0

I have seen similar questions to this on here but none using the hint.

So far I have: Let Dn be the set of points where the jump of f is greater than epsilon = 1/n I need to show that Dn is finite, so that their union, which is the set of all discontinuities is countable.

How do I show they are countable using limits?

Vlad
  • 6,938
Paul
  • 1

2 Answers2

1

If $D_n$ is infinite, then it contains a convergent (infinite) sequence $\{x_k\}$ of points. (Why?) Extracting a subsequence if necessary, you can take $\{x_k\}$ to be monotone. Suppose for example that $x_k$ decreases (strictly) to $c\in[a,b)$. Is this compatible with the existence of the right limit $f(c+)$?

John Dawkins
  • 29,845
  • 1
  • 23
  • 39
  • Using Bolzano Weierstrass theorem: Since f is regulated it is bounded so there is a convergent subsequence D_n_k k≥1? – Paul Nov 01 '15 at 17:04
  • Then I'm not sure what you are trying to say... – Paul Nov 01 '15 at 17:09
  • The set $D_n$ is a subset of $[a,b]$, so it is bounded. If $D_n$ is infinite then you can apply Bolzano-Weierstrass to $D_n$ to get the sequence of points $x_k\in D_n$ mentioned in my answer. – John Dawkins Nov 01 '15 at 17:20
  • Ok so I get eveything you said until the last part. lim y-->c+ f(y) exists from the hint. But I don't get why it doesn't exist for a contradicition using the decreasing sequence – Paul Nov 01 '15 at 17:31
  • Remember that each $x_k$ is taken from $D_n$, so that $|f(x_k+)-f(x_k-)|\ge 1/n$. (I use the notation $f(x+)$ for $\lim_{t\to x,t>x}f(t)$, etc.) It will be helpful to draw a picture. – John Dawkins Nov 01 '15 at 17:41
  • Suppose $f \in R[a,b]$ and $\forall n$, let $D_n$ be the set of points where the jump of $f$ is greater than $\epsilon = 1/n$. How do I put this into maths? We will show that each $D_n$ is finite, so that their union, which is the set of all discontinuities is countable.

    Suppose $D_n$ is infinite. We know that $D_n$ is bounded since it is a subset of $[a,b]$. So applying the BW theorem to $D_n$ we get a convergent subsequence ${x_k} \in D_n$. I know that every sequence has a monotonic subsequence. Let ${x_{kl}$ be the monotonic subsequence. Suppose it strictly decreases to $c \in [a,b]$.

     – Paul Nov 01 '15 at 18:00
  • I don't really get your last comment. How do I show the right limit doesn not exist. Use the definition of limits and negate it ? – Paul Nov 01 '15 at 18:07
1

Pretty well anything you can prove using the Bolzano-Weierstrass theorem on the real line, you can also prove using a nested sequence of intervals, or the Heine-Borel theorem, or the Cousin Covering Lemma and other methods. When you get into a situation like this try one and then try them all. You can't afford to freeze on an exam--be able to use them all.

Here is a sketch for a nested sequence of intervals argument. (Not much different than the BW argument just given). If $D_n$ (as you define it is infinite) then split the interval in half and choose the half containing infinitely many points. Do this over and over again and then claim the existence of a point $c$ belonging to all of the intervals. Since $f(c+)$ exists there is an interval $(c,c+\delta_1)$ where all the values are close together, way closer than $1/n$. Since $f(c-)$ exists there is an interval $(c-\delta_2)$ where all the values are close together, way closer than $1/n$. The interval $(c,c+\delta_1)$ doesn't contain any points from $D_n$, nor does the interval $(c-\delta_2,c)$. That will contradict your definition of the nested intervals. [Of course write it up without saying "close" or "way closer".]

Now somebody post a Heine-Borel argument and a Cousin covering argument.

The hint forced you to use the equivalent characterization of regulated functions as the class of functions with finite one-sided limits. The definition you gave as the class of uniform limits of step functions gives a neater proof: uniform limits preserve continuity at points. Can't get too many discontinuities then since step functions are mostly continuous.

B. S. Thomson
  • 7,717