Re: set proof valid?

In article <D3C0f.196$i%.11@fed1read07>, vsgdp <spam@xxxxxxxx> wrote:
>I have CH(T) is the smallest convex set that contains T (i.e., it is the
>convex hull).
>I have CH(S) is a convex set that also contains T.
>
>CH(T) = CH(T) intersect CH(S) contained in CH(S) ??

In general, it is true that A intersect B is contained in B, so the
second half of this is correct. However, to show that
CH(T) = CH(T) intersect CH(S) you would need to establish what is
apparently your conclusion, namely that CH(T) is contained in
CH(S). So this assertion is exactly what you want to prove (note that
for arbitary sets, A intersect B = A if and only if A is contained in
B).

>Therefore, CH(T) contained in CH(S)?

No. Rather: use the meaning of "smallest" in convex hull. BY
DEFINITION, the convex hull CH(T) is a set that satisfies three
conditions:

(i) CH(T) contains T.
(ii) CH(T) is convex.
(iii) CH(T) is the smallest set that satisfies (i) & (ii);
i..e, if X is ANY set which contains T and is convex, then
_________ is contained in _________.


--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================

Arturo Magidin
magidin@xxxxxxxxxxxxxxxxx

.

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