Increasing function with dense image continuous? Can someone please tell me if the following assertion is true?
Let $f: [a, b] \to \mathbb R$ be a strictly increasing function, such that $f([a,b])$ is dense in dense in $[f(a), f(b)]$. Then $f$ is continuous.
Edit:
Thanks for your answers! I've found this Lemma on my own notes, which i have written down some time ago and wasn't sure anymore, if it is actually true. But I think I've found the solution now. Assume that $f$ is not continuous at a point $c \in (a,b)$. Then we have $$\lim_{x \to c^-} f(x) < \lim_{x \to c^+} f(x) \; .$$ We can conclude that $$ \left( \lim_{x \to c^-} f(x), \lim_{x \to c^+} f(x) \right) \cap f([a,b]) = \left\{ f(c) \right\} \; .$$
Since either $\lim_{x \to c^-} \neq f(c)$ or $\lim_{x \to c^+} f(x) \neq f(c)$ we have either $$\left( \lim_{x \to c^-} f(x), f(c) \right) \cap f([a,b]) = \emptyset$$ or $$\left( f(c), \lim_{x \to c^+} f(x) \right) \cap f([a,b]) = \emptyset$$ which contradicts to the density of $f([a,b])$ in $[f(a), f(b)]$. If $f$ is not continuous at a boundary point $a$ or $b$, a similiar argument leads to a contradiction. This is now correct, right?
 A: Since you didn't really tell us how far you have gotten on your own, I am merely giving you some pointers:
What we want to do is show that for every fixed $x$ and $\varepsilon>0$ there is a $\delta>0$ such that, for every $y: |y-x|<\delta \quad\Rightarrow |f(x)-f(y)|<\varepsilon$.
This is but the pointwise definition of continuity.
Now, let's take a look at your problem. We are given an arbitrary $x$, and a $\varepsilon$>0. 
Do we know that there is a $y\neq x$ such that $f(y)\in [f(x),f(x)+\varepsilon)$? 
What does monotony say about our $y$? What can we conclude about where images of values between $x$ and $y$ are? Can we derive a possible value for a $\delta$ from this?
If you consider these questions I'm fairly confident you'll find a solution on your own. 
Also note that, once we have proven that $f$ is continuous, since continuous functions map compact sets to compact sets, we actually know that $f([a,b])=[f(a),f(b)]$. Which was not your question, just wanted to mention it to be sure.
A: Suppose that $f$ is not continuous at some point $c \in (a,b)$. Since $f$ is monotone its left and right limits both exist, so you must have $$\lim_{x \to c^-}f(x) < \lim_{x \to c^+} f(x).$$ What does this tell you about the range?
