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Show that rays of the form $(-\infty, a)$ and $(b, \infty)$ ; $a,b \in \mathbb R$, are a sub-basis for the topology generated by open intervals of $\mathbb R$ on $\mathbb R$?

I'd just like to know if I am correct.

Proof : Let $S$ be the sub-basis formed with the rays $(-\infty, a)$ and $(b, \infty)$. We have to check that the topology $T_{S}$ generated by the sub-basis $S$ is equal to the topology on $\mathbb{R}$. As $(-\infty, a) \cup ((-\infty, a)\cap(b, \infty)) \cup (b, \infty) = \mathbb{R} $ Then $S$ is a sub-basis for the topology on $\mathbb{R}$.

Am I right?

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    You have only shown that S is a sub-basis of _some_ topology on R. In case you want to show that this is the same topology as the one generated by all open intervals, you should show that: a) all members of S are open in this topology; b) every open interval can be expressed using the sets from S.2011-10-11
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    I don't understand your assertion $(-\infty, a) \cup ((-\infty, a)\cap(b, \infty)) \cup (b, \infty) = \mathbb{R} $. Is it supposed to be true for all $a$ and $b$? Because it isn't.2011-10-11
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    Chris Eagle : I think it is supposed to be true for all $a$ and $b$ because it isn't precised. I think it is better to ask that $b$ < $a$. Martin Sleziak : a) All members of $S$ are open in the topology on $\mathbb{R}$ : it's obvious b) If $a$<$b$ then $(-\infty, a) \cap (b, \infty) = (a, b)$ ; $ \forall B \in $ {open intervals of $\mathbb{R}$}, $B \in (-\infty, a) \cup (a,b) \cup (b, \infty) = \mathbb{R}$ Is it a correct proof?2011-10-11
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    @MTH: That last line in that last comment is ill-formed. $B$ is an open interval, while the elements of $\mathbb R$ are *not* open intervals.2011-10-11

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