying to describe an arbitrary union of intervals, think about a simple case, like $(0,1)\cup (2,3)$...

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  • $\begingroup$ But that tells me that $\mathcal{U}$ is a topology, right? $\endgroup$ – AlejandroL.g Apr 19 '20 at 1:15
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    $\begingroup$ $(0,1)$ and $(2,3)$ are in $\mathcal{U}$. A topology is closed under unions, so $(0,1)\cup (2,3)$ should be in $\mathcal{U}$. Is it? $\endgroup$ – Alex Kruckman Apr 19 '20 at 1:15
  • $\begingroup$ That union is going to be $(0,3)$ isn´t it? and that's in $\mathcal{U}$, isn't it? $\endgroup$ – AlejandroL.g Apr 19 '20 at 1:49
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    $\begingroup$ Really? $1.5\in (0,1)\cup (2,3)$? @JoshuaLópezAraiza $\endgroup$ – Alex Kruckman Apr 19 '20 at 1:50
  • $\begingroup$ And what about $1$ and $2$? These are in $(0,3)$. Are they in $(0,1)\cup (2,3)$? $\endgroup$ – Alex Kruckman Apr 19 '20 at 1:51

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