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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.

1 Answers 1

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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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    @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.)2012-06-03