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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.2012-06-30
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    @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.2012-06-30

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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 $\mathbb{R}$.2012-06-30
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    @Qiaochu: Indeed.2012-06-30
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    @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?2012-06-30
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    @Kamran Also no! It's just as the base of the topology!2012-06-30
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    @John: Huh? What are you talking about?2012-06-30
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    OK thanx I was a little confused. so that means if I define -$I=\{\emptyset,A_\lambda,X:A_\lambda\subset X,\lambda\in\Lambda\}$ where $(\Lambda,\le)$ is any linearly ordered index set and $A_\alpha\subset A_\beta$ whenever $\alpha\le\beta$ then $I$ will be a topology without any restriction on $\Lambda$.2012-06-30
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    @Kamran: Not quite; you may need limiting cases; e.g., if you have an ascending chain in $\Lambda$ that has no maximum. For instance, pick $\Lambda=\mathbb{N}$, $A_{n}=\{k\in\mathbb{N}\mid k\leq n\}$, $X=\mathbb{R}$. Then $\cup A_{n}$ is not in your collection $I$, so $I$ is not closed under arbitrary unions.2012-06-30
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    @Kamran: The question is nonsensical. A given collection either *is* or *is not* a topology. You cannot put conditions on a given collection to make it be or make it cease to be a topology. It either contains the empty set, the whole set, is closed under arbitrary unions and finite intersections; or it fails at least one condition. The collection you give *is* a topology.2012-06-30
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    @Magidin I wanted to know what conditions can we impose in general on the collection $I=\{\emptyset, A_\lambda,X:A_\lambda\subset X,\lambda\in\Lambda\}$ so as to make it a topology.should the index set $\Lambda$ be complete?2012-06-30
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    @Kamran: Completeness is neither here nor there; the problem arises when you have a subset $S$ of $\Lambda$ with no maximum, but the union of the corresponding $A_{\lambda}$ does not equal either an $A_{\mu}$ or $X$. So you need either every subset of $\Lambda$ to have a maximum, or you need to put conditions on the $A_{\lambda}$ corresponding to subsets of $\Lambda$ with no maximum.2012-06-30
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    @Magidin completeness in the sense that every subset of $\Lambda$ must have a supremum, ie.for $\Delta\subset \Lambda$ we have $sup_{\lambda\in\Delta} \lambda\in\Lambda$2012-06-30
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    @Kamran: I *know* what completeness is. That's not enough. $\Lambda=[0,1]$ is complete; but if we let $A_x=[0,x)$ for $x\lt 1$, and $A_1=[0,1]$, then $I = \{\varnothing,A_x,\mathbb{R}\mid x\in[0,1]\}$ is **not** a topology, because $\cup_{0\leq x\lt 1} A_x=[0,1)$ is not an element of $I$. Which is **exactly why** I told you that compleness was neither here nor there. The issue is not suprema, it is **maxima**. If there are sets without maxima, then you need to put conditions on your sets $A_{\lambda}$ corresponding to those sets. *Which I already stated.*2012-06-30
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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?2012-06-30
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(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!