ovoleg said:
Homework Statement
If A is a subset of B and B is a subset of C, then A is a subset of C. But if A = {3}, B={{3},5}, and C = {B, 17}, then A is contained in B and B is contained in C, but A is not contained in C. The set C has exactly two members, and it is easy to see that neither of these members is a set of A
Okay, in the second case, A is a
member of B, not a subset. Also, in that case, B is a
member of C, not a subset.
Homework Equations
Why? if B = {{3}, 5} then shouldn't C be = {{3}, 5, 17} ??
Well, this is an example. C can be anything that demonstrates the point! In the given example, B is a
member of C, not a subset. They you want it, C= {{3},5,17}, B is a
subset of C, not a member. Of course, even if B is a subset of C, since A is not a subset of B, it would not follow that A must be a subset of C, so either way makes the point.
The Attempt at a Solution
I don't see how A is not contained in C
Can someone clarify?
Thanks!
What do YOU mean by "contained in"? Subset or member? "A member of" and "a subset of" are completely different concepts. You should use one of those and not "contained in" which is ambiguous. In the example as given, C= {B, 17}, B is a member of C while A is neither a member of C nor a subset of it. In your example, C= {{3},5,17}, B is a subset of C and A is a member of C but not a subset of C.
Notice, by the way, that in the "theorem" you state: "If A is a subset of B and B is a subset of C, then A is a subset of C" if either of the hypotheses is not true (A is not a subset of B or B is not a subset of C) then the conclusion "A is a subset of C" is not necessarily true. But it still might happen to be true! A really strange example would be with A= {3}, B= {{3},5}, C= {{3},5, 3}. The A is a
member of B, not a subset. B is a subset of C and A is both a member of C (because of the {3} in C) and a subset (because of the 3 in C).