# IVP proof (analysis)

1. Dec 17, 2009

### StarTiger

Tricky problem. Any tips? Thanks SOO much!!

1. The problem statement, all variables and given/known data

Let f be continuous on [a,b] and let c be a real number. If for every x in [a,b] f(x) is NOT c, then either f(x) > c for all x in [a,b] OR f(x) < c for all x in [a,b]. Prove this using a) Bolzano-Wierstrass and b) Heine-Borel property.

2. Relevant equations

B-W property: A set of reals is closed and bounded if and only if every sequence of points chosen fro E has a subsequence that converges to a point in E.
H-B property: A subset of the reals has the HB property if and only if A is both closed and bounded.

(Note the function given fits HB and BW propertis by definition).

3. The attempt at a solution

Some hints:

For WB: suppose false. Explain how there exist sequences {x_n} and {y_n} such that f(x_n) > c, f(y_n) < c and |x_n - y_n| < 1/n
For HB: Suppose false and xplain why there should exist at each point x in [a,b] an open interval I_x centered so that either f(t)>c for all t in intersection of I_x and [a,b] or else f(t)<c for all t in the intersection of I_x in [a,b]

2. Dec 17, 2009

### ystael

This seems like a strange question. The Heine-Borel and Bolzano-Weierstrass theorems each state that a closed and bounded subset of $$\mathbb{R}$$ is compact (each using a different characterization of compactness). However, the intermediate value theorem for closed intervals in $$\mathbb{R}$$ is not a consequence of compactness, but of connectedness.

In your question, one might replace $$[a,b]$$ by $$[0,1] \cup [2,3]$$, a compact but disconnected set. This set possesses the Heine-Borel and Bolzano-Weierstrass properties, but the intermediate value theorem is obviously false for it (consider any function which takes one constant value on $$[0,1]$$ and another on $$[2,3]$$).

While the fact you are asked to prove is true, the theorems you are asked to use to prove it seem totally inappropriate to the task.