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.