show that there exists an increasing function

**wopashui** · October 11th 2011, 05:08 PM

Let D denote the set of rationals in [0,1] and suppose that f: D--> R is increasing. show that there is an increasing function g:[0,1]-->R such that g(x)=f(x) whenever x is rational.

hint:[For x $\in$ [0,1], define g(x)= $sup$ {f(t): 0<=t<=1, t $\in$ Q}]

according to the hint, we need to show that f is equal to its least upper bound, but f is increasing, so is f(1) the least upper bound?

**emakarov** · October 12th 2011, 03:20 AM

Originally Posted by wopashui

hint:[For x $\in$ [0,1], define g(x)= $sup$ {f(t): 0<=t<=1, t $\in$ Q}]

It should say: $g(x)=\sup\{f(t): 0\le t\le x, t\in\mathbb{Q}\}$ .

**SlipEternal** · October 12th 2011, 03:53 AM

From what I see, you are supposed to prove that your function $g$ has the properties stated. So, check. Is $g$ increasing?
Let $x,y \in \[0,1\]$ be chosen arbitrarily, and assume that $x\le y$ . Then, is it true that $g(x)\le g(y)$ ? Why or why not.

Next, does $g(x)=f(x)$ whenever $x$ is rational? Why or why not.

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