Question

Let $D$ denote the set of rationals in $[0,1]$ and suppose that $f: D \rightarrow \mathbb{R}$ is increasing. Show that there is an increasing function $g:[0,1] \rightarrow \mathbb{R}$ such that $g(x)=f(x)$ whenever $x$ is rational. [Hint: For $x \in[0,1]$, define $g(x)=\sup (f(t)$ : $0 \leq t \leq 1, t \in \mathbb{Q}\}$.]

   Let $D$ denote the set of rationals in $[0,1]$ and suppose that $f: D \rightarrow \mathbb{R}$ is increasing. Show that there is an increasing function $g:[0,1] \rightarrow \mathbb{R}$ such that $g(x)=f(x)$ whenever $x$ is rational. [Hint: For $x \in[0,1]$, define $g(x)=\sup (f(t)$ : $0 \leq t \leq 1, t \in \mathbb{Q}\}$.]

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Real analysis

N. L. Carothers 1st Edition

Chapter 1, Problem 50

Instant Answer

Step 1

For any \( x \in [0, 1] \), set \[ g(x) = \sup \{ f(t) : t \in D, 0 \leq t \leq x \} \] where \( D \) is the set of rational numbers in the interval \([0, 1]\). Show more…

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Let $D$ denote the set of rationals in $[0,1]$ and suppose that $f: D \rightarrow \mathbb{R}$ is increasing. Show that there is an increasing function $g:[0,1] \rightarrow \mathbb{R}$ such that $g(x)=f(x)$ whenever $x$ is rational. [Hint: For $x \in[0,1]$, define $g(x)=\sup (f(t)$ : $0 \leq t \leq 1, t \in \mathbb{Q}\}$.]

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