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Strictly increasing $f(x)$ is defined on R. Then for any convergent strictly increasing sequence $a_n$ sequence $f(a_n)$ is also convergent.

The answer is TRUE, but I believe I have a counterexample:

$a_n = 1-\frac{1}{n}; f(x) = x+n \Rightarrow f(a_n) = 1 - \frac{1}{n} + n$

Am I correct?

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    $\begingroup$ You're using $n$ in two different ways - both as a constant and as part of a sequence. If you take $f(x) = x + m$ then $f(a_n) = 1-\frac1n + m$ converges $\endgroup$
    – Mathmo123
    Jul 10, 2014 at 11:21
  • $\begingroup$ OK, how do I create a function that jumps constantly? $\endgroup$
    – Yal dc
    Jul 10, 2014 at 11:23

1 Answer 1

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$f$ is strictly increasing function. $a_n$ is a strictly increasing convergent sequence, then $a_n$ is bounded. Let $|a_n|\le M \quad \forall n\in \mathbb{N}$, then the sequence $f(a_n)$ is strictly increasing, and $|f(a_n)|\le |f(M)|$; i.e $f(a_n)$ is bounded and hence converges by monotone convergence theorem.

That means $\lim_{n\rightarrow \infty}{f(a_n)}$ exists. Now a possible question is, if $a_n\rightarrow l$ whether the follows is true: $$\lim_{n\rightarrow \infty}{f(a_n)}=f(l) \tag{1}$$

No. Consider the function $$ f(x)= \left\{\begin{array}{cl} e^x, &x\lt 0 \\ e^x+1&x \ge 0 \end{array}\right. $$ Clearly $f$ is strictly increasing, has jump discontinuity at $x=0$.

Let $a_n=-\frac{1}{n}$ now $a_n$ is strictly increasing, converges to $0$. But $f(a_n)=1, n\ge 1$ while $f(0)=2$.

That said, if $f$ is continuous $(1)$ holds.

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  • $\begingroup$ Nice, now I'm convinced. When I thought about this problem, I recalled the result that we could "move" limit into the function (i.e. $\lim f(x) = f(\lim x)$) if f(x) is continuous. But in this problem, we are given a function that might be discontinuous. Does that mean that continuity is just sufficient and not necessary condition for such "move" of limit inside the function? $\endgroup$
    – Yal dc
    Jul 10, 2014 at 11:32
  • $\begingroup$ @Yaldc I haven't used any argument of continuity in my proof. $\endgroup$
    – hrkrshnn
    Jul 10, 2014 at 11:37
  • $\begingroup$ I see, my question wasn't about the proof. $\endgroup$
    – Yal dc
    Jul 10, 2014 at 11:40
  • $\begingroup$ @Yaldc. Let $f$ be a mapping of $X$ into $Y$, let $p$ be limit point of $X$, then $f$ is continuous at $p$ if and only if $\lim\limits_{x\rightarrow p}{f(x)}=f(p)$. Is that what you are looking for? $\endgroup$
    – hrkrshnn
    Jul 10, 2014 at 11:44
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    $\begingroup$ @Yaldc I have edited my post, I believe that your query is now answered. $\endgroup$
    – hrkrshnn
    Jul 10, 2014 at 12:45

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