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Let $f$ and $g$ be real-valued continuous functions on $\Bbb R^2$ that satisfy the following condition: $$ x<y \implies x< f(x,y) < g(x,y) <y $$ Assume that there are two sequences $\{a_n\}$ and $\{b_n\}$ that satisfy the following: $$ a_1 < b_1, a_1,b_1 \in \Bbb R,\quad a_{n+1}=f(a_n, b_n),\quad b_{n+1}=g(a_n,b_n). $$ It's easily seen that $$a_1 < a_2 < ...< a_n < ...< b_n < b_{n-1} < ... < b_1.$$ Hence $\{a_n\}$ and $\{b_n\}$ are monotonic and bounded thus convergent. Now I want to prove that $\lim_{n \to \infty} a_n= \lim_{n \to \infty} b_n$.

My attempt: Assume that $\lim_{n \to \infty} a_n=a, \lim_{n \to \infty} b_n=b$. Since $a_n < b_n$, $a \le b$. Since $a_{n+1}=f(a_n, b_n)$, we have $a=f(a,b)$. If $a< b$ then $a < f(a,b)$, a contradiction. It follows that $a=b$.

Does the proof above look fine?

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    $\begingroup$ It looks fine to me. $\endgroup$ – MereMortal47 Jul 11 '16 at 8:26
  • $\begingroup$ @Asemismaiel My only concern is whether $a=f(a,b)$ if $a_{n+1}=f(a_n,b_n)$? $\endgroup$ – user Jul 11 '16 at 10:35
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    $\begingroup$ $\lim f(a_n, b_n) = f(a,b)$ follows from the continuity of $f(x,y)$, for if $f(x,y)$ is continuous at $(a,b)$ then for any sequence converging to $(a,b)$ the sequence of the images of the sequence converge to $f(a,b)$. This is known as the sequential criterion of continuity. $\endgroup$ – MereMortal47 Jul 11 '16 at 10:59
  • $\begingroup$ And taking the limit of both sides is justified by the fact that they are the same sequence, of course. $\endgroup$ – MereMortal47 Jul 11 '16 at 11:01
  • $\begingroup$ @Asemismaiel Thanks for clearing up the confusion. I want to make this post answered. Would you mind adding your comments as an answer? $\endgroup$ – user Jul 11 '16 at 11:08
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It looks fine to me.

$\lim f(a_n, b_n) = f(a,b)$ follows from the continuity of $f(x,y)$, for if $f(x,y)$ is continuous at $(a,b)$ then for any sequence converging to $(a,b)$ the sequence of the images of the sequence converges to $f(a,b)$. This is known as the sequential criterion of continuity.

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