Topologising RP2 using open sets in R#

**Bernhard** · May 19th 2012, 05:39 PM

I am reading Martin Crossley's book - Essential Topology - basically to get an understanding of Topology and then to build a knowledge of Algebraic Topology! (That is the aim, anyway!)

On page 27, Example 3.33 (see attachment) Crossley is explaining the toplogising of $\mathbb{R} P^2$ where, of course, $\mathbb{R} P^2$ consists of lines through the origin in $\mathbb {R}^3$ .

We take a subset of $\mathbb{R} P^2$ i.e. a collection of lines in $\mathbb {R}^3$ , and then take a union of these lines to get a subset of $\mathbb {R}^3$ .

Crossley then defines a subset of $\mathbb{R} P^2$ to be open if the corresponding subset of $\mathbb {R}^3$ is open.

Crossley then argues that there is a special problem with the origin, presumably because the intersection of a number of lines through the origin is the origin itself alone and this is not an open set in $\mathbb {R}^3$ . (in a toplological space finite intersections of open sets must be open) [Is this reasoning correct?]

After resolving this problem by omitting the origin from $\mathbb {R}^3$ in his definition of openness, Crossley then asserts:

"Unions and intersections of $\mathbb{R} P^2$ correspond to unions and intersections of $\mathbb {R}^3$ - {0} ..."

But I cannot see that this is the case.

If we consider two lines $l_1$ and $l_2$ passing through the origin (see my diagram - topologising RP2 using open sets in R3 - attached) then the union of these is supposed to be an open set in $\mathbb {R}^3$ - {0} . But surely this would only be the case if we consider a complete cone of lines through the origin. With two lines - take a point x on one of them - then surely there is no open ball around this point in $\mathbb {R}^3$ - {0} ??? ( again - see my diagram - topologising RP2 using open sets in R3 - attached) So the set is not open in $\mathbb {R}^3$ - {0}?

Can someone please clarify this for me?

Peter

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