[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 

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 

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. 
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 
> 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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