Hi! I've been trying to figure out whether I can put "=" sign between two particular sets.
Assume that I have proved that for a function g: R -> R is always true that
g(A ∩ B) ⊆ g(A) ∩ g(B) for all sets A, B ⊆ R.
But it doesn't seem to be true the other way around.
If C ⊆ Z+, D⊆ Z-, consider g(k)={|k|:k€Z}
∃x € C => ∃y€ D: y=-x.
Thus, ∃A={x}, ∃B={y:y=-x}
g(A)=x, g(B)=x. If t€g(A) and t€g(B) => t€ (g(A) ∩ g(B)). Thus, g(A) ∩ g(B)=x.
Similarly, if t€A and t€B => t€A∩B.
With that said, A∩B=∅.
g(A∩B)=g(∅)=∅ => g(A)∩g(D) is not subset of g(A∩B).
Am I mistaken somewhere? Thanks!
Assume that I have proved that for a function g: R -> R is always true that
g(A ∩ B) ⊆ g(A) ∩ g(B) for all sets A, B ⊆ R.
But it doesn't seem to be true the other way around.
If C ⊆ Z+, D⊆ Z-, consider g(k)={|k|:k€Z}
∃x € C => ∃y€ D: y=-x.
Thus, ∃A={x}, ∃B={y:y=-x}
g(A)=x, g(B)=x. If t€g(A) and t€g(B) => t€ (g(A) ∩ g(B)). Thus, g(A) ∩ g(B)=x.
Similarly, if t€A and t€B => t€A∩B.
With that said, A∩B=∅.
g(A∩B)=g(∅)=∅ => g(A)∩g(D) is not subset of g(A∩B).
Am I mistaken somewhere? Thanks!