# Proofs with epsilon delta (real analysis)

1. Sep 5, 2011

### Wingeer

Hello,
I have stumbled upon a couple of proofs, but I can not seem to get an intuitive grasp on the whats and the whys in the steps of the proofs. Strictly logical I think I get it. Enough talk however.
Number 1.
"Let f be a continuous function on the real numbers. Then the set {x in R : f(x)>c} is open for any c in R.
Proof:
Let:
$$E_c = \{x \in \mathbf{R}:f(x)>c \}.$$
Assume that x is in E sub c, we have f(x)>c since f is continuous at x there exists delta such that:
$$|f(x) - f(y)|<f(x)-c$$
whenever
$$|x-y|< \delta.$$
For such y we have:
$$f(y) \ge f(x) - |f(x) - f(y)| > f(x) - (f(x) - c)=c$$
and y is in E sub c. We have shown that
$$(x- \delta, x + \delta) \subset E_c$$ and there the set is open."

I get why the set is open and the algebra in the previous step. However I am wondering: Are we kind of setting f(x)-c as our epsilon? And to get to the second last line, are we using some kind of triangle inequality?

2.
"Let f be a function on X, for each positive delta we define:
$$\omega_{f,X}(\delta)= sup \{|f(x) - f(y)|: x,y \in X, |x-y|<\delta \}.$$
f is uniformly continuous on X (1) if and only if $$lim_{\delta \to 0} \omega_{f,X}(\delta)=0. (2)$$
Proof:
By definition, f is uniformly continuous on X if for any epsilon greater than zero there exists $$\delta(\epsilon)>0$$ such that $$\omega_{f,X}(\delta(\epsilon))<\epsilon.$$
Now, assume (2). Then for any epsilon greater than zero there exists $$\delta(\epsilon)>0$$ such that $$\omega_{f,X}(\delta(\epsilon))<\epsilon$$ and f is uniformly continuous.
Assume (1). Then for any epsilon greater than zero there exists $$\delta(\epsilon)>0$$ such that $$\omega_{f,X}(\delta(\epsilon))<\epsilon.$$
Clearly, omega is greater than or equal to zero and non-increasing. We have:
$$\omega_{f,X}(\delta)<\epsilon$$
for any $$\delta \le \delta(\epsilon)$$ and $$lim_{\delta \to 0} \omega_{f,X}(\delta)=0$$ which we were supposed to prove."

I get a bit confused about the delta functions. What does those mean? In proving (2)=>(1) how can we "guarantee" that there exists such deltas?
In the last step of (1)=>(2), since it is non-increasing, should there not be $$\omega_{f,X}(\delta) \le \epsilon$$, since $$\delta \le \delta(\epsilon).$$
Or am I missing a point along the way?
Thanks.

Edit:
Why can't I have LaTeX coding and regular text in the same paragraph? It looks horribly messy.

2. Sep 5, 2011

### susskind_leon

It's indeed a tricky exercise, because delta becomes epsilon, which might be very confusing. However, you should always write down what you know first. Since f is continuous, you have
$$\forall \epsilon\exists \delta: |x-y|<\delta\Rightarrow |f(x)-f(y)|<\epsilon$$
And you want to prove that $E_c$ is open, which means
$$\exists \epsilon: |x-y|<\epsilon\Rightarrow f(y) \in E_c$$
which means that $f(y)>c$
Now you choose as $\epsilon$ from the FIRST equation $\epsilon=f(x)-c$
This gives you a $\delta$ such that
$$|x-y|<\delta\Rightarrow |f(x)-f(y)|<f(x)-c$$
So, as you see, $\delta$ becomes $\epsilon$ or vice versa, however you wanna put it. Now, to finish the deal, you do a case-by-case analysis. In case that $f(x)>f(y)$ you get
$$c<f(x)-f(x)+f(y)=f(y)$$
In case that $f(y)>f(x)$ you're done already since $f(x)>c$ by assumption.

3. Sep 5, 2011

### Wingeer

Aha. I think I see. The difference in this from other proofs is that instead of finding a delta so that the conditional is true, we are actually finding an epsilon. If that is what you meant by delta becoming epsilon? And since the function is continuous we are allowed to choose which value for epsilon we want, because we are guaranteed that there exists a delta "corresponding" to it?
The case-by-case analysis is actually way easier than fooling around with some reversed triangle inequality, or whatever. Thank you! Could you do the second as well? :-)

4. Sep 5, 2011

### susskind_leon

2.
To answer your first question "why is there such a delta": It's the definition of the limit.
Limit means that for every epsilon you can find a delta such that for all x<delta you get f(x)<epsilon (assuming the limit is 0 with x->0).
This exercise is a piece of cake if you write down the definition of the limit and uniformly continuity.
uniformly continuity means that
$$\forall \epsilon \exists \delta: |x-y|<\delta\Rightarrow|f(x)-f(y)|<\epsilon$$
$\lim_{x\rightarrow 0} f(x)=0$ means
$$\forall \epsilon \exists \delta: x < \delta \Rightarrow f(x) < \epsilon$$
now take $\omega_{f,X}(\delta)$ and you've got it basically written there. Keep in mind that if $|f(x)-f(y)|<\epsilon$ it follows that $\sup\{|f(x)-f(y)|\}\leq\epsilon$

5. Sep 7, 2011

### Wingeer

Yes, it is the definition. But don't we have to prove that the function will be less than epsilon?
In this and your last post there is no difference in your definition of continuity and uniform continuity. However I know the difference, so it is no harm.
If you take the function with input delta, you get that delta>delta?