I'm trying to understand intuitively why the image ( under some function ) of the intersection of subsets of the domain of that function is only contained ( and not equal ) to the intersection of the images of the corresponding sets.
What's the intuition on to why $f\left(\bigcap_{i\in I}A_i\right)\subseteq\bigcap_{i\in I}f(A_i)$ and not $f\left(\bigcap_{i\in I}A_i\right)=\bigcap_{i\in I}f(A_i)$.
Whats the intuition on to why the intersection of images might contain more elements than the image of intersections ?
I even have one counter-example, suppose $f :\mathbb R \to\mathbb R$ defined as $f(x) = x^2$. Then indeed, if A1 = [-1,0] and A2 = [0,1], then the image of the intersections must be contained ( and not be equal ) to the intersection of images.
But i'm trying to abstract and see what's the general requirement for the function and for the family of sets so that the image of intersection is contained and not equal to the intersection of images.
Best Answer
Intuitively this is because a mapping $f:A\rightarrow B$ may identify elements in the sense of:
$$f(x)=f(y) \text{ in } B \text{ where } x\neq y \text{ in }A$$
but a mapping by definition may not distinguish or separate an element into two in this sense:
$$f(x)\neq f(y)\text{ in } B \text{ where } x=y \text{ in } A$$
Say $A_1, A_2\subset A$. Think of $f(A_1)\cap f(A_2)$ as the "identifications" done by $f$ in $B$ of elements from $A$.
Then the mapping $f$ necessarily "identifies" all elements from $A_1\cap A_2$ - i.e.: $$f(A_1\cap A_2)\subset f(A_1)\cap f(A_2)$$
All that can happen additionally during $f$ mapping is that there are more points from $A_1$ and $A_2$ that are not from $A_1\cap A_2$ that are identified by $f$ in $B$ in which case we have $$f(A_1\cap A_2)\not\supset f(A_1)\cap f(A_2)$$