# Issue with factor space topology for $Gr_1(\mathbb{R}^2)$

From the definition of factor space topologies, given a topological space $$(X, \tau)$$, the factor space topology for $$X/\sim$$ is given by:

$$\tau’ = \{ U \subset Y | \pi^{-1}(U) \in \tau \}$$

where $$\pi: X \mapsto X/\sim$$ is the factor map, assigning each $$x \in X$$ to its equivalence class.

Take the metric space $$\mathbb{R}^2$$ with metric $$d(x, y) = \sqrt{(x - y) \cdot (x - y)}$$ as an example with the metric topology given by all open sets $$V \in \tau$$ which contain an open ball $$B_{\epsilon}(x) := \{y \in \mathbb{R}^2 | d(x, y) < \epsilon \}$$ for all points $$x \in V$$, for some $$\epsilon > 0$$. Considering the space $$X / \sim$$ where $$x \sim y$$ when $$\text{span} \{ x \} = \text{span}\{ y \}$$, i.e. for two vectors spanning the same subspace, then $$X / \sim$$ is equivalent to $$Gr_1(\mathbb{R}^2)$$, which is the set of lines passing through the origin.

The problem is that I cannot come up with any non-trivial open subsets in the factor space, based on the definition of the factor space topology. For example, considering all lines $$U$$ with a positive slope, $$\pi^{-1}(U) = \{ (x, y) | x, y \neq 0 \wedge y/x > 0 \}$$, meaning that open balls contained in $$\pi^{-1}(U)$$ exist for all points except for $$(0,0)$$. This issue arises in any construction of an attempted open set in the factor space - I cannot come up with any set whose corresponding set in $$\mathbb{R}^2$$ is open, except for the empty set and $$\mathbb{R}^2$$ itself.

The point $$(0,0)$$ is the only point which cannot be uniquely assigned to an equivalence class. Is this related to the problem? There was some mention of disjoint unions in the literature, but it wasn’t elaborated much.

• In your definition of $d(x,y)$ do you have $x,y \in \mathbb R^2$? So is $\cdot$ the standard scalar product? So $d$ is the standard metric? Jul 6, 2021 at 8:54
• @principal-ideal-domain Yes! As you’re saying, $d$ is the standard metric based on the standard scalar product and $x,y \in \mathbb{R}^2$.
– Max
Jul 6, 2021 at 9:10

What about the set $$A:=\{ x \in \mathbb R^2 : x_1x_2>0 \}?$$ It is clearly an open subset of $$\mathbb R^2$$ and for any element of $$A$$ all equivalent points are also in $$A$$. So you have $$\pi^{-1}(\pi(A))=A,$$ hence $$\pi(A)$$ is open in $$X / \sim$$ and non-trivial.

Supplement: You can further define $$B:=\{ x \in \mathbb R^2 : x_1x_2<0 \}$$ and $$C:=\{ x \in \mathbb R^2 : x_1x_2=0 \}.$$ Then you have $$\pi^{-1}(\pi(B))=B\quad\text{and}\quad\pi^{-1}(\pi(C))=C.$$ So this gives rise to a disjoint union of $$X/\sim$$: $$X/\sim=\pi(A) \cup \pi(B) \cup \pi(C).$$ The sets $$\pi(A)$$ and $$\pi(B)$$ are open, where $$\pi(C)$$ is closed.

• Thanks for the response! My only confusion is when you say that all equivalent points are also in $A$. If we take the point $(1,1)$, then mapping it into the factor space we get the line with slope $1$, right? But this line includes $(0,0)$, so wouldn’t the inverse mapping of this set not be in $A$? When mapping all points where $x_1 x_2 > 0$, don’t you get a set of straight lines that fill up the open first and third quadrants along with the point $(0,0)$? After further thought, maybe my idea of the inverse map is incorrect?
– Max
Jul 6, 2021 at 9:18
• Wait… is it the case that $\pi^{-1}$ does not include $(0,0)$? I.e, mapping $(1,1)$ to the factor space and then back gives the set of points on the straight line, except for $(0,0)$?
– Max
Jul 6, 2021 at 9:25
• @Max $(0,0)$ and $(1,1)$ lie in different classes because they span different spaces. The span of $(0,0)$ is a $0$-dimensional, the span of $(1,1)$ is $1$-dimensional. So they can't be equal. It's no problem that one of them contains the other. You are looking for equalitiy, that's what you need to get an equivalence relation! Otherwise it wouldn't be symmetric if you would look for inclusion. Jul 6, 2021 at 9:27
• That makes sense! Thank you so much!
– Max
Jul 6, 2021 at 9:52