# proof that an alternative definition of limit is equivalent to the usual one

How to prove the following theorem?

Let $I \in R$ be an open interval, let $c \in I$, and let $f: I-{c} \rightarrow R$ be a function. Then $\lim \limits_{x \rightarrow c} {f(x)}$ exists if for each $\epsilon > 0$, there is some $\delta > 0$ such that $x,y \in I$ and $\vert x-c \vert < \delta$ and $\vert y-c \vert < \delta$ implies $\vert f(x)-f(y) \vert < \epsilon$.

I think this needs standard limit definition, sup and inf properties to prove. And I came up with a following scratch of proof:

(1) Suppose for each $\epsilon >0$, there is some $\delta >0$ such that $\vert x-c \vert < \delta$ and $\vert y-c \vert < \delta$ implies $\vert f(x)-f(y) \vert < \epsilon$. For each $r>0$, let $A_r$ = $I \bigcap (c-r,c+r)$. Then, for each $\epsilon>0$, there is some $\delta>0$, such that $x,y \in A_\delta$ implies $\vert f(x)-f(y) \vert < \epsilon$. Then I want to show there is some $a>0$ such that $f(A_a)$ is bounded.

(2) If $f(A_a)$ is bounded, then for each $s \in (0,a)$, $f(A_s) \subseteq f(A_a)$, and thus $f(A_s)$ is bounded. Define $a_s = \mathrm{glb}~f(A_s)$ and $b_s = \mathrm{lub}~f(A_s)$. Let $A = \{a_s \mid s \in (0,a)\}$ and $B=\{b_s \mid s \in (0,a)\}$. Then we know that A has a least upper bound and B has a greatest lower bound, and $\mathrm{lub}(A) \leq \mathrm{glb}(B)$. Now I want to show $\mathrm{lub}(A) = \mathrm{glb}(B)$

(3) If I could show $\mathrm{lub}(A) = \mathrm{glb}(B)$, let $M=\mathrm{lub}(A) = \mathrm{glb}(B)$, I want to show $\lim \limits_{x \rightarrow c} {f(x)} = M$.

Can someone give me some help on how to prove (1) (2) (3), $f(A_a)$ is bounded, $\mathrm{lub}(A) = \mathrm{glb}(B)$, and $\lim \limits_{x \rightarrow c} {f(x)} = M$? Thanks!

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Do you mean prove that this is equivalent to the standard definition? Because otherwise a definition doesn't need a proof. –  Brian Mar 17 '11 at 0:49
In case you are not given this as a definition already, what standard definition do you refer to? –  shuhalo Mar 17 '11 at 0:54
Don't you need to suppose also that $x$ and $y$ are not equal to $c$? Otherwise, it'll not work when $f$ is discontinuous at $c$. –  lhf Mar 17 '11 at 1:42
@lhf: yes that's true, thanks! –  Lindsay Duran Mar 17 '11 at 1:43
Or, as Calle noted, your function is actually continuous. But it should work with the restriction I suggested; the limit could then be something other than $f(c)$, of course. –  lhf Mar 17 '11 at 1:49
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(1) To show there is an $a$ such that $f(A_a)$ is bounded (and therefore, $f(A_b)$ is bounded for all $b\leq a$), simply pick $\delta$ that "works" for $\epsilon=1$. Now fix $x\in A_{\delta}$, and let $M = f(x)$. For all $y\in A_{\delta}$, you know that $|f(x)-f(y)|\lt 1$, hence $$|f(y)| - |f(x)| \leq |f(y)-f(x)| \lt 1$$ so $|f(y)|\lt 1 + |f(x)| = 1+M$. This shows that $f(A_{\delta})\subseteq [-1-M, 1+M]$, hence $f(A_{\delta})$ is bounded.

I confess that I don't really see how to handle (2) easily.

Alternative way.

Here is a possible way of attacking the problem which is closely related to what you are trying to do, but perhaps a bit easier: try constructing a sequence of points $x_1,x_2,\ldots$, with $x_1\to c$, and such that $f(x_1),f(x_2),\ldots$ is a Cauchy sequence. This will give you a "target" value for the limit, and then you can prove that the limit equals that target.

