# Thread: Pretty confusing proof: I don't even know what it is I am supposed to prove...

1. ## Pretty confusing proof: I don't even know what it is I am supposed to prove...

Let f-->R and xo is a cluster point of D. Prove that f has a limit at xo if for each epsilon>0 there is a neighborhood Q of xo such that for all x,y belonging to QintersectD x=/=xo, y=/=xo, we have |f(x)-f(y)|<epsilon.

I have an exam coming up in a week and I don't even know where to start with such weird problems.
I found a proof of this statement on the web (Since we don't have a solution manual), but it is very long and complicated..using many things that we have not used in this course. I was hoping someone could help me get an easier proof so that if such a problem appeared on my exam, I might actually have a chance because there is no way I could come up with the proof I found. Thanks

2. Originally Posted by zhupolongjoe
Let f-->R and xo is a cluster point of D. Prove that f has a limit at xo if for each epsilon>0 there is a neighborhood Q of xo such that for all x,y belonging to QintersectD x=/=xo, y=/=xo, we have |f(x)-f(y)|<epsilon.

I have an exam coming up in a week and I don't even know where to start with such weird problems.
I found a proof of this statement on the web (Since we don't have a solution manual), but it is very long and complicated..using many things that we have not used in this course. I was hoping someone could help me get an easier proof so that if such a problem appeared on my exam, I might actually have a chance because there is no way I could come up with the proof I found. Thanks
Couple of clarification questions:

What do you mean by:

f-->R

(Last I checked, "Big Grin" was not a set...)

And what set are you denoting as D?

3. Oops lol, didn't notice that. It's just any set D

4. Originally Posted by zhupolongjoe
Let f: D-->R and xo is a cluster point of D. Prove that f has a limit at xo if for each epsilon>0 there is a neighborhood Q of xo such that for all x,y belonging to QintersectD x=/=xo, y=/=xo, we have |f(x)-f(y)|<epsilon.

I have an exam coming up in a week and I don't even know where to start with such weird problems.
I found a proof of this statement on the web (Since we don't have a solution manual), but it is very long and complicated..using many things that we have not used in this course. I was hoping someone could help me get an easier proof so that if such a problem appeared on my exam, I might actually have a chance because there is no way I could come up with the proof I found. Thanks
Let me clean this up a bit:

Let $f: D\longrightarrow\mathbb{R}$ be a function and let $x_0$ be a cluster point of $D$. Prove that $\lim_{x_n\to x_0}f(x_n)$ exists if $\forall~\epsilon>0$, $\exists~\delta$ such that $x,y\in Q_{\delta}(x_0)$ implies that $|f(x)-f(y)|<\epsilon$.

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Proof: Let $\{x_n\}\subset D$ be a sequence that converges to $x_0$. (We know such a sequence exists because $x_0$ is a cluster point.) We want to prove that the sequence $\{f(x_n)\}\subset\mathbb{R}$ converges.

We know that given a $\delta$, all but finitely many $x_n$ are contained in $Q_{\delta}(x_0)$, that is, $n>N$ implies $x_n\in Q_{\delta}(x_0)$. So for $n,m>N$ we know that $x_m,x_n\in Q_{\delta}(x_0)$ and therefore $|f(x_n)-f(x_m)|<\epsilon$.

But this means that $\{f(x_n)\}$ is Cauchy, and therefore converges because $\mathbb{R}$ is complete. In other words, $\lim_{x_n\to x_0}f(x_n)$ exists. $\square$

5. Thanks a bunch! I understood that
a lot more than the other proof I found. BTW: I love your David Hilbert quote...