Let A be a nonempty set and fix the set B where B is a subset of A. Define the relation R on the power set of A as follows: for all subsets X, Y of A, X R Y if and only if B intersect X = B intersect Y.
Suppose we make some sets...
A={1,2,3} and B={1,2}
How do i find the partition of the power set of A? Do I need to know what X and Y are first?
A relation partitions a set if and only if it is an equivalence relation. The partitions are known as equivalence classes. If you've worked with congruence classes (mod n), then you will have a good idea of this concept from experience.
Wikipedia states it thus: "For any equivalence relation on a set X, the set of its equivalence classes is a partition of X. Conversely, from any partition P of X, we can define an equivalence relation on X by setting x ~ y precisely when x and y are in the same part in P. Thus the notions of equivalence relation and partition are essentially equivalent." (source)
So we know beforehand from the problem statement that we are dealing with an equivalence relation, but we could verify it if we wanted.
But anyway, I'd like to consider a larger example to better illustrate how this applies.
A = {1,2,3,4,5,6}
B = {1,2,3}
Now notice the powerset of B has 2^3 elements, and each of these sets defines an equivalence class. So for example, {1,2} is in the powerset of B. Consider
C = {1,2,5}
D = {1,2,5,6}
E = {1,2,4}
F = {1,2}
These are all equivalent under the relation R.
Does that make sense?
"To know X,Y first"? you need to check which subsets of A are related according to the given relation...for example, , since
. From here, both will be in the same partition set (i.e., equivalence class) of the power set of A, since they're r-related.
Well, now you find all the partitions sets (=equiv. classes) one by one. After all, there're only 8 subsets here...
Tonio
I'm not sure what you mean
{3} is in the powerset of B.
So it is in the partition described completely by
{3} 000
{3,6} 001
{3,5} 010
{3,5,6} 011
{3,4} 100
{3,4,6} 101
{3,4,5} 110
{3,4,5,6} 111
where I wrote binary numbers alongside to show how I was choosing elements from A that are not in B. So the above 8 sets all have the same intersection with B, so are equivalent under R, and there are no other sets in this partition.
Am I wrong?
Let A be a nonempty set and fix the set B where B⊆A. Define the relation R on ℘(A), the power set of A, as follows: for all subsets X,Y of A, X R Y iff B∩X = B∩Y.
Suppose we make some sets...
A={1,2,3}
B={1,2}
How do i find the partition of the power set of A? Do I need to know what X and Y are first?
well we can find that:
℘(A) = {{},{1},{2},{3},{1,2},{1,3},{2,3},{1,2,3}}
therefore the partition of the power set of A would be:
[{},{3}],
[{1},{1,3}],
[{2},{2,3}],
[{1,2},{1,2,3}]
reasoning:
[{},{3}]
if X={}, Y={} and B={1,2}
B∩X = B∩Y = {} = {} which is true
if X={}, Y={3} and B={1,2}
B∩X = B∩Y = {} = {} which is true
if X={3}, Y={3} and B={1,2}
B∩X = B∩Y = {} = {} which is true
if X={3}, Y={} and B={1,2}
B∩X = B∩Y = {} = {} which is true
~~~~~~~~~~~~~~~~~~~~~~~~~~
[{1},{1,3}]
if X={1}, Y={1} and B={1,2}
B∩X = B∩Y = {1} = {1} which is true
if X={1}, Y={1,3} and B={1,2}
B∩X = B∩Y = {1} = {1} which is true
if X={1,3}, Y={1} and B={1,2}
B∩X = B∩Y = {1} = {1} which is true
if X={1,3}, Y={1,3} and B={1,2}
B∩X = B∩Y = {1} = {1} which is true
~~~~~~~~~~~~~~~~~~~~~~~~~~
similarly for [{2},{2,3}] and [{1,2},{1,2,3}]