The circle as a set closed and bounded

1. Oct 31, 2012

dapias09

Hi guys,

I would like to understand why a circle (and in general a n-sphere) as a subset of R^2 (in general R^(n+1)) with the standard topolgy is considered a closed and a bounded set.

I think that this can be a closed set because its complement (the interior of the circle and the rest of the plane) is open. And could be bounded because it has a finite extension (but ths is very intuitive). I cannot figure out what is the interior, the closure or the boundary of this set.

2. Oct 31, 2012

micromass

Staff Emeritus
Consider the function

$$f:\mathbb{R}^2\rightarrow \mathbb{R}:(x,y)\rightarrow \sqrt{x^2+y^2}$$

Then the circle is $f^{-1}(1)$ and thus closed.
Now, the closure is easy to find, no?

For the interior, you have to find out for which points p on the circle, there exists an open ball (=disk) that is contains in the circle. What are those points?

Last edited: Nov 2, 2012
3. Nov 1, 2012

dapias09

Hi micromass.
I think you mean the function $(x,y) → x^{2}+y^{2}-a^{2}$, being a the radius of the circle. I think you said that this is a closed set because is the preimage of the zero point that is a closed set on the reals.
I don't figure out why you said that now the closure is easy to find, I guess you mean that if the closure of the image is zero, the closure of the preimage is the circle itself.
I think that this set hasn't interior, because there isn't a open ball contained in the circle, all the disks that intersect the circle has elements that don't belong to the circle.
I would prefer to consider the circle as a subset of R^2, I think that it isn't needed to introduce the function "f". What do you think?
Thank you

4. Nov 1, 2012

lavinia

The maximum distance between two points on a circle or a sphere is the diameter. So it is bounded.

Your statement that the circle is the complement of an open set is correct an is not just intuitive. Try to find a proof using this idea.

The function X^2 + y^2 is continuous and the circle is the inverse image of a point, other than zero. E.G. the circle of radius 1 is the inverse image of 1. But a point on the real line is closed and the inverse image of a closed set under a continuous map is also closed. This proof buys you nothing over the direct proof that the circle is the complement of an open set because the proof that a point is closed is the same.

5. Nov 1, 2012

dapias09

Thank you lavinia.
I think I got it.

6. Nov 2, 2012

Bacle2

Edit: there are other approaches/arguments that make the proof for a single point easier than that for S^1.

Last edited: Nov 2, 2012
7. Nov 2, 2012

micromass

Staff Emeritus
What exactly was wrong in lavinia's explanation?

8. Nov 2, 2012

Bacle2

I'm not saying the explanation is wrong, only that the statement is not accurate IMHO:

S/he said that the proof of a point being closed is just as hard, or almost the same as

the proof of S1 being closed. I understood that the proof consisted of showing

the complement is open (in which case both arguments would be equivalent).

But if we use the limit point approach, the fact that a point is closed is immediate, as

any ball about a point is not contained in the point.

Basically, I object to the statement " .......because the proof that a point is closed is the same" . It is the same if we use the complement-is-open

approach, but not otherwise.