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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.

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1 Answer 1

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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 $m_n$.

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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

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