# [FOM] Theorem on Convex sets

joeshipman@aol.com joeshipman at aol.com
Thu Oct 6 10:00:06 EDT 2005

```Jeremy, there is a lot hidden in "We can assume that the surface of A
is made up of 'small' (hyper-)polygons A_i."

I believe this idea can be extended to a rigorous proof (it is easier
if you project from a point in the interior of A than along lines
parallel to a normal to A_i). But the hard part is explaining why you
can make this assumption for an arbitrary convex set.

I regard this as being foundationally relevant because a rigorous proof
(in either 2 or N dimensions) seems to require lots of "machinery". I
believe this is related to the difficulty the ancient Greek geometers
had showing that the circumference of a circle was less than that of a
circumscribed polygon -- they required an additional axiom or an
"exhaustion" argument.

(So who was the first to show rigorously that the boundary of a convex
set can by approximated by hyperpolygons, with appropriate attention to
epsilons?)

This came up because I was showing my 13-year-old daughter how to
estimate pi by inscribed and circumscribed polygons. She accepted the
lower bound because she accepted that a straight line is shorter than
an arc between the same endpoints; but that principle is not enough to
get the upper bound, you need something extra, as she quite properly
noted.

The foundational question: "How much extra" do you need?

-- Joe Shipman

-----Original Message-----
From: Jeremy Clark <jeremy.clark at wanadoo.fr>
To: joeshipman at aol.com
Cc: FOM at cs.nyu.edu
Sent: Thu, 6 Oct 2005 15:18:50 +0200
Subject: Re: [FOM] Theorem on Convex sets

Here is a proof for N dimensions:

We can assume that the surface of A is made up of "small"
(hyper-)polygons A_i. Project each of these outwards onto the surface
of B (which needn't be convex) along lines parallel to a normal to A_i.
You get shadows B_i on B which do not intersect (convexity of A) and
must be of area not smaller than that of A_i (because we chose normals
to A_i), so the surface area of A must be smaller than that of B.

Regards,

Jeremy Clark

On Oct 5, 2005, at 9:40 pm, joeshipman at aol.com wrote:

> If A contained in B are convex sets in the plane, the boundary of A
is
> no larger than the boundary of B.
>
> Is this true in N dimensions, and if so, who proved it?
>
> In 2 dimensions, I have trouble proving the theorem without rather
> advanced tools. Does anyone know a simple proof?
>
> -- Joe Shipman
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