# Identifying compactness and connectedness of subspace P = $\{(x, y, z)\in \mathbb{R}^3 : x^2+y^2+z^2 = 1 ,~ x^2+y^2\neq 0\}$

I have to check for compactness and connectedness of subspace P = $\{(x, y, z)\in \mathbb{R}^3 : x^2+y^2+z^2 = 1 ,~ x^2+y^2\neq 0\}$

Intuitively it is clear to me that subspace P is not compact as it is not closed.

Thank you very much.

-
Please ignore my previous faulty answer. Sorry! –  Potato Jun 7 '12 at 6:26
@Potato It's ok. It happens. –  srijan Jun 7 '12 at 6:30

Let $S=\{\langle x,y,z\rangle:x^2+y^2+z^2=1\}$; this is the surface of the sphere of radius $1$ centred at the origin. What points must be removed from $S$ to get $P$? You have to remove the points the points $\langle x,y,z\rangle\in S$ such that $x^2+y^2=0$. Which points are these?

If $\langle x,y,z\rangle\in S$, then $x^2+y^2+z^2=1$, so if in addition $x^2+y^2=0$, it must be that $z^2=1$, and hence $z=\pm 1$. That is, the only points of $S$ that are removed to get $P$ are $\langle 0,0,1\rangle$ and $\langle 0,0,-1\rangle$. You could think of these two points as the north and south poles of $S$; $P$ is then everything except these two poles. You should have little trouble showing that $P$ is connected.

For example, notice that if $p$ and $q$ are distinct points of $P$, you can travel from $p$ to $q$ without leaving $P$: just follow a line of constant longitude from $p$ to the equator, then go round the equator until you reach the longitude of $q$, and finally follow a line of constant longitude from the equator to $q$.

-
Thank you very much sir. I have one doubt; Here two points are removed and still set is connected. How many points can we removed so that set still remains connected? I mean can we removed countably infinite number of points and still set is connected? –  srijan Jun 7 '12 at 6:04
@srijan, note that if you remove half the equator, the remainder is still connected. –  Lubin Jun 7 '12 at 6:10
If you remove the entire equator except for one point it will remain path connected. Just creates a very busy point :-). –  copper.hat Jun 7 '12 at 6:25
@srijan: Thank you. As you may have guessed, I like explaining things. :-) –  Brian M. Scott Jun 7 '12 at 6:27
@Potato: Not to worry: I once squared $2$ and came up with $16$. –  Brian M. Scott Jun 7 '12 at 6:31

You should try to visualize the set $P$. The only points that are eliminated are those for which $x^2+y^2 = 0$, or in other words, the points for which $x=y=0$. There are only two such points on the sphere corresponding to the north and south poles $(0,0,1), (0,0,-1)$.

It should be clear that the set is not closed, since I can take the points $(\sqrt{\frac{t(2-t)}{2}}, \sqrt{\frac{t(2-t)}{2}}, 1-t)$ as $t \to 0$. The limit point is the north pole, which is not in $P$.

It should also be clear that $P$ is connected, as I get from anywhere on earth to anywhere else (barring the poles, of course) without going through the poles. In fact, it is path connected. Take any two points in $P$. Then draw paths from either point to the equator, then connect the points on the equator.

-
thank you very much. I understand now. +1 for your answer. –  srijan Jun 7 '12 at 6:06
Glad to help. I would strongly encourage you to spend more time trying to visualize the spaces you work with. It is not always easy (or possible), but often there are useful analogs. –  copper.hat Jun 7 '12 at 6:13
Thank you very much for valuable comments and answers."Take any two points in P. Then draw.." this line is self contained. Indirectly we can say If A is connected to B(equator) and B is connected to C. then we have path from C. Am i right? –  srijan Jun 7 '12 at 6:18
Yes indeed, you are. –  copper.hat Jun 7 '12 at 6:24
Sometimes it is really hard to accept one answer when you have more than one answers of equal importance. –  srijan Jun 7 '12 at 6:27