# Exercise involving the quotient topology

Define a partition of $X = \mathbb{R}^2 − \{0\}$ by taking each ray emanating from the origin as an element in the partition. Which topological space appears topologically equivalent to the quotient space that results from this partition?

So a set $U$ is open in $X/\sim$ if the union of the equivalence classes contained in $U$ is open in the original topology, correct? In such a case, wouldn't the quotient topology simply be the trivial topology? My reasoning is largely based on intuition and simple geometric arguments, so it could very well be flawed. Any help would be greatly appreciated!

• Hint: Circle. – Berci Feb 20 '16 at 23:43
• I'm afraid I don't quite understand. Could you elaborate a little more please? – user316319 Feb 20 '16 at 23:52
• Please read this tutorial on how to typeset mathematics on this site. – N. F. Taussig Feb 20 '16 at 23:58

For each $x\in X$, let $[x]$ be the equivalence class of $x$ and let $p:X\to X/\sim$ be the map which maps $x\to[x]$. Consider $g:X\to S^1$ given by $g(x)=x/\lVert x\rVert$. Since $g^{-1}(x)=[x]$, you can show that $g$ is a quotient map and it induces a homeomorphism $f:(X/\sim)\to S^1$ such that $g=f\circ p$.
your description seems to be the definition of the real projective line. This is the usual real axis with an extra point "attached", the infinity: $\infty$. Topologically it is homeomorphic to $S^1$ i.e. the circle. Physicists often describe it as a "single-point compactification of the real line".
It can also be equivalently thought of as the set of $1d$ subspaces of a $2d$ vector space $V$ and is commonly denoted as $\mathbf P^1(\mathbf R)$.
• Topologists also refer to the space as the "one-point compactification of $\mathbb R$". Also known as the Alexandroff compactification. – Tim Raczkowski Feb 21 '16 at 0:14