1

Let $f$ be a function $\mathbb R \to \mathbb R$ continuous on the right at every point of $\mathbb R$. Show that the set of points of continuity of $f$ is dense in $\mathbb R$.

Does anyone have a hint ? Using only basic results.

  • Doesn't it suffice to show that any point of discontinuity can be approached by a sequence with terms in the points of continuity of $f$? And that's possible from the right because it's continuous on the right? – stressed out Dec 07 '17 at 18:39

1 Answers1

1

For any $f:\Bbb R \to \Bbb R$ and any $q\in \Bbb Q^+$ let $D(q,f)\subset \Bbb R$ where $x\in D(q,f)$ iff every open set containing $x$ contains $y$ and $z$ with $|f(y)-f(z)|\geq q.$ (Remark: $y$ or $z$ may, or may not, be equal to $x.$)

(i). The set $D$ of points of discontinuity of $f$ is $\cup_{q\in \Bbb Q^+}D(q,f).$

(ii). Prove that every $D(q,f)$ is closed.

(iii). Use (ii) to show that every $D(q,f)$ has empty interior if $f$ is continuous on the right.

(iv). If $f$ is continuous on the right then (ii) and (iii) imply by the Baire Category Theorem that $D=\cup_{q\in \Bbb Q^+}D(q,f)$ is co-dense.

DanielWainfleet
  • 59,529
  • Consider this converse, for which I do not have an answer: If D is any first-category $F_{\sigma}$ subset of $\Bbb R,$ does there exist $f:\Bbb R\to \Bbb R$ that is continuous on the right, where $D$ is the set of discontinuities of $f$ ?..... I know how to prove that $D$ is the set of discontinuities of $some$ $g:\Bbb R\to \Bbb R$ but not by a method that makes $g$ continuous on the right. – DanielWainfleet Dec 08 '17 at 22:36