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Let $A\subseteq R,$ let $f : A\to R,$ and suppose that $(a,\infty)\subseteq A$ for some $a \in R.$ Then the following statements are equivalent:

(i) $L=\lim_{x\to \infty}f(x)$

(ii) For every sequence $(x_n)$ in $A \cap (a,\infty)=(a,\infty)$ such that $\lim x_n=\infty$, the sequence $f(x_n)$ converges to $L.$

I tried proving this in the following way:

First we try to show, $(i)\implies (ii).$ If, $\lim_{x\to\infty}f(x)=L$ then this means,

For every $\epsilon\gt 0$ there exists a $K>a$ such that if $x\geq K,$ we have $|f(x)-L|\lt \epsilon.$

Now, if $(x_n)$ is a sequence in $A \cap (a,\infty)=(a,\infty)$ such that $\lim x_n=\infty$ then, $\exists M\in\Bbb N$ such that $x\gt K(\gt a),$ when $n\geq M.$ But if, $x_n>K$ we have, $|f(x_n)-L|\lt \epsilon.$ To sum, we have shown that, if $n\geq M$, then, $|f(x_n)-L|\lt \epsilon.$ But $\epsilon \gt 0$ and $(x_n)$ was arbitrary and this implies, that for every sequence $(x_n)$ in $A \cap (a,\infty)=(a,\infty)$ such that $\lim x_n=\infty$, the sequence $f(x_n)$ converges to $L.$


However, while to show the converse, I think I make my argument faulty:

Next, we try to show, that $(ii)\implies (i).$

If for every sequence $(x_n)$ in $A \cap (a,\infty)=(a,\infty)$ such that $\lim x_n=\infty$, the sequence $f(x_n)$ converges to $L,$ then we fix, an arbitrary $\epsilon\gt 0.$

Now, we denote, each sequence as $X_i=(x_n^{(i)})$, where $i\in \Bbb N$ and $X_i\in A \cap (a,\infty)=(a,\infty)$. If $\lim X_i=\infty$ for a particular $i\in Bbb N$ then $\exists M_i\in \Bbb N$ such that, $|f(x_n^{(i)}-L|\lt \epsilon$ for all $n\geq M.$ For each, $X_i$ we obtain, an $M_i$ and if, $M=\max(M_i)_{i\in \Bbb N},$ we may take, $x\geq M$ and then, $|f(x)-L|\lt\epsilon.$ Hence, we claim that, $L=\lim_{x\to \infty}f(x).$


I doubt my proof about the converse statement as, $\max\{\text{infinite set}\}$ has no meaning at all. We can take $\max$ element in a finite set and not in an infinite one. Secondly, in the part, where I wrote, "$x\geq M$ and then, $|f(x)-L|\lt\epsilon$" might be faulty as well. This is because, if $x\geq M$ then, $|f(x)-L|\lt\epsilon$ will be true, only if, $x$ is in some properly diverging sequence $X_i$ tending to $\infty.$ But, how can we gurantee, that $x$ will be in such a sequence $X_i$ just like that?

For these two reasons, I feel my argument breaks down.

All in all, I don't understand how to show, $(ii)\implies (i)$ ?

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    $\begingroup$ For $(ii) \implies (i)$, try the contrapositive. $\endgroup$
    – terran
    Commented Aug 26, 2023 at 7:29
  • $\begingroup$ @terran Contrapositive of what ? $(i)\to (ii)$ ? $\endgroup$ Commented Aug 26, 2023 at 7:38
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    $\begingroup$ The contrapositive of $(ii) \implies (i)$ is $\text{not } (i) \implies \text{not } (ii)$. $\endgroup$
    – terran
    Commented Aug 26, 2023 at 7:41
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    $\begingroup$ Yes, that's the approach I meant! :-) $\endgroup$
    – terran
    Commented Aug 26, 2023 at 16:08
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    $\begingroup$ @terran Thanks a ton for confirming ! $\endgroup$ Commented Aug 26, 2023 at 16:12

2 Answers 2

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We can prove that $(ii)\implies (i)$ in a much simpler way. We show this, by proving that the contrapositive of the statement i.e $\neg (i)\implies \neg (ii)$ is true.

The contrapositive translates to the below statement:

If $\lim_{x\to\infty}f(x)\neq L$ then $\exists (x_n)\in A\cap (a,\infty)$ satisfying, $\lim x_n=\infty$ such that, $\lim f(x_n)\neq L\tag 1$

This is true, for if $\lim_{x\to\infty}f(x)\neq L$ then, $\exists \epsilon\gt 0$ such that, $\forall K_i\gt a$ $\exists x'_i\gt K_i$ we have, $|f(x'_i)-L|\geq \epsilon.$ We consider the sequence $(x_i')\in A\cap (a,\infty)$ and note that, $\lim x_i' =\infty.$

This can readily be verified if, $\alpha\in \Bbb R$ then, $\exists K_{i_0}\in\Bbb N$ such that $K_{i_0}\gt \alpha$ then, $x'_{i_0}>K_{i_0}>\alpha$ and hence, $\forall i\geq i_0$ we have, $x'_i\gt K_i\gt K_{i_0}\gt\alpha$ so that $x'_i\gt \alpha$ for all $i\geq i_0.$ Hence, $\lim x'_i=\infty$ follows.

But we note that, $|f(x'_i)-L|\geq \epsilon$ for all $i\in \Bbb N.$ So, the required sequence in $(1)$ is $(x_i').$ So, $(1)$ is indeed true and so, $(ii)\implies (i).$

The portion, $(i)\implies (ii)$ in OP seems correct.

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As hinted in the comments, you can show that if $\lim_{x \to \infty} f (x) \ne L$, then there is some sequence $(x_n)$ (where $x_n > a$ for each $n$) with $\lim_{n \to \infty} x_n = \infty$ such that $(f (x_n))$ does not converge to $L$.

Since $\lim_{x \to \infty} f (x) \ne L$, then there is some open ball $C = B (L, \varepsilon)$ such that $f^{-1} (C)$ does not contain any set of the form $A \cap (r, \infty)$, where $r \in \mathbb{R}$. Since $(2 |a|, \infty) \subseteq (a, \infty) \subseteq A$, this implies there is some element $x_1 > 2 |a| \ge 0$ of $A$ such that $f (x_1) \notin C$. Now, since $(2 x_1, \infty) \subseteq (a, \infty) \subseteq A$, there is some $x_2 > 2 x_1$ contained in $A$ such that $f (x_2) \notin C$. Continue inductively so that we get a sequence of points of $A$

$$x_1 < x_2 < \dots$$

each of which is positive and none of whose images is contained in $C$. We claim that this sequence satisfies the desired properties.

It is easy to see that $(f (x_n))$ does not converge to $L$, since the neighborhood $C$ of $L$ contains no points of this sequence (by design).

To show that $\lim_{n \to \infty} x_n = \infty$, assume for the sake of contradiction that $x_n < s$ for each $n$ for some $s \in A$. Note that $s$ must be positive since each $x_n$ is positive. By design, $x_{n + 1} > 2 x_n$ for each $n$, so that $x_n > x_1 (2)^{n - 1}$. So, we have that $2^{n - 1} < \frac{x_n}{x_1} < \frac{s}{x_1} \implies n < 1 + \log_2 \left( \frac{s}{x_1} \right)$ for each positive integer $n$, clearly a contradiction.

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