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I was looking through a proof that $\mathbb P^n$ is Hausdorff (take it here as the set of all one dimensional subspaces in $\mathbb{R}^{n+1}$, but it's unclear to me why this isn't nearly trivial.

If I have two points $x,y \in \mathbb{R}^{n+1}$, aren't the images of the disjoint open balls also disjoint open sets in $\mathbb P^n$?

That is, if $x,y \in \mathbb{R}^{n+1}$, and $B_x,B_y$ are disjoint balls, aren't $\pi(B_x)$ and $\pi(B_y)$ disjoint open sets in $\mathbb P^n$? ($\pi : \mathbb{R}^{n+1} \to \mathbb{P}^n$.) Namely, aren't these images the sort of cones in $\mathbb R^{n+1}$ consisting of lines less than $\epsilon$ angle away from the line through $x$ and $0$?

edit: does it work to do the following: let $N$ be sufficiently large

take $L_y$, the line through $y$, and consider the compact subset $\overline{B}_N(x) \cap L_y$. Then, minimize $$\|x-z\|$$ over this subset for $z \in \overline{B}_N(x) \cap L_y$. Repeat for the minimum distance from $y$ to $L_x$. Then use half the smaller of these radii.

Does this work? (i.e. the length of the line segment perpendicular to $L_y$ through $x$)


marked as duplicate by colormegone, Daniel McLaury, amd, Alex M., hardmath Mar 18 '16 at 21:35

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  • $\begingroup$ You can find a ball around $\mathbf{p}\in\mathbb{R}^{n+1}$ that is disjoint from a ball around $-\mathbf{p}$ and yet they have the same image under what I assume $\pi$ is. $\endgroup$ – Derek Elkins Mar 16 '16 at 4:52
  • $\begingroup$ Sorry, I meant $y \neq \lambda x$, $-x$ just gets glued to $x$ anyways $\endgroup$ – Anthony Peter Mar 16 '16 at 4:57
  • $\begingroup$ It is in fact, nearly trivial! $\endgroup$ – Cheerful Parsnip Mar 16 '16 at 6:02
  • $\begingroup$ @GrumpyParsnip Does the edit I've written work? $\endgroup$ – Anthony Peter Mar 16 '16 at 6:23
  • 1
    $\begingroup$ Two things. First, you are not being careful about the origin: the domain of the projection map is $\mathbb{R}^{n+1}-\{0\}$, not $\mathbb{R}^{n+1}$; and you are not being careful about choosing your radii so that your balls do not contain $0$. Second, instead of using metric arguments based on distance in $\mathbb{R}^{n+1}$, it would work better to use metric arguments based on angle between lines; then you would not run into so much trouble regarding the bad behavior of radii of Euclidean balls with respect to the projection map $\pi$. $\endgroup$ – Lee Mosher Mar 16 '16 at 13:56

In fact $P^n$ is a metric space, as the quotient of the sphere by an isometric involution. The distance between two lines is just the measure $\in [0, \pi/2]$ of the un-oriented angle between these two lines. More generally the quotient of a metric space by a finite (compact) group of isometry is also a metric space. The distance between two orbits being the Hausdorff distance between tow finite (compact) subsets of a metric space.


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