1. Sep 26, 2008

JG89

The question seemed simple enough, but something feels funny about my proof. I would appreciate if someone could please check it.

Question: Prove that if f(x) is monotonic on [a,b] and satisfies the intermediate value property, then f(x) is continuous.

Proof: Let e denote epsilon and d denote delta. For a point p in [a,b], we wish to prove |f(x) - f(p)| < e whenever |x - p| < d, for e > 0 and d > 0. Suppose we have a neighborhood about p such that p - d < p < p + d. Also suppose there is an x in the interval, so we have p - d < x < p + d. By the intermediate value property, and since the function is monotonic, we have:
f(p-d) < f(x) < f(p+d). Since the function satisfies the intermediate value property and is monotonic, then we can find an e such that f(p-d) < f(p) - e < f(x) < f(p) + e < f(p+d) for some neighborhood about f(p). The inequality f(p) - e < f(x) < f(p) + e implies |f(x) - f(p)| < e. We also have the inequality p - d < x < p + d, so that implies |x-p| < d. Therefore the function is continuous in the closed interval [a,b]. QED.

Last edited: Sep 26, 2008
2. Sep 26, 2008

jostpuur

There is some problems with your proof attempt.

There's not really a mistake here, but your language is vague. We don't want to prove something for all $$\epsilon >0$$ and $$\delta >0$$. We want to find a suitable $$\delta >0$$ for each fixed $$\epsilon > 0$$.

This is true, but it is sufficient to assume f being monotonic (and increasing) here. The intermediate value property is not used here yet.

This is something useless. We don't want to find any particular $$\epsilon >0$$. First we assume $$\epsilon >0$$ to be fixed to some arbitrary value, and then we want to find $$\delta >0$$.

3. Sep 26, 2008

JG89

I tried to clear it up a bit:

Let e > 0. Then we wish to prove that f(x) is continuous at a point p in the interval [a,b] if for every positive e we can find a positive number d such that |f(x) - f(p)| < e for all values x in the domain of f for which |x - p| < d. So, we suppose we have a neighborhood about f(p) such that f(p) - e < f(p) < f(p) + e. Since f is monotonic, then
A < p < B, where f(A) = f(p) - e and f(b) = f(p) + e. There is then an x value in this domain such that f(A) < f(x) < f(B). Or, f(p) - e < f(x) < f(p) + e, which implies |f(x) - f(p)| < e. Taking the interval A < p < B, we can then find a suitable positive number d such that A < p - d < x < p + d < B, for all x in the open interval (p-d,p+d). That last inequality however implies |x-p| < d. Thus we have proved that |f(x) - f(p)| < e whenever |x-p| < d.

Last edited: Sep 26, 2008
4. Sep 26, 2008

JG89

See Above

Last edited: Sep 26, 2008
5. Sep 28, 2008

sutupidmath

Hmmm... I am not quite sure that i absolutely know how to do this one. However, what i tried looks logical, at least to me.. but waiting for other's input is the best idea.

For the sake of defeniteness, lets suppose that f is strictly monotono increasing on [a,b].

Let $$\epsilon>0$$ such that the interval $$(f(p)-\epsilon,f(p)+\epsilon)$$ is contained in (f(a),f(b)). So,

$$f(p)-\epsilon<f(p)<f(p)+\epsilon$$

Now, since $$f(a)<f(p)-\epsilon<f(p),\exists A \in(a, p)$$ such that

$$f(A)=f(p)-\epsilon$$ this isby IVT. Also, since

$$f(p)<f(p)+\epsilon<f(b), \exists B \in (p,b)$$ such that

$$f(B)=f(p)+\epsilon$$ this again by IVT. In other words,

the set $$(f(A),f(B))$$ is contained into $$(f(a),f(b))$$

Now, there exists some $$\delta$$ such that

$$A<p-\delta<p<p+\delta<B$$ Now again, since f monotonoincreasing

$$f(A)<f(p-\delta)<f(p)<f(p+\delta)<f(B)$$ now by IVT $$\exists x \in (p-\delta, p+\delta)$$ such that

$$f(p-\delta)<f(x)<f(p+\delta)$$ but notice that the set

$$(f(p-\delta),f(p+\delta))$$ is contained within $$(f(p)-\epsilon,f(p)+\epsilon)$$

Indeed, what this is telling us is that, the set

$$(p-\delta, p+\delta)$$ is mapped completely into the set $$(f(p)-\epsilon,f(p)+\epsilon)$$ by f.

