# Show that the unit sphere is closed

• Oct 23rd 2013, 12:10 PM
sakuraxkisu
Show that the unit sphere is closed
(note: please ignore the version of this I posted in the pre-university geometry section)

Hi, my question is:

Show that the unit sphere with center 0 in R3, namely the set S2 :={x in R3 : (x1)^2+(x2)^2+(x3)^2=1} is closed in R3.

What I've done so far is say that to prove it's closed, I need to show that the complement of S2, XS2, is open, so:

XS2 :={x in R3 : (x1)^2+(x2)^2+(x3)^2 is not equal to 1}. I then stated the definition of an open set:

XS2 is open if for all x in S2 there exists r>0 such that B(x, r) is a subset of S2, B(x, r)={y : d(x, y)<r}.

What I'm stuck on in this question is the choice of r, and how to show that, for any y in B(x,r), y is in S2.

Any help would be appreciated :)
• Oct 23rd 2013, 12:31 PM
HallsofIvy
Re: Show that the unit sphere is closed
Do you understand that if "[tex]x_1^2+ x_2^2+ x_3^2\ne 1[tex]", then either [tex]x_1^2+ x_2^2+ x_3^2> 1[tex] or [tex]x_1^2+ x_2^2+ x_3^2< 1[tex]? So do this in two parts:

1) If $x_1^2+ x_2^2+ x_3^2> 1$, let $d= x_1^2+ x_2^2+ x_3^2- 1$.

2) If $x_1^2+ x_2^2+ x_3^2< 1$, let $d= 1- x_1^2+ x_2^2+ x_3^2$.
• Oct 23rd 2013, 12:57 PM
Plato
Re: Show that the unit sphere is closed
Quote:

Originally Posted by sakuraxkisu
Show that the unit sphere with center 0 in R3, namely the set S2 :={x in R3 : (x1)^2+(x2)^2+(x3)^2=1} is closed in R3.

What I've done so far is say that to prove it's closed, I need to show that the complement of S2 is open, so:
XS2 :={x in R3 : (x1)^2+(x2)^2+(x3)^2 is not equal to 1}. I then stated the definition of an open set:
X`S2 is open if for all x in S2 there exists r>0 such that B(x, r) is a subset of S2, B(x, r)={y : d(x, y)<r}.
What I'm stuck on in this question is the choice of r, and how to show that, for any y in B(x,r), y is in S2.

For notation say that $C=(S^2)^c$ the complement.

If $P: (a,b,c)\in C$ then find the distance. $\delta>0$ from $P$ to $S^2$.
That can be done using analytic geometry. Now let $r=\tfrac{\delta}{2}$.
The ball $\frak{B}(P;r)\cap S^2=\emptyset.$
• Oct 23rd 2013, 01:04 PM
sakuraxkisu
Re: Show that the unit sphere is closed
Yeah, having two different cases makes sense. But I was wondering, for case 2), did you mean d=1-((x1)^2+(x2)^2+(x3)^2) or d=1-(x1)^2+(x2)^2+(x3)^2 ? Thank you.
• Oct 23rd 2013, 01:14 PM
SlipEternal
Re: Show that the unit sphere is closed
Quote:

Originally Posted by HallsofIvy
Do you understand that if "[tex]x_1^2+ x_2^2+ x_3^2\ne 1[tex]", then either [tex]x_1^2+ x_2^2+ x_3^2> 1[tex] or [tex]x_1^2+ x_2^2+ x_3^2< 1[tex]? So do this in two parts:

1) If $x_1^2+ x_2^2+ x_3^2> 1$, let $d= x_1^2+ x_2^2+ x_3^2- 1$.

2) If $x_1^2+ x_2^2+ x_3^2< 1$, let $d= 1- x_1^2+ x_2^2+ x_3^2$.

Suppose $x_1=x_2=2, x_3=1$. Then you would be setting $d=4$, and the distance from $(2,2,1)$ to $(0,0,0)$ is $\sqrt{5}<4$. So that ball would have points inside and outside the sphere.

