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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.
But I am not sure about connectedness of P. Please help me with this.

Thank you very much.

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    @Potato It's ok. It happens.2012-06-07

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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 $q$.

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    @Ananda: It really depends enormously on the specific space involved. You might be interested in the notions of [cut point](http://en.wikipedia.org/wiki/Cut-point) and [dispersion point](http://en.wikipedia.org/wiki/Dispersion_point); the Wikipedia articles are stubs, but they give you a starting point.2012-06-07
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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.

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    copper.hat$ $Oh yes . In the end best regards to you.2012-06-07