Finding a function and a sequence of recursively defined sets that gives a counterexample. Let X be a set and $f:X\rightarrow X$. Define the sequence $(A_n)$ recursively by $A_1=X$ and $A_n=f(A_{n-1})$ for $n>1$. Let $A=\bigcap_{n\in N}A_n$. My problem is asking me to show $f(A)⊊A$. I have already shown $f(A)\subset A$, but I'm having a hard time finding a function $f$ and sequence $(A_n)$ such that $f(A)\neq A$.
One of my attempts:
Let $X=A_1=[0,2]$, and for $n>1$ take $f(A_n)=\left[0,\frac{1}{2}+\frac{1}{n}\right]$. *(not sure if I can do this next step) "Take $f\left(\left[0,\frac{1}{2}\right]\right)={0}$". We have $A=\left[0, \frac{1}{2}\right]$. Therefore, $f(A)={0}\neq [0,2]=A$
All of my counterexamples led to a similar argument above, but I'm not sure if it is a valid argument. My only hope is that is that I am taking $f$ to be a dilation fixing 0 each time.
 A: I think you have to define an $f:X\to X$, and not define only the images (because not necessarily such an $f$ exists).
But in you example, write $[0,2]=\left[0,\frac{1}{2}\right]\cup\left(\bigcup_{n\in\mathbb{N}}\left(\frac{1}{2}+\frac{1}{n+1},\frac{1}{2}+\frac{1}{n}\right]\right)\cup \left(\frac{3}{2},2\right]$, and define $f:X\to X$:


*

*$f(x)=0,\forall x\in\left[0,\frac{1}{2}\right];$ (That is, $f$ sends $\left[0,\frac{1}{2}\right]$ to $\{0\}$)

* For each $n\in\mathbb{N}$,
$$f(x)=\left(\frac{1}{2}+\frac{1}{n+1}\right)\left(\frac{x-\left(\frac{1}{2}+\frac{1}{(n+1)}\right)}{\frac{1}{n}-\frac{1}{(n+1)}}\right),\forall x\in\left(\frac{1}{2}+\frac{1}{n+1},\frac{1}{2}+\frac{1}{n}\right]$$
Note: That is, $f$ sends $\left(\frac{1}{2}+\frac{1}{n+1},\frac{1}{2}+\frac{1}{n}\right]$ to $\left(0,\frac{1}{2}+\frac{1}{n+1}\right]$

* $f(x)=3\left(\frac{x-3}{2}\right),\forall x\in\left(\frac{3}{2},2\right]$ (That is, $f$ sends $\left(\frac{3}{2},2\right]$ to $\left(0,\frac{3}{2}\right]$)


So, $f$ satisfies your requirements, and then $f(A)\ne A$.
