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Please check proof on continuity

  1. Sep 26, 2008 #1
    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. jcsd
  3. Sep 26, 2008 #2
    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 [tex]\epsilon >0[/tex] and [tex]\delta >0[/tex]. We want to find a suitable [tex]\delta >0[/tex] for each fixed [tex]\epsilon > 0[/tex].

    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 [tex]\epsilon >0[/tex]. First we assume [tex]\epsilon >0[/tex] to be fixed to some arbitrary value, and then we want to find [tex]\delta >0[/tex].
     
  4. Sep 26, 2008 #3
    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
  5. Sep 26, 2008 #4
    See Above
     
    Last edited: Sep 26, 2008
  6. Sep 28, 2008 #5
    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 [tex] \epsilon>0[/tex] such that the interval [tex] (f(p)-\epsilon,f(p)+\epsilon)[/tex] is contained in (f(a),f(b)). So,

    [tex] f(p)-\epsilon<f(p)<f(p)+\epsilon[/tex]

    Now, since [tex] f(a)<f(p)-\epsilon<f(p),\exists A \in(a, p)[/tex] such that

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

    [tex] f(p)<f(p)+\epsilon<f(b), \exists B \in (p,b)[/tex] such that

    [tex]f(B)=f(p)+\epsilon[/tex] this again by IVT. In other words,


    the set [tex] (f(A),f(B))[/tex] is contained into [tex](f(a),f(b))[/tex]


    Now, there exists some [tex]\delta[/tex] such that


    [tex] A<p-\delta<p<p+\delta<B[/tex] Now again, since f monotonoincreasing

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


    [tex] f(p-\delta)<f(x)<f(p+\delta)[/tex] but notice that the set

    [tex] (f(p-\delta),f(p+\delta))[/tex] is contained within [tex] (f(p)-\epsilon,f(p)+\epsilon)[/tex]


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

    [tex] (p-\delta, p+\delta)[/tex] is mapped completely into the set [tex] (f(p)-\epsilon,f(p)+\epsilon)[/tex] by f.

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

    (i.e, whenever [tex]|x-p|<\delta=> |f(x)-f(p)|<\epsilon[/tex] ) 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.
     
  7. Sep 28, 2008 #6
    It looks like to me that our proofs are very similar
     
  8. Sep 28, 2008 #7
    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
  9. Sep 28, 2008 #8

    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?
     
  10. Sep 28, 2008 #9
    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.
     
  11. Sep 28, 2008 #10
    "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?
     
  12. Sep 28, 2008 #11
    Yeah, like i said in my post!


    "Now, there exists some [tex]\delta[/tex] such that


    [tex] A<p-\delta<p<p+\delta<B[/tex] Now again, since f monotonoincreasing

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


    [tex] f(p-\delta)<f(x)<f(p+\delta)[/tex] but notice that the set

    [tex] (f(p-\delta),f(p+\delta))[/tex] is contained within [tex] (f(p)-\epsilon,f(p)+\epsilon)[/tex]


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

    [tex] (p-\delta, p+\delta)[/tex] is mapped completely into the set [tex] (f(p)-\epsilon,f(p)+\epsilon)[/tex] by f."
     
  13. Sep 28, 2008 #12
    Thanks for the help! I'll be sure to make my proofs more specific from now on.
     
  14. Sep 28, 2008 #13
    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.....
     
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