Question about basis of the euclidean topology.

Question

I know that a subset $S$ of $\mathbb{R}$ is open if and only if it is a union of open intervals (I know how to prove this). So $B_1=\{(a,b):a,b\in \mathbb{R},a

Answer 1

Yes, $B_1$ is also a base for the Euclidean topology. There's no contradiction: a topology can have more than one base.

(In fact, let $A\subseteq\mathbb{R}$ and let $I_A=\{(a, b): a, b\in A\}.$ Then $I_A$ is a base for the Euclidean topology iff $A$ is dense. So, for instance, we can replace the rationals with the dyadic rationals, or similar.)

However, your reasoning is a bit over the place. You write:

$B_1$ . . . should also be a base for the euclidean topology since it is an open interval and any open subset can be thought as the union of open intervals.

This isn't right.

• $B_1$ isn't an open interval, it's a set of open intervals.

• What this tells us is that the topology generated by $B_1$ is a subset of the Euclidean topology, so now we just need to show the other direction.

• It's not enough to say that "any open set can be thought of as the union of open intervals" - you need to say that any open set is the union of open intervals from $B_1$, and this takes proof: how would you write $(\sqrt{2}, \pi)$ as a union of intervals with rational endpoints?

• That almost finishes the proof - now we know that $B_1$ generates the Euclidean topology. But all this shows is that $B_1$ is a subbase for the Euclidean topology - to show that it's a base, we need to show that it's closed under finite intersections.