So: let $\epsilon_1 = \frac{1}{2}$. Then there exists a $\delta_1$, and we may assume $\delta_1\lt \frac{1}{2}$, such that $|x-c|\lt \delta_1$ and $|y-c|\lt \delta_1$ implies $|f(x)-f(y)|\lt \epsilon_1$. Take $x_1 = c-(\delta_1/2)$.

Now let $\epsilon_2 = \frac{1}{4}$. There exists a $\delta_2$, and we may assume $\delta_2\lt \min(\frac{1}{4},\delta_1)$, such that $|x-c|\lt \delta_2$ and $|y-c|\lt\delta_2$ implies $|f(x)-f(y)|\lt \epsilon_2$. Take $x_2 = c-(\delta_2/2)$.

Let $\epsilon_3 = \frac{1}{8}$. Then there exists a $\delta_3$, which we may assume satisfies $\delta_3\lt \min(\frac{1}{8},\delta_2)$, such that $|x-c|\lt\delta_3$ and $|y-c|\lt \delta_3$ implies $|f(x)-f(y)|\lt \epsilon_3$. Take $x_3 = c-(\delta_3/2)$.

Continuing this way, let $\epsilon_{n+1} = \frac{1}{2^{n+1}}$, and let $\delta_{n+1}\lt \min(\frac{1}{2^{n+1}},\delta_{n})$ be such that if $|x-c|\lt\delta_{n+1}$ and $|y-c|\lt\delta_{n+1}$ then $|f(x)-f(y)|\lt \epsilon_{n+1}$. Let $x_{n+1} = c - (\delta_{n+1}/2)$.

Now we have a sequence of points $\{x_n\}$ in $I$, with $x_n\to c$, and such that $|x_n - c| \lt \frac{1}{2^n}$ for all $n$.

Consider now the sequence $\{ f(x_m)\mid m=1,2,\ldots\}$. I claim this is a Cauchy sequence.

Indeed: let $\epsilon\gt 0$. Then there exists a natural number $N\gt 0$ such that $1/2^N \lt \epsilon$. Let $n,m\geq N$. Then $|x_n-c| \lt \delta_{N}$ and $|x_m-c|\leq \delta_N$, so we know that $$|f(x_n)-f(x_m)|\leq \epsilon_N = \frac{1}{2^N}\lt \epsilon.$$ Thus, for every $\epsilon\gt 0$ there exists $N\gt 0$ such that for all $n,m\geq N$, $|f(x_n)-f(x_m)|\lt \epsilon$. Hence, the sequence $\{f(x_n)\}$ is a Cauchy sequence, and therefore has a limit, $L$.

Now, since $x_n\to c$, then if there is a limit for $f(x)$ as $x\to c$, then it will have to be equal to $L$. Prove that this is indeed the limit.

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Thanks about the alternative way! I sorta figure about the boundness proof too and got $f(x) < \epsilon + \vert f(y) \vert$ but I'm not sure I got it correct. Thanks for the help! –  Lindsay Duran Mar 17 '11 at 4:42
@Lindsay Duran: Since $f(x)\leq |f(x)|$, $|f(x)|\lt \epsilon+|f(y)|$ holds (whenever $|x-c|\lt \delta$ and $|y-c|\lt\delta$), yes. That works. –  Arturo Magidin Mar 17 '11 at 4:54
I think to prove lubA=glbB, we need to use the lemma saying "lubA = glbB iff for any $\epsilon>0$, there exists $a \in A$ and $b \in B$ such that $b-a <\epsilon$". Let $a_s = f(c) - \epsilon /2$ and $b_s = f(c) +\epsilon /2$, if we could show $a_s = glb f(A_s)$ and $b_s = lub f(A_s)$, then we could prove lubA = glbB –  Lindsay Duran Mar 17 '11 at 5:23
@Lindsay: As I understand your problem, you cannot talk about $f(c)$, because you cannot assume that $f$ is defined (or continuous) at $c$. –  Arturo Magidin Mar 17 '11 at 13:14
@Lindsay Duran: As I see the problem, the "real issue" (that is, the only difficult part) is finding a "target" limit $L$. If you know what the limit "should be", then it is not hard to figure out how to use the given condition to show the limit exists using the usual definition of limit. So they key is really to find that possible target. My "alternative way" finds (sort of; using the fact that Cauchy sequences must converge) a potential target. If we assume this is going to work out, then that potetial target has to be the target, so we may as well try it on for size. –  Arturo Magidin Mar 17 '11 at 17:28