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Set of continuities

  1. Aug 3, 2009 #1
    I've been trying to answer the following question:
    Let f be an arbitrary function of $(-\infty,+\infty)$ and let L be the set of point where f is right-continuous but not left-continuous. Show that L is countable.
    Any help?
  2. jcsd
  3. Aug 3, 2009 #2
    Start with the definitions... if f is right continuous at p, that means that for any e>0, there is a d>0 such that at every point q in the interval (p,p+d), |f(p)-f(q)| < e.

    Conversely, if f is not left continuous at a point p, that means that there exists an e>0 such that there is no d>0 such that at every point q in (p-d,p), |f(p)-f(q)| < e. At each point p, pick some e=e(p) that works for the definition of left continuous at that p. (choose e(p) = 0 if p is not in L, for convenience)

    Now, here is a method you can count L by: first count {p in L : e(p) > 1}. Then count {p in L : e(p) > 1/2}. Then count {p in L : e(p) > 1/4}. And so forth. L is the union of these sets, and the countable union of countable sets is itself countable, so this gives a way to count L.

    To show that this works, you first need to show that for any k>0, {p in L : e(p) > k} is countable. You can do this by first proving a lemma that at each such p, there is some f>0 such that for every point q in the interval (p, p + f), e(q) <= k. (Prove it by contradiction using the definition of right continuous). Now all you have to do is show that the lemma implies {p in L : e(p) > k} is countable, which shouldn't be too hard (just devise some way to count a bunch of isolated points).
    Last edited: Aug 4, 2009
  4. Aug 4, 2009 #3
    Thank you very much. Given your guidance, I think I managed to prove the claim.

    Let p be the set of point of f where f is right-continuous but not left-continuous and let p be an element of L.

    Since f is not left-continuous in p, there is \epsilon(p)>0 such that, for all \delta>0, there is q \in (p-\delta,p) such that |f(p)-f(q)|>\epsilon(p).

    Now define M_n = \{p \in L: \epsilon(p)>\frac{1}{n}\}. If we show that M_n is countable then we are done since L=\cup_n M_n.

    Now we use your lemma. For any p \in M_n there is a k(p) such that if m in (p,p+k), \epsilon(m)<1/n

    Suppose this is not true. Then f is not right-continuous, since for any k, there exists m,n s.t. p
    < n < m \leq p+k and |f(m)-f(n)|>1/n. Since k is arbitrary, f(p) is not right-continuous.

    Finally, note that we can pick k so that p+k is rational. So there is a one-to-one mapping between points p in M_n and p+k(p) in a subset of the rational numbers, showing that M_n is countable. Since the countable union of countable sets is also countable, L is countable.
  5. Aug 4, 2009 #4
    I just realized I botched the definitions in my last post. Fixed in the edit
    You could use some more detail here. (Also, you used n twice). It would go more like this:

    Suppose the lemma is not true. Then there is some p in M_n such that for any k, there is an m in (p,p+k) where e(m) >= 1/n. Now since p is right-continuous, for any e' there is a d such that if q is in (p,p+d) then |f(p)-f(q)| < e'. Choose e'=1/(2n), and k=d. Using the fact that e(m) >= 1/n, that means we can pick any d' and there will be a point q' in (m-d',m) where |f(m)-f(q')| > e(m) >= 1/n. Choose d' so that p < p+d' < m, which means q' is in (p,p+d). Summarizing the situation, we have:
    |f(p)-f(q')| < 1/(2n)
    |f(p)-f(m)| < 1/(2n)
    |f(m)-f(q')| >= 1/n
    You can show there's a contradiction with these three inequalities.

    Heh, nice.
    Last edited: Aug 4, 2009
  6. Aug 4, 2009 #5
    Yes, the triangular inequality yields the contradiction. thanks
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