I am solving some problems from the "Problems in real analysis" by aliprantis et. el. (it is available in the pdf format on Google).
I am looking at the first chaper's problems in elementary set theory. Problems 1.1, problems: 7) and 9). I have two questions. First one (relating to 7):
The problem (to validate the claim):
$$f \left(\bigcap_{i \in I} A_i \right) \subseteq \bigcap_{i \in I} f(A_i)$$
Would there ever be a case where the set on the left is a subset of the set on the right. Because I can only think of the cases when the two are equal. So I would rather state the problem as, validate:
$$f \left(\bigcap_{i \in I} A_i \right) = \bigcap_{i \in I} f(A_i)$$
Second question (relating to problem 9): Why do we now have the equals sign instead of the subset or equals sign:
$$f^{-1} \left(\bigcap_{i \in I} B_i \right) = \bigcap_{i \in I} f(B_i)$$
and I would prove this second case in the exacttly same way I have proved the first one, which is:
$$f^{-1} \left(\bigcap_{i \in I} B_i \right) \subseteq f(B_j) \ \ \text{for some $j \in I$} \iff f^{-1} \left(\bigcap_{i \in I} B_i \right) \subseteq \bigcap_{i \in I} f(B_i)$$
Would this be incorrect? (I guess I have three questions then).
Let $f(x)=x^2$. Then $f([0,1]\cap [-1,0])\ne f([0,1])\cap([-1,0])$.