Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Problem: Prove that for any function f from a countable subset M of $\mathbb{R}$ there exists a sequence of continuous functions {f_n} that converges pointwise to f on M.

Context: This was put forward in lecture as a way to deduce that in the Baire-Osgood Theorem the condition of completeness was needed to get at least one point of continuity.

I have been struggling on a way to make headway on this problem. We are using Carothers Real Analysis and we have studied up to Chapter 12, this includes Baire Category Theorem, Arzela-Ascoli Theorem, and Stone-Wierstrass.

I feel that I am missing something something obvious that would simplify the problem significantly but I do not see how I can use any of the results we have found in class or in the text.

I would appreciate any insight or suggestions on how to go about the proof.

share|cite|improve this question
up vote 1 down vote accepted

Just enumerate the elements of $M$ as $m_k$. Then define $f_n$ to be equal to $f$ on $m_1$ through $m_n$. On all other elements $x$ define $f_n$ to have the value given by taking a straight line segment from $(m_i, f(m_i))$ to $(m_j, f(m_j))$ where $m_i$ is the immediate predecessor of $x$ in $m_1, ... , m_n$ and $m_j$ is the immediate successor. If $x$ is a lower bound for $m_1, ... , m_n$ define $f_n$ equal to it's value on $m_1$. Similarly if $x$ is an upper bound, define $f_n$ equal to it's value on $m_n$.

share|cite|improve this answer
Thank you for your reply. I have a quick question; I understand this extension is continuous on $\mathbb{R}$ but how do we know f_n is continuous on M? – aawaldrop Nov 15 '12 at 1:51
Since f_n is continuous on $\mathbb{R}$ it is continuous on any subset of $\mathbb{R}$ in the subspace topology. – Seth Nov 15 '12 at 1:55
Thank you you have been very helpful. – aawaldrop Nov 15 '12 at 1:57

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.