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What does it mean for a topology to be generated? For example $X=\mathbb{R}$ be topology generated by $[a,b)$. Isn't the topology a collection of open sets? $[a,b)$ is not open though.

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Let $X$ be a set, and let $\mathscr{S}$ be any family of subsets of $X$. Then there is a unique topology $\mathscr{T}$ on $X$ such that

  1. $\mathscr{S}\subseteq\mathscr{T}$, and
  2. if $\mathscr{T}'$ is any topology on $X$ such that $\mathscr{S}\subseteq\mathscr{T}'$, then $\mathscr{T}\subseteq\mathscr{T}'$.

In other words, $\mathscr{T}$ is the smallest topology containing each set in the collection $\mathscr{S}$. We say that $\mathscr{T}$ is the topology generated by $\mathscr{S}$. $\mathscr{S}$ is said to be a subbase (or subbases) for the topology $\mathscr{T}$.

In your example I strongly suspect that you’ve incorrectly paraphrased the statement that’s bothering you: I’m reasonably sure that what was actually described was the topology on $\Bbb R$ generated not by a single set $[a,b)$, but by the collection of all sets of the form $[a,b)$, where $a,b\in\Bbb R$ and $a<b$. In other words,

$$\mathscr{S}=\Big\{[a,b):a,b\in\Bbb R\text{ and }a<b\Big\}\;.$$

It turns out that this $\mathscr{S}$ is already a base (or basis) for a topology on $\Bbb R$, so the non-empty open sets in the topology that it generates are simply the sets that are unions of intervals of the form $[a,b)$. This topology is called the Sorgenfrey or lower-limit topology on $\Bbb R$, and the resulting space is known as the Sorgenfrey line; it’s a fairly important example in general topology.

It’s true that sets of the form $[a,b)$ are not open in the usual topology on $\Bbb R$, but that’s completely irrelevant: you’re defining a new topology on $\Bbb R$, one that is in many ways very different from the usual one.

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So, in this case $\mathscr{S}$ will be family of open subsets of $X$ – James Bond Oct 1 '13 at 0:29
@JamesBond: No, the family of open subsets of $X$ is equal to the collection of all unions of members of $\mathscr{S}$. For example, the set $[0,1)\cup[2,3)$ is open in the Sorgenfrey topology, but it’s not in $\mathscr{S}$. For that matter, $(0,1)$ is open in the Sorgenfrey topology: it’s the union of the intervals $[x,1)$ with $0<x<1$. And it’s not in $\mathscr{S}$ either. – Brian M. Scott Oct 1 '13 at 0:39
I see. If I understand this correctly is it true that: $[0,1]\cup[2,3)$ is open in Sorgenfrey topology because it's the union of open sets $[0,1]$ and $[2,3)$ which are in $\mathscr{I}$ and for (0,1) its the infinite union of open sets of the form [x,1) – James Bond Oct 1 '13 at 0:57
@JamesBond: No, $[0,1]$ isn’t open in the Sorgenfrey topology: there is no set of the form $[a,b)$ that contains $1$ and is a subset of $[0,1]$. But $[0,1)$ is open in the Sorgenfrey topology. – Brian M. Scott Oct 1 '13 at 0:59
That was my mistake. I meant $[0,1)$ and not $[0,1]$ – James Bond Oct 1 '13 at 1:11

It is open if you declare it to be :) So you have to look at all unions and finite intersections of such sets (for all $a<b$) to get the topology.

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Open set is another name for an element in the family of sets that is called topology. Any collection of subsets of $\Bbb R$ can form a topology as long as certain properties are satisfied. On the other hand, any collection of sets $\mathcal C$ generates a topology $\tau$, namely the smallest topology such that $\mathcal C\subseteq\tau$. It can be constructed by taking arbitrary unions of finite intersections of the sets in $\cal C.$

So if you want $[a,b)$ to be an open set, first take finite intersections of such intervals. But these are again of the form $[a,b)$. So the open sets, i.e. the elements of the topology generated by the family $\{[a,b)\mid a,b\in\Bbb R\}$ will simply be the unions of intervals of the form $[a,b)$.

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