Instead, consider the point $(x_1,x_2,x_3)$ as a vector. Let $\vec{x} = (x_1,x_2,x_3)$. Let $r = \left|\dfrac{|\vec{x}|-1}{2}\right|$ where $|\vec{x}|$ is the magnitude of $\vec{x}$. That radius will guarantee you remain in the respective "inside" or "outside" of the sphere.

Edit: This is actually the method that Plato just posted.
• Oct 23rd 2013, 02:22 PM
sakuraxkisu
Re: Show that the unit sphere is closed
Thank you for your help. I'm having a bit of trouble showing that, for case 1, where (x1)^2+(x2)^2+(x3)^2 >1, any y in B(x, r) is in S2. I know that I need to get:
y=sqrt((y1)^2+(y2)^2+(y3)^2) > 1
I'm not sure how to though, would you be able to help me? Thank you.
• Oct 23rd 2013, 02:28 PM
SlipEternal
Re: Show that the unit sphere is closed
Quote:

Originally Posted by sakuraxkisu
Thank you for your help. I'm having a bit of trouble showing that, for case 1, where (x1)^2+(x2)^2+(x3)^2 >1, any y in B(x, r) is in S2. I know that I need to get:
y=sqrt((y1)^2+(y2)^2+(y3)^2) > 1
I'm not sure how to though, would you be able to help me? Thank you.

If you follow Plato's advice and take the minimum distance from your point to a on the sphere and divide it by 2, there is no need to break down cases. The formula I gave you does just that.
• Oct 23rd 2013, 02:36 PM
sakuraxkisu
Re: Show that the unit sphere is closed
So then how would you show that, for any y in B(x, r), y is in S2, using the formula for r that you gave me instead of the 2 cases?
• Oct 23rd 2013, 02:44 PM
SlipEternal
Re: Show that the unit sphere is closed
I would say, "Because we chose a radius that is smaller than the minimum distance between the point and the sphere, all points in our ball are in $\mathbb{R}^3\setminus S^2$".

Edit: If you need something beyond that, then yeah, you should break it down into cases:

If $\vec{x}\in\mathbb{R}^3\setminus S^2$ and $|\vec{x}|<1$ then $d(\vec{0},\vec{y}) \le d(\vec{0},\vec{x}) + d(\vec{x},\vec{y}) < |\vec{x}| + \left|\dfrac{1-|\vec{x}|}{2}\right| = \dfrac{1+|\vec{x}|}{2} < 1$.

If $\vec{x}\in\mathbb{R}^3\setminus S^2$ and $|\vec{x}|>1$ then $d(\vec{0},\vec{x})\le d(\vec{0},\vec{y}) + d(\vec{y},\vec{x})$. Subtracting the distance from y to x from both sides we get: $d(\vec{0},\vec{y}) \ge d(\vec{0},\vec{x}) - d(\vec{y},\vec{x}) = |\vec{x}| - \left|\dfrac{1-|\vec{x}|}{2}\right| = |\vec{x}| + \dfrac{1-|\vec{x}|}{2} = \dfrac{1+|\vec{x}|}{2} > 1$.
• Oct 23rd 2013, 03:22 PM
SlipEternal
Re: Show that the unit sphere is closed
Oh, and I should add, another way to describe the points of the sphere: $S^2 = \{\vec{x}\in\mathbb{R}^3: |\vec{x}| = 1\}$
• Oct 26th 2013, 09:16 AM
Rebesques
Re: Show that the unit sphere is closed
Ofcourse, you could also define the (continuous) map $f(x_1,x_2,x_3)=x_1^2+x_2^2+x_3^2-1$ and use the fact that
$S^2=f^{-1}(\{0\})$.
• Oct 26th 2013, 05:29 PM
ayrcast
Re: Show that the unit sphere is closed
you could also prove it proving that is a compact, in other words that for every open cover you could find a finite open subcover. If you show this (is not difficoult) you can prove that is closed (compacts in T2 spaces are closed).
• Oct 30th 2013, 11:23 AM
sakuraxkisu
Re: Show that the unit sphere is closed
Thank you everyone for all your help! :)