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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 $f(A_n)=[0,1/2+1/n]$. *(not sure if I can do this next step) "Take f([0,1/2])={0}". We have $A=[0, 1/2]$. 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.

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1 Answer 1

up vote 1 down vote accepted

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 $[0,2]=[0,1/2]\cup\left(\bigcup_{n\in\mathbb{N}}(1/2+1/(n+1),1/2+1/n]\right)\cup (3/2,2]$, and define $f:X\to X$:

  1. $f(x)=0,\forall x\in[0,1/2];$ (That is, $f$ sends $[0,1/2]$ to $\{0\}$)
  2. For each $n\in\mathbb{N}$, $$f(x)=\left(\frac{1}{2}+\frac{1}{n+1}\right)\left(\frac{x-(1/2+1/(n+1))}{1/n-1/(n+1)}\right),\forall x\in(1/2+1/(n+1),1/2+1/n]$$ (That is, $f$ sends $(1/2+1/(n+1),1/2+1/n]$ to $(0,1/2+1/(n+1)]$)
  3. $f(x)=3(x-3/2),\forall x\in(3/2,2]$ (That is, $f$ sends $(3/2,2]$ to $(0,3/2]$)

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

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