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While studying the properties of ordinal utility functions, I came across the following question.

Given a strictly increasing function $f : D \rightarrow \mathbb{R}$, where $D$ is an arbitrary non-empty subset of $ \mathbb{R} $, can one always find a strictly increasing function $g : \mathbb{R} \rightarrow \mathbb{R}$ that is defined everywhere on $\mathbb{R}$ and is equal to $f$ everywhere in the set $D$?

I feel that the answer should be positive, however, there might be some counterexample I'm unaware of.

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    Given the counterexamples found below, it makes only sense to ask for such a continuation if $f$ is bounded on every bounded subset of $D$ and one drops the condition that the monotonicity should be *strict*. But then $g(x):=\sup\{f(x)\mid x\in D\cap (-\infty,x]\}$ for x>\inf D and $g(x)=\inf\{f(x)\mid x\in D\}$ for $x\le\inf D$ will work.2012-09-15

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No, consider for instance the case where $D=(0,1)$ and $f$ is the function given by $f(x)=-\frac1x$.

There cannot exist a stricly increasing extension of $f$ to $\mathbb R$ because $\lim\limits_{x\to 0^+}f(x)=-\infty$.

(That is, the value of the extension at any non-positive point would have to be lower than any real number. That is not possible.)

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    True. So being extendible is a relatively strong requirement.2012-09-15
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Another counterexample : set $D=(-\infty,0[\cup[1,+\infty)$, and define $f$ on $D$ like this : $f(x)=\begin{cases}\begin{array}{ccc}x& \text{if}& x<0 \\ x-1 & \text{if} & x\geq 1\end{array}\end{cases}$

Assume that $g$ exists and choose $x\in[0,1)$. For all negative $y$, you'll get : $g(y)=f(y)=y so $g(x)$ should be greater than any negative number yet less than $0$. This is absurd.

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    That a nice example, especially with regard to Asaf's comment to my answer.2012-09-15