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In $\mathbb{R}^3$ with the usual topology, let:

\begin{align*} V &= \{(x, y, z)\in \mathbb{R}^3 : x^2+y^2+z^2 = 1,\,y\neq 0\} \\ W &= \{(x, y, z)\in\mathbb{R}^3 : y = 0\} \end{align*}

I have to check for connectedness and compactness of $V \cup W$.

Here is my approach: Subspace $V$ being closed and bounded and hence compact while $W$ is closed but unbounded and hence not compact. This implies that $V\cup W$ is not compact.

For connectedness I know that union of two non-disjoint connected sets is connected. Intutively I think that both are non intersecting. But I am not sure with this.

Is there any other way to solve this problem?

Thanks for helping me.

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1 Answer 1

up vote 6 down vote accepted

You’re right that $V\cup W$ is not compact, since $W$ is unbounded, but wrong about $V$ being closed. $V$ is the surface of a sphere minus the equator of the sphere, so it’s not closed: every point on the equator is a limit point. $V$ is also not connected: it’s the disjoint union of the two open hemispheres. However, the missing equator is part of the plane $y=0$, so the union $V\cup W$ is connected.

It may help to let $S=\{\langle x,y,z\rangle\in\Bbb R^3:x^2+y^2+z^2=1\}$; this is the closure of $V$ in $\Bbb R^3$ and is both compact and connected, being just a unit sphere. Note that $$S\cap W=\{\langle x,0,z\rangle:x^2+z^2=1\}=S\setminus V\;;\tag{1}$$ that is, $S$ intersects $W$, the $xz$-plane, exactly in the equator that’s missing from $V$. Thus, $V\cup W=S\cup W$, and that’s the union of two connected sets whose intersection $(1)$ is also connected, so $V\cup W$ is connected.

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Sir, I am not getting this point: $V$ is the surface of a sphere minus the equator of the sphere. Could you explain me please? –  srijan Jun 3 '12 at 5:30
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@srijan: Do you see that $S$ is the surface of a sphere of radius one with centre at the origin? $V$ contains every point $\langle x,y,z\rangle\in S$ except those which $y=0$. When $y=0$, the point is in the $xz$-plane. Thus, $V$ contains every point of $S$ except the ones in the $xz$-plane. Those points lie on the circle $x^2+z^2=1$ in the $xz$-plane; that’s a great circle around the sphere $S$, so I called it the equator. Maybe the picture would be clearer if you changed $V$ to $\{\langle x,y,z\rangle\in\Bbb R^3:z\ne 0\}$: now the missing circle is in the $xy$-plane, where we ... –  Brian M. Scott Jun 3 '12 at 5:41
    
... usually visualize the equator of a sphere. –  Brian M. Scott Jun 3 '12 at 5:41
    
Now I got your point sir. The equator is defined as the intersection of the surface of the sphere with a plane (the equatorial plane) that passes through the center of the sphere and is orthogonal to its rotation axis. The equator is a great circle, that is, its center coincides with the center of the sphere. And that is missing from $V$ so $V$ can't be closed. And due to this we can write $V$ as a disjoint union of two hemispheres. Am i right sir? –  srijan Jun 3 '12 at 5:46
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@srijan: Yes, now you’re visualizing it correctly. (I was using the term equator in a loose sense, meaning any great circle around the sphere.) –  Brian M. Scott Jun 3 '12 at 5:48

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