# Thread: Function with Darboux's Property

1. ## Function with Darboux's Property

Let $\displaystyle f: [0,1] \rightarrow \mathbb{R}$ be a function with Darboux's property. $\displaystyle \forall \ y \in \mathbb{R}$, we have $\displaystyle f^-1({{y}})$ is closed. Prove that $\displaystyle f$ is continuous.

So I have an idea of how to approach this, but I'm not entirely sure on the closed part - does that necessary entail that the function is surjective? (If not, the pre-image wouldn't exist?)

2. Originally Posted by h2osprey
Let $\displaystyle f: [0,1] \rightarrow \mathbb{R}$ be a function with Darboux's property. $\displaystyle \forall \ y \in \mathbb{R}$, we have $\displaystyle f^-1({{y}})$ is closed. Prove that $\displaystyle f$ is continuous.

So I have an idea of how to approach this, but I'm not entirely sure on the closed part - does that necessary entail that the function is surjective? (If not, the pre-image wouldn't exist?)
Read this.... What you want is a corollary on the 2nd page.

http://tt.lamf.uwindsor.ca/314folder...ns/Darboux.pdf

3. Originally Posted by TheEmptySet
Read this.... What you want is a corollary on the 2nd page.

http://tt.lamf.uwindsor.ca/314folder...ns/Darboux.pdf
4. Ah I think I've solved it, but I'm not entirely sure if this is correct. Basically we prove by contradiction: assume that the function is not continuous, then for some $\displaystyle x_0 \in\ [0,1],\ \exists\ \epsilon_0 > 0$ such that $\displaystyle \forall\ \delta > 0, \exists\ x_\delta$ such that $\displaystyle |x_\delta\ - x_0| < \delta\$ but $\displaystyle |f(x_\delta)\ - f(x_0)| \ge \epsilon_0$. Then we can find a sequence $\displaystyle \{x_n\}\$ which converges to $\displaystyle x_0$, but $\displaystyle f(x_n) = f(x_0) - \epsilon/2,\ \forall\ n \ge 1$ (this is due to Darboux's property, and we're assuming without loss of generality that $\displaystyle f(x_\delta)\ - f(x_0) \ge -\epsilon_0$).
Now this sequence is in the pre-image of $\displaystyle f(x_0) - \epsilon/2$, which is closed by assumption. Hence it contains all its accumulation points $\displaystyle \implies f(x_0) \in\ \{f(x_0) - \epsilon / 2\}$, which is a contradiction. Hence, $\displaystyle f$ is continuous.