The circle as a set closed and bounded

[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.

Thank you for your help.

 

[micromass]

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?

 

[dapias09]

Hi micromass.
Thanks for your reply.
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

 

[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.

Rien n'est vrai que le beau

 

[dapias09]

Thank you lavinia.
I think I got it.

 

[Bacle2]

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.

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

 

[micromass]

Not so; you can show that a singleton contains all of its limit points (none)

What exactly was wrong in lavinia's explanation?

 

[Bacle2]

What exactly was wrong in lavinia's explanation?

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.