Read this.... What you want is a corollary on the 2nd page.
http://tt.lamf.uwindsor.ca/314folder...ns/Darboux.pdf
Let be a function with Darboux's property. , we have is closed. Prove that 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
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 such that such that but . Then we can find a sequence which converges to , but (this is due to Darboux's property, and we're assuming without loss of generality that ).
Now this sequence is in the pre-image of , which is closed by assumption. Hence it contains all its accumulation points , which is a contradiction. Hence, is continuous.
Is that okay?