The usual definition I have seen for a quotient set of some set $S$ is with respect to some equivalence relation $\sim$, i.e. $S / \sim$, which means all elements $x$ and $y$ where $x \sim y$ are collapsed into a single element of $S / \sim$. Quotient sets can also be defined over certain algebraic sets with respect to some subset of the set, where it is generally understood to mean that two elements are considered equivalent if their difference is a member of this subset. For example $\mathbb{Z} / 10 \mathbb{Z}$ represents the integers modulo 10, where $x \sim y \Leftrightarrow x - y \in 10 \mathbb{Z}$. I have recently come across this particular notation: $$ \mathbb{R}P^n \cong (\mathbb{R}^{n+1} \backslash \{0_v\}) / (\mathbb{R} \backslash \{0\}) $$ Here, $(\mathbb{R}^{n+1} \backslash \{0_v\}) / (\mathbb{R} \backslash \{0\})$ is meant to represent a set of nonzero real $n+1$ dimensional vectors which are considered equivalent iff one is a real nonzero multiple of the other. The author insists that this is standard mathematical notation that does not depend on any mathematical structure other than basic set theory, but I have never seen a quotient set defined in this way before. They then go on to use this notation to derive several other set equivalences using what they say are basic properties of quotients. Can anyone point me to some literature that uses this type of notation for properties of quotient sets?
2026-03-21 20:25:02.1774124702
Quotient set with respect to another set
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This is a case of a quotient by a group action. In this case, $\mathbb{R}\backslash\{0\}$ (as a group under multiplication) acts by scalar multiplication on $\mathbb{R}^{n+1}\backslash\{0\}$.
Generally, if a group $G$ acts on a set $X$, we denote the set of orbits by $X/G$ (see here). If $X$ is additionally a topological space (and the group action is continuous), like in this case, the set of orbits $X/G$ can be equipped with the quotient topology (see here).
You could phrase this with basic set theory if you wanted, but fundamentally it's an idea from group theory. (Although I'd be interested to hear someone tell me otherwise!)