The following space is a version of example 71 from "Counterexamples in Topology". Let $X = \mathbb R$ be the real line with the Euclidean topology $\tau$. We define a new topology $\tau*$ on $\mathbb R$ to be the topology generated from the following basic open sets.
- Each point of the form $k+2^{-n}$, where $k,n \in \mathbb N$, is an open basic set.
- Open intervals of the form $(a,b)$ where $a<b$ are basic open sets.
I Claim that $\mathcal B$ is a basis for a topology, which implies that $\tau*$ is a topology on $\mathbb R$
Proof:
- $\bigcup \mathcal B = X$. This is obvious since, $\tau*$ is an extention of $\tau$.
If $U,V \in \mathcal B$, and $x \in U \cap V$, then:
a. If $x = k+2^{-n}$ for some $k,n \in \mathbb N$, then $x$ itself is a basic open set.
b. If $x$ is not of the form $k+2^{-n}$ for some $k,n \in \mathbb N$, then, there are two intervals, $(a_u,b_u) \subset U,(a_v,b_v) \subset V$ which contains $x$, and the intersection $(a_u,b_u) \cap (a_v,b_v)$ is an open basic neighborhood, which contains $X$.
What do you think? Is the new proof correct??
Thank you!