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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 $\mathbb{R}^3$.

Cartesian Bear
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  • I am guessing they take planes as basis and find the topology generated by the basis – tryst with freedom Mar 06 '22 at 13:24
  • https://www.google.com/url?sa=t&source=web&rct=j&url=https://math.stackexchange.com/questions/907885/what-is-the-topology-generated-by-a-basis&ved=2ahUKEwjn2uzsy7H2AhVFIMUKHXa7Cf4QFnoECCIQAQ&usg=AOvVaw2ZnJht_xjaw7rXJY4QLtMN – tryst with freedom Mar 06 '22 at 13:30

1 Answers1

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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.

Roselyn van Lauwe
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  • Welcome to the site! Why do we need the condition that intersection of basis elements must be writable as Union of basis elements? – tryst with freedom Mar 06 '22 at 14:45
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    @Buraian A basis for a topology $\mathcal{T}$ is a collection of open sets $\mathcal{B}$ such that every open set $U$ of $\mathcal{T}$ may be written as a union of basis sets $B \in \mathcal{B}$. Because finite intersections of open sets are open it follows in particular that intersections of basis sets must themselves be able to be written as unions of basis sets. – Roselyn van Lauwe Mar 06 '22 at 14:49
  • Ohhh thank you. That clears it up now. – tryst with freedom Mar 06 '22 at 14:57
  • @RoselynBaxter So should the original challenge/problem then be worded as "determine whether some version of the collection is a basis for some topology of $\mathbb{R}^3$. Furthermore determine what topology this is"? – Cartesian Bear Mar 06 '22 at 15:57
  • @SickSeries I'd interpret the problem as 'find the topology for which this set is a sub-basis' and not worry too much as to wether it is a basis. In this case the answer would be 'the discrete topology'. – Roselyn van Lauwe Mar 06 '22 at 16:02