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This came up in class once:

Suppose we have a set $R \subset \mathbb{R}^2$ and $x \in \mathbb{R}^2$. If we define a function $A$ on all the lines $l$ going through $x$ such that $A(l)$ gives how much of the area of $R$ is "above" (or to the right left of) the line $l$, must $A$ be continuous?

I do think it is continuous. If we take a horizontal line through $x$ and rotate it by $\theta$, we have a continuous function giving the area of the part of $R$ whose pre-image (under this rotation) is the part of $R$ initially under the horizontal line.

However, I'm at loss how to rigorously show continuity using pre-image of open sets is an open set. In particular, how should an open set be defined on the set of lines through $x$, and thus allowing me to rework a proof.

EDIT: What we mean by the ``above'': For a non-vertical line $l$ passing through $x$, we rotate $S$ clockwise until $l$ becomes horizontal. Then $A(l)$ gives the area of $R$ which is above this horizontal line (the image of $l$ after the rotation). If $l$ were vertical, we take the area to the left of $l$,

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  • $\begingroup$ "above" (or to the right of) what does this mean exactly? $\endgroup$ – Mikhail Katz May 30 '13 at 13:57
  • $\begingroup$ What we mean by the ``above'': For line $l$ passing through $x$, we rotate $S$ clockwise until $l$ becomes horizontal. Then $A(l)$ gives the area of $R$ which is above this horizontal line (the image of $l$ after the rotation). $\endgroup$ – lucas May 30 '13 at 14:24
  • $\begingroup$ You can make this a well-defined question with a bit more care. Look at oriented lines (with direction vector $\mathbf v$) and take the normal vector $\mathbf n$ so that $\mathbf v,\mathbf n$ is right-handed. Now the normal vector $\mathbf n$ determines a unique half-plane. Make $\mathbf v$ be a unit vector and it moves on a circle. The standard topology on the circle gives you the topology on the set of oriented lines. (For non-oriented lines, it is also a circle: The angle $\theta$ with the horizontal varies over $[0,\pi]$ with $0$ and $\pi$ identified to a single point.) $\endgroup$ – Ted Shifrin May 30 '13 at 15:31
  • $\begingroup$ @TedShifrin If I understand it correctly, my problem deals with unoriented lines, whose topology corresponds with the standard topology of the circle and thus also with the standard topology on $[0,1]$. An open set of unoriented lines consists of lines spanned by vectors inside a cone over an arc of the circle. Do I understand correctly? $\endgroup$ – lucas May 30 '13 at 16:37
  • $\begingroup$ Yes, you do. But because of all the hullabaloo that everyone's raised about making your problem make sense (i.e., which side is which), just work with oriented lines. Otherwise when you go around $180^\circ$ you'll find that $A(\ell)$ has two different values. $\endgroup$ – Ted Shifrin May 30 '13 at 16:40
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Of course, the arbitrary convention about vertical lines does not save continuity there: if a vertical line is rotated slightly counterclockwise, you switch from area to the left to area above -- which is area to the right. Ted Shifrin pointed out the only way out of this conundrum, which is to use oriented line.

There is another issue if the area of $R$ is infinite. Suppose $R$ is the lower half-plane: then the area of $R$ above $x$-axis is $0$, but any rotation will give area $\infty$. This is not continuous.

So, you should assume that the area of $R$ is finite. Leaving the vertical controversy aside, continuity holds. The reason is that the difference of areas is the area of intersection of $R$ with a thin sector. As one line converges to the other, this intersection shrinks to the empty set. And the following is true for quite general measures: $$\operatorname{area}\left( \bigcap_{n=1}^\infty A_n\right ) = \lim_{n\to \infty } \operatorname{area} A_n \tag1 $$ provided that $A_1\supseteq A_2\supseteq\dots$ and $\operatorname{area} A_1 <\infty$. In your case both sides of (1) are $0$, which yields continuity. (By $A_n$ I mean the intersection of $R$ with an aforementioned thin sector.)

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If the line is vertical, do you take the area in the plane to the right? Apparently for all other lines you take the area "above". Then the region whose area you are calculating does not depend continuously on the parameter. Therefore there is no reason to think that this will define a continuous function.

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  • $\begingroup$ If the line is vertical, I take it to the left. I realized just now that I had another orientation for the vertical line. Let me edit the question. $\endgroup$ – lucas May 30 '13 at 14:48
  • $\begingroup$ Left or right, the region whose area you are calculating still does not depend continuously on the parameter! $\endgroup$ – Mikhail Katz May 30 '13 at 14:54
  • $\begingroup$ Allow me a question to help me understand it better. Let $l_0$ be a horizontal line through $x$ and define $f:[0,\pi] \rightarrow \mathbb{R}$ as $f(\theta)$ gives the area of $P$ under $l_0$ after rotating $R$ clockwise by an angle $\theta$ about $x$. How different is $f$ from the function $A$ (in the problem)? $\endgroup$ – lucas May 30 '13 at 15:27

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