Basis of Euclidean topology on $\mathbb{R}$ such that no element is contained in another [closed]

Question

What is an example of a topological base ${\cal B}$ for $\mathbb{R}$ with the Euclidean topology such that for every $B_1\neq B_2 \in {\cal B}$ we have $B_1\not\subseteq B_2$?

Answer 1

A topological space has a basis which is an antichain w.r.t. set inclusion if and only if its Kolmogorov quotient (ie $T_0$-fication) is discrete.

The reason is that in this case, any two basis elements need to have empty intersection (otherwise their intersection would need to have a basis element as a subset, which is then included in the original sets).

If you only want a subbasis with this property, you can built one as follows:

Include every open interval $(n,n+2)$ for $n \in \mathbb{Z}$.

Pick a basis $(B_n)_{n \in \mathbb{N}}$ where each $B_n$ is included in some interval $(k,k+2)$ with $k+2 < n$. Now include all $B_n \cup (n,n+1)$.

We can then recover the original $B_n$ by intersecting with a suitable interval, so we have a subbasis. Moreover, the sets we put in are incomparable by $\subseteq$ by construction.