# Is there an example of family of open intervals in $\mathbb R$ …?

Is there an example of family of open intervals in $\mathbb R$ such that any arbitrary union of such open intervals is again an open interval? In other words can we define a topology on $\mathbb R$ with open intervals only?, (i.e. $A$ is open in $\mathbb R$ if and only if $A$ is an open interval) However we have such a topology when we take such collection as base.

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For a family of such open intervals, just make them nested. –  André Nicolas Jun 30 '12 at 5:11
@Kamran: Yes, it is an open interval, namely the open interval $(-\infty,\infty)$. It's open, and it is an interval because given any two $a,b$ in the set, $[a,b]$ is contained in the set. That's the definition of an interval. –  Arturo Magidin Jun 30 '12 at 5:46

Your two questions are not really equivalent; under the usual topology, the collection of open intervals $\{(-r,r)\mid r\in\mathbb{R}\}$ and $\mathbb{R}$ satisfies your first condition (interpret $(-r,r)$ with $r\lt 0$ as empty): an arbitrary union of such open intervals is again an open interval. It even defines a topology, but it is not a topology such all open intervals are open.

No topology can consist exactly of the open intervals, since it would necessarily contain $(0,1)$ and $(2,3)$ as open sets, hence contain its union as an open set, but $(0,1)\cup(2,3)$ is not an interval.

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Well, we also need $\mathbb{R}$. –  Qiaochu Yuan Jun 30 '12 at 5:12
@Qiaochu: Indeed. –  Arturo Magidin Jun 30 '12 at 5:12
@Kamran: "such intervals" refers to the intervals in the collection I specified: the intervals of the form $(-r,r)$ together with the entire line. The union of all such intervals is the entire real line, which is an interval, namely $(-\infty,\infty)$. What is not clear? –  Arturo Magidin Jun 30 '12 at 5:45
@Kamran Also no! It's just as the base of the topology! –  Paul Jun 30 '12 at 5:49
@John: Huh? What are you talking about? –  Arturo Magidin Jun 30 '12 at 5:51

For your second question, the answer is NO. If it satisfies your condition, it can't generate a topology. Just as Arturo Magidin said, $(0,1)$ and $(2,3)$ as open sets, hence contain its union as an open set, but $(0,1)\cup(2,3)$ is not an interval:)

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Aren't you just repeating what I said? –  Arturo Magidin Jun 30 '12 at 6:23

(I'll approach this somewhat differently.)

A family of open intervals in $\mathbb{R}$ such that the union of any subfamily is again an open interval certainly exists: take, for example, the family of all intervals of the form $(r,s)$ where $r < 0 < s$.

Suppose now that $\mathcal{I}$ is such a family of open intervals which moreover forms a topology on $\mathbb{R}$. I claim that this topology is not T$_1$:

Assume that $\{ x \}$ is closed in this topology for some $x \in \mathbb{R}$. Then there must be an $I \in \mathcal{I}$ such that $x-1 \in I \subseteq ( - \infty , x )$, and similarly there must be an $J \in \mathcal{I}$ such that $x+1 \in J \subseteq (x , + \infty )$. However it easily follows that $I \cup J$ is not an interval!

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