# Let $D$ be the union of all lines through $P=(0,0,2)$ and the open ball $B((0,0,0),1)$. Show that $D$ is not open nor closed.

I am studying multivariable calculus and would like to ask if I have answered the following exercise correctly:

1. Let $D$ be the union of all lines through $P=(0,0,2)$ and the open ball $B((0,0,0),1)$ (that is, all lines passing through both $P$ and that ball). Show $D$ is not open nor closed.

2. Describe the closure of $D$, the boundary points of $D$, and the interior points of $D$ (there's no need to prove the correctness of the description).

I will not write the whole proof, just the general details. Please tell me if I have the right idea.

For 1., I show $P$ is a boundary point that is contained in $D$ (thus $D$ is not open), and that $(0,1,0)$ is a limit point that is not in $D$ (thus $D$ is not closed).

For 2., the closure is the union of all lines passing through the closed ball $B((0,0,0),1)$ and $P$, the interior points are the points in $D - P$, and the boundary points are the union of lines passing through a boundary point of the open ball $B((0,0,0),1)$, minus $B((0,0,0),1)$ itself and $B((0,0,3),1)$.

Thanks!

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I'm having trouble parsing your question. Do you mean D is the union of all the lines through P, plus the ball as well? Or do you mean D is the union of lines that pass through both P and the ball? – user22805 Mar 24 '12 at 4:26
@DavidWallace Surely the second, as otherwise it would be the whole set (presumably he's working in $\mathbb R^3$) which is open and closed. – Alex Becker Mar 24 '12 at 4:50
@DavidWallace: Yes, Alex is correct, I mean the second. It is the union of all lines that pass both through P, and through a point in the ball. – ro44 Mar 24 '12 at 5:07
I don't think your claim about $(0,1,0)$ is right. The idea is right, but that's not the right point. You need something like $(0,a,b)$. with $b>0$. – alex.jordan Mar 24 '12 at 5:42
@alex.jordan: I've chosen it because it's a boundary point of the ball, basically. Since the ball is contained within D, a series inside D that approximates $(0,1,0)$ is $(0,1-1/n,0)$. On the other hand, $(0,1,0)$ doesn't seem to be in D. Let's assume it is, so we have a point $R$ inside the ball such that a line exists which passes through $(0,1,0)$, $R$ and $P$. This means that for some $t$, $R=((0,1,0)-(1-t)P)/t=(0,1,2-2/t)$ - but this is impossible because that would mean R is outside the ball. Does this seem incorrect? – ro44 Mar 24 '12 at 5:45

Fix $p \in B((0, 0, 1), 1)$ and denote the line joining $(0, 0, 2)$ and $p$ by $L_p$. Let $(a, b, 0)$ be the intersection of $L$ and the $xy$-plane, then

$$L_p = \{(0, 0, 2) + t[(a, b, 0) - (0, 0, 2)] \mid t \in \mathbb{R}\} = \{(ta, tb, 2 - 2t) \mid t \in \mathbb{R}\}.$$

By assumption, $L$ intersects the ball $B((0, 0, 1), 1)$ so it intersects the sphere $\partial B((0, 0, 0), 1)$ twice. Therefore there are two real solutions to $(ta)^2 + (tb)^2 + (2 - 2t)^2 = 1$ which simplifies to $(a^2 + b^2 + 4)t^2 - 8t + 3 = 0$. In order for this quadratic to have two real solutions, we must have

\begin{align*} 0 &< \Delta\\ &= (-8)^2 - 4(a^2 + b^2 + 4)(3)\\ &= 64 - 12a^2 -12b^2 - 48\\ &= 16 - 12a^2 - 12b^2\\ &= 4(4 - 3(a^2+b^2)). \end{align*}

Therefore, $a^2 + b^2 < \frac{4}{3}$. Conversely, if $a^2 + b^2 < \frac{4}{3}$, then the line joining $(0, 0, 2)$ and $(a, b, 0)$ intersects the ball $B((0, 0, 1), 1)$. So

$$D = \bigcup_{p \in B((0, 0, 1), 1)}L_p = \bigcup_{a^2 + b^2 < \frac{4}{3}} L_{(a, b, 0)}.$$

Let $(x, y, z) \in D$, then $(x, y, z) = (at, bt, 2 - 2t)$ for some $a, b, t \in \mathbb{R}$ with $a^2 + b^2 < \frac{4}{3}$. If $t = 0$, $(x, y, z) = (0, 0, 2)$. If, $t \neq 0$, that is $(x, y, z) \in D\setminus\{(0, 0, 2\}$, we have

$$x^2 + y^2 = a^2t^2 + b^2t^2 = (a^2 + b^2)t^2 < \frac{4}{3}t^2 = \frac{4}{3}\left(1 - \frac{1}{2}z\right)^2 = \frac{1}{3}(2 - z)^2.$$

Conversely, given $(x, y, z) \in \mathbb{R}^3$ such that $x^2 + y^2 < \frac{1}{3}(2 - z)^2$, note that $2 - z\neq 0$, so

\begin{align*} \left(\frac{x}{2 - z}\right)^2 + \left(\frac{y}{2-z}\right)^2 &< \frac{1}{3}\\ \left(\frac{2x}{2 - z}\right)^2 + \left(\frac{2y}{2-z}\right)^2 &< \frac{4}{3}. \end{align*}

Furthermore, $(x, y, z) = (x_0t, y_0t, 2 - 2t)$ for $t = \frac{1}{2}(2 - z)\neq 0$ where $x_0 = \dfrac{2x}{2 - z}$ and $y_0 = \dfrac{2y}{2-z}$. So $(x, y, z) \in D\setminus\{(0, 0, 2)\}$.

Therefore,

$$D = \left\{(x, y, z) \in \mathbb{R}^3 \mid x^2 + y^2 < \frac{1}{3}(2-z)^2\right\}\cup\{(0, 0, 2)\}.$$

That is, $D$ is the point $(0, 0, 2)$ together with the set of points enclosed by the (double) cone defined by $x^2 + y^2 < \frac{4}{3}(2-z)^2$, but not including the points on the cone itself.

With this description of the set $D$ at hand, the problem becomes much simpler. I will answer the second question first as it is easier. The closure, interior and boundary of $D$ are

\begin{align*} \overline{D} &= \left\{(x, y, z) \in \mathbb{R}^3 \mid x^2 + y^2 \leq \frac{1}{3}(2-z)^2\right\}\\ D^{\circ} &= \left\{(x, y, z) \in \mathbb{R}^3 \mid x^2 + y^2 < \frac{1}{3}(2-z)^2\right\}\\ \partial D &= \left\{(x, y, z) \in \mathbb{R}^3 \mid x^2 + y^2 = \frac{1}{3}(2-z)^2\right\}. \end{align*}

As $D \neq \overline{D}$, $D$ is not closed and as $D^{\circ} \neq D$, $D$ is not open.

You can give a more complete proof of the fact that that $D$ is neither open nor closed, but once you understand what $D$ is geometrically, I think it is pretty clear.

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