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

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. –  Martin Sleziak Oct 11 '11 at 11:40
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. –  Chris Eagle Oct 11 '11 at 11:45
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? –  MTH Oct 11 '11 at 12:07
@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. –  Asaf Karagila Oct 11 '11 at 19:39

What you should check is the following:

Let's call the topology generated by the sets of the form $(a,b)$ $\mathcal T$.

Given $(a,b)$, with $a<b$, then you immediately see $(a,b)=(-\infty,b)\cap (a,+\infty)\in \mathcal T_S$, since a topology is closed under finite intersection. Hence $\mathcal T_S$ contains the basis of $\mathcal T$, hence

$$\mathcal T\subseteq\mathcal T_S.$$

The other inclusion is as follows: write $$(-\infty,a)=\bigcup_{n=0}^\infty(a-n-2,a-n)$$ and $$(b,+\infty)=\bigcup_{n=0}^\infty(b+n,b+n+2).$$ In this way you immediately see that $\mathcal T$ contains a basis of $\mathcal T_S$ (topologies are closed under arbitrary unions) hence $$\mathcal T_S\subseteq \mathcal T,$$ and you are done.

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Given a topological space $(X,\tau)$, we say that $\mathcal A$ is a sub-basis for $\tau$ if the finite intersections of elements from $\mathcal A$ form a basis for $\tau$.

That is to say, we want to show that every open set is a union of finite intersection of rays. It is sufficient to show that every open interval is such union of finite intersection of rays. Why? Since the union of open intervals forms the standard topology, so if we can create all open intervals we can create all the standard open sets.

And sure enough, given an interval $(a,b)$ we can write it as the intersection of $(-\infty,b)\cap(a,\infty)$.

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