# Let D be the union of all lines through $P=(0,0,2)$ and the open ball $B((0,0,0),1)$, show 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