Boundedness in extended complex plane

[matness]

what does it mean to be bounded in extended complex plane (say C')?
can we say all subsets of C' is bdd because C' itself bdd?

 

[matt grime]

The extended complex plane, in my opinion, is not bounded. Bounded ought to mean, for a metric space, contained in some open ball of *finite* radius.

 

[matness]

to be more explicitly
1)if i use spherical metric( i.e. i think on the sphere ) then C equivalently sphere without north pole is bdd .
why adding one point namely inf. changes boundedness
2)another point of view : i can prove C' is compact so it is closed and bdd

maybe iam missing somethings but i can't find them

 

[matt grime]

You are missing something. Just because the extended complex plane is homeomorphic to a sphere does not mean it is bounded. The real line is homeomorphic to (-1,1), and I doubt you think that R is bounded.

Compact is not the same as closed and bounded. Cu{oo} is not a subset of R^3. It is homeomorphic to a subset of R^3, but 'size' is not invariant under homeomorphism.

 

[matness]

i didnt say it is bdd because S^2 is in R^3. for example an open ball with center at inf. and radius some small quantiity (say eps.) contains
{ z€C s.t. |z|>eps } U {inf} by the way it is defined .
here what i stuck : why adding only one point(o.k. now it is a ball)
changes boundedness.
if you say i can't take inf. as a center i won't insist on that
but i am pretty sure i saw somewhere compactness implies bddness.

 

[matt grime]

It is not bounded. Boundedness applies to subsets of metric spaces. What is the metric you're using? How far from 2 is infinity? Compact is not the same as closed and bounded. The point is you cannot have balls of infinite radii. I said nothing about infinity being or not being the centre of an open set, but as it happens, you cannot have infinity as the centre of a ball defined by the metric because there is no metric given.

 

[matness]

now there is a new thing that comes to my mind.
the reason that i cannot say such a thing because a set is bdd whenever that this set is a subset (proper)?.otherwise we would say C is bdd in C
OR R is bdd in R e.t.c.
still wrong?

 

[matt grime]

You can't say it is bounded because you need to use a metric to define the size of your sets. You haven't put a metrc on the extended complex plane.

 

[matness]

:) i got it with one second difference . so i stop bothering.
thanks