- #36
EonsNearby
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- 0
Okay, well since these are countably finite sets, I can verify that A and C have a cardinality of 3 and that B and D have a cardinality of 5. Since A and B do not have any elements that match, the intersection of these two sets is the empty set (the same goes for C and D). Anyway, a bijection for A -> C would be the f(x) = x + 8 where x is an element from A. This is one-to-one because every element in A will be matched to a unique element in C. This is also onto because the range of f is every element in C. To visually demonstrate:
Set A++++++Set C
1 -----------> 9
2 -----------> 10
3 -----------> 11
A bijection for B -> D would be g(x) = x + 8 where x is an element from B. This is one-to-one because every element in B will be matched to a unique element in D. This is also onto because the range of g is every element in D. To visually demonstrate:
Set B+++++++Set D
4 -----------> 12
5 -----------> 13
6 -----------> 14
7 -----------> 15
8 -----------> 16
Am I correct so far?
Set A++++++Set C
1 -----------> 9
2 -----------> 10
3 -----------> 11
A bijection for B -> D would be g(x) = x + 8 where x is an element from B. This is one-to-one because every element in B will be matched to a unique element in D. This is also onto because the range of g is every element in D. To visually demonstrate:
Set B+++++++Set D
4 -----------> 12
5 -----------> 13
6 -----------> 14
7 -----------> 15
8 -----------> 16
Am I correct so far?