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Prove: Let $A \subset \mathbb{R}$ compact, $f: A \rightarrow \mathbb{R}$ continuous, increasing monotone and $(x_n) \subset A$. Consider. Show that $\lim \sup f (x_n) = f(\lim \sup (x_n))$ and $\lim \inf f (x_n) = f(\lim \inf (x_n))$.

My attempt: Firstly, the image $f(A)$ is a compact set because A is compact and f is continuous. Being so, the numbers $y_o = \inf f(A)$ and $y_1 = \sup f(A)$ belong to $f(A)$. Also, $f:A \rightarrow \mathbb{R}$ is increasing monotone, meaning $x<y \Rightarrow f(x) < f(y)$. Now, since $(x_n) \subset A$ and f continuous, for all $a \in A$, $xn \rightarrow a \Rightarrow f(x_n) \rightarrow f(a).$ That way, letting $m_0 =\lim \inf f (x_n)$, it follows that $x_n \rightarrow m_0 \Rightarrow f(x_n) \rightarrow f(m_0)$, and a similar result for the $sup$.

I got stuck at this point, and would like some hints on what I should do next, or a different approach that not by the definition of continuity. How do I use the fact that $f$ is an increasing monotone function?

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By intermediate value theorem, since $f$ is increasing, continuous and $A$ compact, $$\sup f(A)=f(\sup A).$$ The continuity allowed you to conclude. It's easy to adapt the proof to a set $\{x_n\}_{n\in\mathbb N}\subset A$.

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  • $\begingroup$ Why do I need to have an increasing f (or decreasing)? I don't recall it being a condition inside the IVT. Is it only to guarantee it is not a constant function? $\endgroup$ – DrHAL Jun 4 '16 at 17:00
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    $\begingroup$ The fact that $f$ is increasing gives you only that $\sup f(A)=f(\sup A)$. To simplify, let $A=[a,b]$. IVT tels you that if $y\in f([a,b])$ then $y=f(x)$ for an $x\in [a,b]$. Now, if $y=\sup f([a,b])$ and $\sup f([a,b])\in f([a,b])$ (since the range of a compact is a compact), then, $y=f(x)$ for an $x\in [a,b]$. But if $f$ is strictly increasing, the only possibility is $y=f(b)$, and if $f$ is only increasing and there is a $c\in ]a,b[$ s.t. $y=f(c)$, then, you also have $y=f(a)$. $\endgroup$ – Surb Jun 4 '16 at 17:19
  • $\begingroup$ I don't get the last sentence. Why do I also have $y = f(a)$? Also, if I only have that for the $sup$ part, how can I do it for the $inf$, and using the sequence part? Please clarify. $\endgroup$ – DrHAL Jun 4 '16 at 17:36
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    $\begingroup$ I'm sorry, I wanted to say $y=f(b)$. If $f$ is increasing and that $y=\sup\{f(x)\mid x\in [a,b]\}$ and that $y=f(c)$ where $c\in ]a,b[$, then $f$ is constant on $[c,b]$, and thus, $f(x)=y$ for all $x\in [c,b]$, in particular $f(b)=y$. I hope it's more clear now. $\endgroup$ – Surb Jun 4 '16 at 17:59
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    $\begingroup$ Yes, increasing or deacreasing, both work ! Notice that if $f$ increasing on $[a,b]$ then $\inf f(x)=f(a)$, and if it's decreasing, then $\inf f(x)=f(b)$. $\endgroup$ – Surb Jun 4 '16 at 19:56

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