Why Are f(f^{-1}(E)) and E Not Always Equal in Set Theory?

[henryN7]

From page 89 of baby Rudin:

"Theorem Suppose f is a continuous mapping of a compact metric space X into a metric space Y. Then f(x) is compact.

(truncated)

Note: We have used the relation f(f$$^{-1}$$(E))$$\subset$$E, valid for E $$\subset$$ Y. If E$$\subset$$X, then f$$^{-1}$$(f(E))$$\supset$$E; equality need not hold in either case."

Why is this note true? I'm have never taken set theory, so someone please explain to me why they are not equal, but have this relationship as described. I guess it has something to do with onto and the definition of a function? iono.

 

[HallsofIvy]

First, you need to understand that the notation "$f^{-1}(A)$" where A is a set does NOT imply that f is invertible! (I once presented a proof on something similar to this assuming that f was invertible- very embarassing!) If f is invertible (one-to-one and onto) then $f^{-1}(f(A))= A$ and $f(f^{-1}(B))= B$ but not if f is not invertible.

For set A, f(A) is the set of all y values that we can get by applying f to all members of A: $f(A)= \{y | y= f(x)\}$.

$f^{-1}(B)$ is the set of all x values such that f(x) in in B. $f^{-1}(B)= \{x| f(x)\in B\}$

Look at an example: $f(x)= x^2$ where X and Y are both R.

Let A be the closed interval from -1 to 2: [-1, 2].

Then f(A)= [0, 4], the set of all y-values for x from -1 to 2. But applying f to any number between -2 and 2 will also give y-values in [0, 4].
$f^{-1}(f(A))= f^{-1}([0, 4])= [-2, 2]$ and $[-1, 2]\subset [-2, 2]$.

Let B be the closed interval from -1 to 4: [-1, 4].

Then $f^{-1}(B)= [-2, 2]$, the same as $f^{-1}([0, 4])$ because NO values of x given negative y.

And $f(f^{-1}(B)= f([-2, 2])= [0, 4]\subset [-1, 4]$

 

[henryN7]

Awesome explanation, thank you. I knew it was something about bijections, but I couldn't think of any example for the second inequality for some reason.