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I know that Dirichlet function has uncountable many discontinuities. I think they are removable, because the discontinuities can be removed by redefining the function values of the rational numbers as 0.

So Dirichlet function is a function that has uncountable many removable discontinuities, then my question is can we construct a function with uncountable many jump discontinuities? If not, how do we prove it is impossible? Thank you.

An odd but similar question is can we have a function that has uncountable many infinite discontinuities?

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    $\begingroup$ The discontinuities of the Dirichlet function are not removable. The one sided limits at any point do not exist. As to the question on jump discontinuities, the answer is negative. See "Elementary Real Analysis by Thompson, Bruckner, and Bruckner, Theorem 5.64. (This section is available on Google Books). $\endgroup$ – matt biesecker May 10 '15 at 4:57
  • $\begingroup$ @mattbiesecker Thank you for the reply. Now I am puzzled what type the discontinuities of Dirichlet function are? They seem not to fall into any category of the removable, jump or infinite discontinuities? $\endgroup$ – Tony May 10 '15 at 13:14
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$f$ has a jump discontinuity at $x=a$ if $f(a-) = \lim_{x \to a-} f(x)$ and $f(a+) = \lim_{x \to a+} f(x)$ exist, but $f(a-) \ne f(a+)$. Let $S(a) = |f(a+) - f(a-)$ be the size of the jump discontinuity at $a$. Let $J(n)$ be the set of jump discontinuities $a$ with $S(a) > 1/n$. Then I claim $J(n)$ is a discrete set. Since a discrete subset of $\mathbb R$ is countable, and the union of countably many countable sets is countable, this implies that the set of jump discontinuities of $f$ is countable.

To prove the claim, suppose $a \in J(n)$, and take $\epsilon < 1/(2n)$. There is $\delta > 0$ so that for $a -\delta < x < a$, $|f(x) - f(a-)| < \epsilon$ and for $a < x < a + \delta$, $|f(x) - f(a+)| < \epsilon$. Thus if $x$ and $y$ are both in $(a-\delta, a)$ or both in $(a, a+\delta)$, $|f(x) - f(y)| < 2 \epsilon < 1/n$. This implies that if $x$ is a jump discontinuity in one of these intervals, $S(x) \le 2 \epsilon < 1/n$.

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  • $\begingroup$ hi , can we have a function has uncountable removable discontinue points? $\endgroup$ – Idele Oct 17 '16 at 4:53
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    $\begingroup$ No. A removable discontinuity of $f$ is a point $a$ where $\lim_{x \to a} f(x)$ exists but is not $f(a)$. By a similar argument to my answer above, the set of removable discontinuities where $|f(a) - \lim_{x \to a} f(x)| > 1/n$ is discrete. $\endgroup$ – Robert Israel Oct 19 '16 at 7:05

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