- #1

- 662

- 1

## Main Question or Discussion Point

Hi, I am trying to show that RP^n is the

compactification of R^n. I have some , but not all I

need:

I have also heard the claim that every compact n-manifold is

a compactification of R^n, but I cannot find a good general

argument . I can see, e.g., for n=2, we can construct a manifold

by using an n-gon, and identifying some edges. Then the interior

seems clearly to be homeo. to R^n.

In some other cases, e.g., S^n, this seems easy to show

using the stereo projection, and some work with the topology

defined on the (Alexandroff) 1-pt. compactification ( use

as open sets in S^n, all open sets in C --or in R^n --together

with complements of compact subsets K of C.) Then R^n is open

in S^n, and its closure is necessarily S^n itself, i.e., R^n is

embedded densely in the compact space S^n ( S^n is

compact by construction.)

For the sake of practicing my pointset topology and some

basic diff. geometry, I tried to show that RP^n is also a

compactification of R^n, before trying a more general argument:

i) I know RP^n is compact, as the quotient of the

compact space S^n by a continuous (quotient) map.

ii) I think I can show R^n is dense in RP^n, since

the identification of S^n is made only in the

boundary of S^n, and the interior of S^n is homeo.

to R^n, i.e., missing some boundary points. The

interior of the disk (homeo. to R^n ) is dense in the

disk. Is the image

(under the quotient map) also dense in RP^n?

Do I need something else?

Let me be more specific:

The quotient map

q:S^n-->S^n/~ is continuous, by construction. Then

the continuous image of S^n is compact in the quotient

topology. Right?

Maybe more rigorously, we should start by giving

S^n the subspace topology of R^(n+1). Then, as a closed

subset/subspace of a compact space, it is

itself compact.

Now: How you have R^n in RP^n (i.e. which embedding) and

why it is

dense in the quotient topology (without hand waving)?

Let's see: if we consider RP^n as the set of

points in S^n/~ ( x~y iff x=ty ; t non-zero ),

then R^n would be everything except for the equator

D^n_ /\D_n+ , i.e., in a "standard" coordinate system

X1,..,Xn, R^n={x=(x1,...,xn);||x||=1 and xn>0 }

Then R^n is a saturated open subset of S^n , under ~

(i.e., the map q(x)=x/~ )so that (since quotient maps

send saturated open sets to open sets) the image

p(R^n) is open in S^n/~

But Cl(R^n)={x=(x1,..,xn);||x||=1, xn>=0 }

is a saturated closed set re q(x)=x/~. So

q(Cl(R^n)) is closed in RP^n.

Now we need to show that Cl(q(R^n))=RP^n

But for any point z in q(Cl(R^n)), where

xn=0 for z, we have:

q^-1( q(Cl(R^n))-{z} ) is not closed in

(S^n, subspace) (though I don't have a good

argument to this effect)

Then q((Cl(R^n)) =RP^n is the closure of

(the embedded image of ) R^n in RP^n, and

so R^n is dense in RP^n.

Thanks For any Comments.

compactification of R^n. I have some , but not all I

need:

I have also heard the claim that every compact n-manifold is

a compactification of R^n, but I cannot find a good general

argument . I can see, e.g., for n=2, we can construct a manifold

by using an n-gon, and identifying some edges. Then the interior

seems clearly to be homeo. to R^n.

In some other cases, e.g., S^n, this seems easy to show

using the stereo projection, and some work with the topology

defined on the (Alexandroff) 1-pt. compactification ( use

as open sets in S^n, all open sets in C --or in R^n --together

with complements of compact subsets K of C.) Then R^n is open

in S^n, and its closure is necessarily S^n itself, i.e., R^n is

embedded densely in the compact space S^n ( S^n is

compact by construction.)

For the sake of practicing my pointset topology and some

basic diff. geometry, I tried to show that RP^n is also a

compactification of R^n, before trying a more general argument:

i) I know RP^n is compact, as the quotient of the

compact space S^n by a continuous (quotient) map.

ii) I think I can show R^n is dense in RP^n, since

the identification of S^n is made only in the

boundary of S^n, and the interior of S^n is homeo.

to R^n, i.e., missing some boundary points. The

interior of the disk (homeo. to R^n ) is dense in the

disk. Is the image

(under the quotient map) also dense in RP^n?

Do I need something else?

Let me be more specific:

The quotient map

q:S^n-->S^n/~ is continuous, by construction. Then

the continuous image of S^n is compact in the quotient

topology. Right?

Maybe more rigorously, we should start by giving

S^n the subspace topology of R^(n+1). Then, as a closed

subset/subspace of a compact space, it is

itself compact.

Now: How you have R^n in RP^n (i.e. which embedding) and

why it is

dense in the quotient topology (without hand waving)?

Let's see: if we consider RP^n as the set of

points in S^n/~ ( x~y iff x=ty ; t non-zero ),

then R^n would be everything except for the equator

D^n_ /\D_n+ , i.e., in a "standard" coordinate system

X1,..,Xn, R^n={x=(x1,...,xn);||x||=1 and xn>0 }

Then R^n is a saturated open subset of S^n , under ~

(i.e., the map q(x)=x/~ )so that (since quotient maps

send saturated open sets to open sets) the image

p(R^n) is open in S^n/~

But Cl(R^n)={x=(x1,..,xn);||x||=1, xn>=0 }

is a saturated closed set re q(x)=x/~. So

q(Cl(R^n)) is closed in RP^n.

Now we need to show that Cl(q(R^n))=RP^n

But for any point z in q(Cl(R^n)), where

xn=0 for z, we have:

q^-1( q(Cl(R^n))-{z} ) is not closed in

(S^n, subspace) (though I don't have a good

argument to this effect)

Then q((Cl(R^n)) =RP^n is the closure of

(the embedded image of ) R^n in RP^n, and

so R^n is dense in RP^n.

Thanks For any Comments.