Which now expressed explicitly, means that $$\forall x \in (p-\delta, p+\delta)=>f(x) \in (f(p)-\epsilon,f(p)+\epsilon)$$

(i.e, whenever $$|x-p|<\delta=> |f(x)-f(p)|<\epsilon$$ ) SO, this actually means that f is continuous at a.

This is as far as i was able to take this one. SOrry, if i have confused you even more.

6. Sep 28, 2008

JG89

It looks like to me that our proofs are very similar

7. Sep 28, 2008

sutupidmath

Very simmilar, indeed, with some exceptions! I think in your proof you took some things for granted. Like when you let f(A)=... or f(B)=.... etc. My personal attitude is that when you write a proof you shall not let anything unaddressed, that is you must justify every single step, otherwise that wouldn't be a proof.

Last edited: Sep 28, 2008
8. Sep 28, 2008

JG89

There must be a value in [a,b] such that f(at that value) = f(p) - e, for example, because of the intermediate value property. That's how I let f(A) = f(p) - e and f(B) = f(p) + e. You're right, I should address these things when others are going to be checking my proof. It just seemed clear to me (obviously because I am the one writing it though).

What other things did I seem to take for granted?

9. Sep 28, 2008

sutupidmath

When you say that:"Taking the interval A < p < B, we can then find a suitable positive number d such that A < p - d < x < p + d < B, for all x in the open interval (p-d,p+d). That last inequality however implies |x-p| < d."
I agree here, but you need to point out, like i did in my proof as to why all values that are in the open interval (p-d,p+d) will be mapped into the interval (f(p)-e,f(p)+e) by the function f. You take this one for granted, which as far as i am concerned you shouldn't do, since this is the core of the problem i think.
In other words, you don't tell us as why there is no value, call it c, in the interval (p-d,p+d), such that f(c) is not in (f(p)-e,f(p)+e)???
Other than this, i think the rest is ok.

10. Sep 28, 2008

JG89

"you need to point out, like i did in my proof as to why all values that are in the open interval (p-d,p+d) will be mapped into the interval (f(p)-e,f(p)+e) by the function f."

if A < p - d < x < p + d < B, then by the monotonicity of f, we have f(A) < f(p-d) < x < f(p +d)
< f(B), but since f(A) = f(p) - e and f(B) = f(p) + e, we have f(p) - e < f(p -d) < f(x) < f(p+d) < f(p) + e, thus all values in the interval (p-d,p+d) are mapped into the interval (f(p) - e, f(p) + e).

Do you mean like that?

11. Sep 28, 2008

sutupidmath

Yeah, like i said in my post!

"Now, there exists some $$\delta$$ such that

$$A<p-\delta<p<p+\delta<B$$ Now again, since f monotonoincreasing

$$f(A)<f(p-\delta)<f(p)<f(p+\delta)<f(B)$$ now by IVT $$\exists x \in (p-\delta, p+\delta)$$ such that

$$f(p-\delta)<f(x)<f(p+\delta)$$ but notice that the set

$$(f(p-\delta),f(p+\delta))$$ is contained within $$(f(p)-\epsilon,f(p)+\epsilon)$$

Indeed, what this is telling us is that, the set

$$(p-\delta, p+\delta)$$ is mapped completely into the set $$(f(p)-\epsilon,f(p)+\epsilon)$$ by f."

12. Sep 28, 2008

JG89

Thanks for the help! I'll be sure to make my proofs more specific from now on.

13. Sep 28, 2008

sutupidmath

I don't have that much of an experience in proofs either, since i am only a second-semester freshman, so i guess other people would give better advice as what you can take for granted and what not. But, up to this stage i always write my proofs long and with as many details as possible, but latter on, who knows, i might change my mind....lol.....