Please check proof on continuity

[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.

 

[jostpuur]

There is some problems with your proof attempt.

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.

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

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

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).

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

 

[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.

 

[JG89]

See Above

 

[sutupidmath]

See Above

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, let's suppose that f is strictly monotono increasing on [a,b].

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

f(p)-\epsilon&lt;f(p)&lt;f(p)+\epsilon

Now, since f(a)&lt;f(p)-\epsilon&lt;f(p),\exists A \in(a, p) such that

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

f(p)&lt;f(p)+\epsilon&lt;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&lt;p-\delta&lt;p&lt;p+\delta&lt;B Now again, since f monotonoincreasing

f(A)&lt;f(p-\delta)&lt;f(p)&lt;f(p+\delta)&lt;f(B) now by IVT \exists x \in (p-\delta, p+\delta) such that


f(p-\delta)&lt;f(x)&lt;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)=&gt;f(x) \in (f(p)-\epsilon,f(p)+\epsilon)

(i.e, whenever |x-p|&lt;\delta=&gt; |f(x)-f(p)|&lt;\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.

 

[JG89]

It looks like to me that our proofs are very similar

 

[sutupidmath]

It looks like to me that our proofs are very similar

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.

 

[JG89]

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.

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?

 

[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.

 

[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?

 

[sutupidmath]

"

Do you mean like that?

Yeah, like i said in my post!


"Now, there exists some \delta such that


A&lt;p-\delta&lt;p&lt;p+\delta&lt;B Now again, since f monotonoincreasing

f(A)&lt;f(p-\delta)&lt;f(p)&lt;f(p+\delta)&lt;f(B) now by IVT \exists x \in (p-\delta, p+\delta) such that


f(p-\delta)&lt;f(x)&lt;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."

 

[JG89]

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

 

[sutupidmath]

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

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...