Given a set $A$ and a function $f: A\rightarrow A$, denote by $f_{1}(A)=f(A)$, $f_{2}(A)=f(f_{1}(A))$, $f_{3}(A)=f(f_{2}(A))$, and so on, ($f_{n}(A)=f(f_{n-1}(A))$, where the notation $f(B)$ means the set $\{ f(x) \ : \ x\in B\}$ of images of points from $B$).
Denote also by $f_{\infty}(A)=f_{1}(A)\cap f_{2}(A)\cap \ldots = \bigcap_{n\geq 1}f_{n}(A)$.

a) Show that if $A$ is finite, then $f(f_{\infty}(A))=f_{\infty}(A)$.

b) Determine if the above is true for $A=\mathbb{N}\times \mathbb{N}$ and the function 
\[f\big((m,n)\big)=\begin{cases}(m+1,n) & \mbox{if }n\geq m\geq 1 \\ (0,0) & \mbox{if }m>n \\ (0,n+1) & \mbox{if }n=0. \end{cases}\]
### Part (a)

1. **Understanding the Problem:**
   We need to show that if \( A \) is finite, then \( f(f_{\infty}(A)) = f_{\infty}(A) \). 

2. **Sequence of Sets:**
   Define the sequence of sets \( f_n(A) \) as follows:
   \[
   f_1(A) = f(A), \quad f_2(A) = f(f_1(A)), \quad f_3(A) = f(f_2(A)), \quad \ldots, \quad f_n(A) = f(f_{n-1}(A))
   \]
   Also, define:
   \[
   f_{\infty}(A) = \bigcap_{n \geq 1} f_n(A)
   \]

3. **Finite Set Property:**
   Since \( A \) is finite, the sequence of sets \( f_n(A) \) is a descending chain of subsets of \( A \):
   \[
   A \supseteq f_1(A) \supseteq f_2(A) \supseteq f_3(A) \supseteq \ldots
   \]
   Because \( A \) is finite, this descending chain must stabilize at some point. That is, there exists some \( k_0 \) such that:
   \[
   f_{k_0}(A) = f_{k_0+1}(A) = f_{k_0+2}(A) = \ldots
   \]

4. **Stabilization:**
   Let \( B = f_{k_0}(A) \). Then:
   \[
   f(B) = f(f_{k_0}(A)) = f_{k_0+1}(A) = B
   \]
   Hence, \( B \) is a fixed point of the function \( f \).

5. **Intersection:**
   Since \( f_{\infty}(A) = \bigcap_{n \geq 1} f_n(A) \) and the sequence stabilizes at \( f_{k_0}(A) \), we have:
   \[
   f_{\infty}(A) = f_{k_0}(A)
   \]
   Therefore:
   \[
   f(f_{\infty}(A)) = f(f_{k_0}(A)) = f_{k_0+1}(A) = f_{k_0}(A) = f_{\infty}(A)
   \]

Thus, we have shown that if \( A \) is finite, then \( f(f_{\infty}(A)) = f_{\infty}(A) \).

\(\blacksquare\)

### Part (b)

1. **Understanding the Function:**
   Given \( A = \mathbb{N} \times \mathbb{N} \) and the function \( f \) defined as:
   \[
   f((m,n)) = 
   \begin{cases} 
   (m+1, n) & \text{if } n \geq m \geq 1 \\
   (0, 0) & \text{if } m > n \\
   (0, n+1) & \text{if } n = 0 
   \end{cases}
   \]

2. **Behavior of \( f \):**
   - If \( n \geq m \geq 1 \), \( f((m,n)) = (m+1, n) \).
   - If \( m > n \), \( f((m,n)) = (0, 0) \).
   - If \( n = 0 \), \( f((m,0)) = (0, n+1) = (0, 1) \).

3. **Iterating \( f \):**
   - For \( (m,n) \) with \( n \geq m \geq 1 \), iterating \( f \) will increase \( m \) until \( m > n \), at which point \( f((m,n)) = (0, 0) \).
   - For \( (m,n) \) with \( m > n \), \( f((m,n)) = (0, 0) \).
   - For \( (m,0) \), iterating \( f \) will eventually lead to \( (0, 0) \).

4. **Fixed Point:**
   The only fixed point of \( f \) is \( (0, 0) \), since for any other point, iterating \( f \) will eventually lead to \( (0, 0) \).

5. **Intersection:**
   Since \( f_n(A) \) will eventually contain only \( (0, 0) \) for large \( n \), we have:
   \[
   f_{\infty}(A) = \bigcap_{n \geq 1} f_n(A) = \{(0, 0)\}
   \]

6. **Verification:**
   \[
   f(f_{\infty}(A)) = f(\{(0, 0)\}) = \{f((0, 0))\} = \{(0, 0)\} = f_{\infty}(A)
   \]

Thus, the statement \( f(f_{\infty}(A)) = f_{\infty}(A) \) is true for \( A = \mathbb{N} \times \mathbb{N} \) and the given function \( f \).

\(\blacksquare\)

The final answer is \( \boxed{ f(f_{\infty}(A)) = f_{\infty}(A) } \) is true for both parts (a) and (b).