b***e
发帖数: 1419 | 1 For the first question, we first prove the following:
pr1(F) = X and pr1(G) = Y.
These are pretty straightforward by showing contradiction
otherwise. If there exist x in X such that x is not in
pr1(F), then F(x) = \phi, and G(F(x)) = \phi != {x}.
For any x in X, suppose F(x) = B (!= \phi), then G(B) = {x}.
This implies for any y in B, G(y) = {x}. Pick any y0 in B.
We know G(y0) = {x}. Dually, we can show that F(x) = {y0}.
In summary, we know for any x in X, there exists one and only
one y0, s |
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