Function with Darboux's Property

h2osprey · #1 $Old$ May 23rd, 2009, 11:33 AM

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

TheEmptySet · #2 $Old$ May 23rd, 2009, 12:06 PM

Quote:

Originally Posted by h2osprey $View Post$

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

TheEmptySet · #3 $Old$ May 23rd, 2009, 12:08 PM

Quote:

Originally Posted by TheEmptySet $View Post$

Read this.... What you want is a corollary on the 2nd page.

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

Sorry I misread your question, The answer is not in this

h2osprey · #4 $Old$ May 24th, 2009, 01:58 PM

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 $x_0 \in\ [0,1],\ \exists\ \epsilon_0 > 0$ such that $\forall\ \delta > 0, \exists\ x_\delta$ such that $|x_\delta\ - x_0| < \delta\$ but $|f(x_\delta)\ - f(x_0)| \ge \epsilon_0$ . Then we can find a sequence $\{x_n\}\$ which converges to $x_0$ , but $f(x_n) = f(x_0) - \epsilon/2,\ \forall\ n \ge 1$ (this is due to Darboux's property, and we're assuming without loss of generality that $f(x_\delta)\ - f(x_0) \ge -\epsilon_0$ ).

Now this sequence is in the pre-image of $f(x_0) - \epsilon/2$ , which is closed by assumption. Hence it contains all its accumulation points $\implies f(x_0) \in\ \{f(x_0) - \epsilon / 2\}$ , which is a contradiction. Hence, $f$ is continuous.

Is that okay?