I am reading Rudin's "Real and Complex Analysis" and came across the statement,
"In the real line $R^1$ a set is open if and only if it is a union of open segments (a,b)."
I have two questions:
1. A subset $A$ of $X$ may be open with respect to a topology $\tau_1$ in $X$, but not open with respect to another topology $\tau_2$ in $X$. Therefore, what does it mean to say that, a set is open in the real line $R^1$, without specifying underlying topology?
2. I see that a collection of all sets which are arbitrary unions of open interval consists a topology $\tau$ in $R^1$. So arbitrary union of open interval is open in X with topology $\tau$.(if part, which is straight forward.) But how can I show that arbitrary open set in X with respect to some random topology $\tau'$ can be expressed as a union of open intervals?(only if part) I think this question is related to the first question since I need clarification regarding the meaning of an open set in X means without any topology specified.
I do realize that $R^1$ is a metric space, and the statement is perfectly clear with the definition of open set in a metric space. But is there a way to understand the above statement without relying on the concept of metric space?
Rudin is saying that in the standard topology on $\Bbb R$ induced by the metric $d(x,y) = |x-y|$, every open set can be written as a union of intervals of the form $(a,b)$.