# show that there exists an increasing function

• Oct 11th 2011, 05:08 PM
wopashui
show that there exists an increasing function
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?
• Oct 12th 2011, 03:20 AM
emakarov
Re: show that there exists an increasing function
Quote:

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: $g(x)=\sup\{f(t): 0\le t\le x, t\in\mathbb{Q}\}$.
• Oct 12th 2011, 03:53 AM
SlipEternal
Re: show that there exists an increasing function
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.