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 $\displaystyle \in$ [0,1], define g(x)=$\displaystyle sup${f(t): 0<=t<=1, t$\displaystyle \in$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?

Re: show that there exists an increasing function

Quote:

Originally Posted by

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

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

Re: show that there exists an increasing function

From what I see, you are supposed to prove that your function $\displaystyle g$ has the properties stated. So, check. Is $\displaystyle g$ increasing?

Let $\displaystyle x,y \in \[0,1\]$ be chosen arbitrarily, and assume that $\displaystyle x\le y$. Then, is it true that $\displaystyle g(x)\le g(y)$? Why or why not.

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