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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?2012-03-24
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    @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.2012-03-24
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    @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.2012-03-24
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    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 $b>0$.2012-03-24
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    @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?2012-03-24
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    I don't agree with your description of the boundary points of $D$. Remember that boundary points of the open ball $B((0,0,0),1)$ form an entire sphere. So the union of lines from $(0,0,2)$ to points somewhere on this sphere will make a solid cone.2012-03-24
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    @r044 $(0,1,0)$ _is_ in $D$. There are points on the line connecting this point to $(0,0,2)$ that are in the interior of $B((0,0,0),1)$. Take for example $(0,0.9,0.2)$.2012-03-24
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    @alex.jordan: I see your point about the boundary points. Do you have any suggestions? Thanks!2012-03-24
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    @alex.jordan: Ah, you're right about (0,1,0), I'll rethink this through.2012-03-24
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    Draw a picture in 2D of a cross-section. I think you can find that the point which would play the role that you wanted $(0,1,0)$ to play is something like $(0,\sqrt{3}/2,1/2)$. As for the boundary points, they make the surface of this cone. If you want to describe them in with lines, find the circle on the ball where lines to $P$ are tangent to the ball. The boundary points are the union of all such lines.2012-03-24
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    @alex.jordan: Thanks for the tip about the boundary points, that's a nice way to describe it. Unfortunately, my lack of spatial imagination (and drawing talent) makes it difficult for me to understand what you're saying about $(0,\sqrt{3}/2,1/2)$ - can you describe how you arrived at that in more detail?2012-03-24
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    I guess I can just solve this algebraically, e.g. looking for some vector $(0,a,b)$ such that the norm of the resulting vector will be bigger than 1.2012-03-24

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