Re: equality of finite spaces
- From: sikari.h@xxxxxxxx (Sikari)
- Date: 14 Apr 2005 08:25:16 -0700
Th
For finite sets A,B of size n and A is a subset of B implies B is a subset of A
or B=A
I thank Arturo Magidin,Bill Dubuque and Keith Lewis for their contributions.
I write hier my proof,where I use induction.
Base:
n=1, a_1 in A_1 => a_1 in B_1 because A_1 is a subset of B_1 but then a_1=b_1;then
B_1 is a subset of A_1 or B_1=A_1
I show arguments for n=2 also
n=2,a_1 in A_2 => a_1 in B_2 because A_2 is a subset of B_2 but then a_1=b_1 or
a_1=b_2 and then a_2=b_2 or a_2=b_1 i.e. a_1=b_1,a_2=b_2 or a_1=b_2,a_2=b_1;
in each case B_2 is a subset of A_2 or B_2=A_2.
Step:
Assume theorem is true for n=k i.e.
if A_k is a subset of B_k, both of size k then B_k is a subset of A_k or B_k=A_k
then for n=k+1 the sets A_{k+1}=A_k U {a_[k+1]},B{k+1}=B_k U {b_[k+1]},
but this means that the only possibility is that a_[k+1]=b_[k+1] because
A_{k+1} is a subset of B_{k+1} but then B_{k+1} is a subset of A_{k+1} or
B_{k+1}=A_{k+1} and we are done.
Do you think my proof is correct? Could it be done without using all
indexes for sets and their elements?
.
- References:
- equality of finite spaces
- From: Sikari
- equality of finite spaces
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