I am reading a text on non-standard analysis. I need to prove the following: Suppose that $f$ is non-decreasing on the real interval $[a,b]$ and that $f$ satisfies the intermediate value property. Then $f$ is continuous on $[a,b]$.
I need to show this by showing if $y-x$ is infinitesimal, then so is $f(y)-f(x)$.
My first attempt was to pick $c$ so that $f(c)-f(x)\leq y-x$ (for $y>x$). If $c \geq y$, then $$ 0\leq f(y)-f(x) \leq f(c) - f(x) \leq y-x. $$ Since $y-x$ is infinitesimal, so is $f(y)-f(x)$. If $c \leq y$, that argument doesn't work and I am unable to see how to fix it.
My second thought was to use the fact, for all $\varepsilon > 0$ there are $x=x_1, \ldots, x_n=y$ so that $$ f(x_{k+1})-f(x_k) \leq \varepsilon. $$ Then $$ f(y)-f(x) = \sum_{k=1}^n f(x_{k+1})-f(x_k) \leq n\varepsilon. $$ The transfer principle applies, but then I am left with a sum in $^*\mathbb{N}$, and I don't see why it should be infinitesimal.
Are there any ways to complete either of my arguments? Or, would someone give me a hint in the right direction? Thanks.
If $f(t)$ is not continuous on $[a,b]$ then there are infinitely close points $x<y\in{}^{\ast}[a,b]$ where $f(x)-f(y)$ is appreciable (i.e., noninfinitesimal). Let $c=st(x)=st(y)\in [a,b]$, let $K=st(f(x))$ and $L=st(f(y))$ so that $K<L$. Then $f(t)\leq K$ for all (real) points $t\in [a,c)$ and $f(t)\geq L$ for all $t\in (c,b]$. For the intermediate value theorem to hold, $f(c)$ would have to take all the values in the interval $(K,L)$, contradiction.