One of the footnotes in my reading material is the following:
As a challenge, try to determine the topology of $\mathbb{R}^3$ spanned by the set of all planes of $\mathbb{R}^3$
And beyond verifying that the said collection is a basis for some topology (presented below), I'm not sure how to proceed with the challenge. Namely, it seems to me that the one is to find a correspondence between some known topology of $\mathbb{R}^3$, but as the discussed topologies have been the one induced by i.) standard metric on $\mathbb{R}$, ii.) the powerset of $\mathbb{R}$, iii.) the collection of half-open intervals $[x, y)$ in $\mathbb{R}$, I'm quite puzzled on what I'd have to say, since out of those three examples I only see that any plane element can be spanned by an arbitrary union of $[x_1, y_1)\times [x_2, y_2) \times [x_3, y_3)$.
(Showing that the collection of planes does indeed span a topology): Let $\mathcal{B}$ be the set of all planes of $\mathbb{R}^3$. One can easily verify that $\mathbb{R}^3$ can be written as a union of the elements of $\mathcal{B}$, and if $B_1, B_2 \in \mathcal{B}$ such that $B_1 \cap B_2 \neq \varnothing$, then their intersection is simply another element of $\mathcal{B}$, corresponding to the line of intersection between two planes. Thus we are confident that $\mathcal{B}$ is indeed a basis for some topology $\mathcal{T}$ of $\mathbb{R}^3$.
Best Answer
The set $\mathcal{B}$ of all planes in $\Bbb{R}^3$ does not form a basis for a topology, because an intersection of two planes may be a line, which cannot be written as a union of planes. If we let $\mathcal{B}$ be a sub-basis for the topology $\mathcal{T}$ then we find that every singleton subset $\{\vec{x}\} \subset \Bbb{R}^3$ must be open, because a point is the intersection of three planes. This implies that every subset of $\Bbb{R}^3$ is open, so this is the discrete topology.