# Intermediate Value Theorem Proof & Sign-Preserving Lemma

Studying Calculus currently in my Analysis I course, and I am stuck on this lemma given by my professor, and thus the general proof. Here are my questions:

1. Is my interpretation of the lemma listed below correct? If not, where am I going wrong?
2. In the lemma, why do we choose $$\epsilon = f(a)$$?
3. Is there a simpler version of the proof of Intermediate Value Theorem that may be a more concise proof?

The lemma is a claim that:

Suppose $$f : [a,b] \rightarrow \mathbb{R}$$ is continuous at $$a \in X$$ with $$f(a) > 0$$. Then there exists a $$\delta > 0$$ such that for all $$x \in X$$ with |x-a| < $$\delta$$, one has that $$f(x) > 0$$. Similarily, if $$f(a) < 0$$.

Proof. Let $$\epsilon = f(a)$$. By definition of continuity, there exists $$\delta > 0$$ such that for all $$x \in X$$, |f(x) - f(a)| < $$\epsilon$$. In particular, $$f(x) > f(a) - \epsilon = 0$$. The proof for $$f(a) < 0$$ is similar.

Here is my interpretation of this claim, and let me know where I am going wrong with my thinking. There is a continuous function with the $$lim_{x \rightarrow a} f(x) = f(a)$$. When f(a) is positive, f(x) is positive. I am interpreting this correctly? Similarly for when f(a) is negative, f(x) is negative. I've heard this called the sign-preserving function.

Then, we use this lemma in our proof of the Intermediate Value Theorem and this is where my confusion lies. The proof of the Intermediate Value Theorem as given by our prof goes like so:

Replacing $$f$$ with $$-f$$ if needed, we may assume $$f(a) < f(b)$$. Replacing $$f$$ with $$f-y_0$$, we may assume $$y_0 = 0$$. Thus, we will show that if $$f : [a,b] \rightarrow \mathbb{R}$$ is continuous, with f(a) < 0, f(b) > 0, then there exists $$x_0 \in (a,b)$$ with $$f(x_0) = 0$$.

This is the first part of the proof. The second part sets up the structure for a contradiction by a proven claim P13 that if an upper-bound exists, then a least upper-bound exists, called a supremum.

Let A = {$$x \in [a,b] : f_{|[a,x]} < 0$$}. The set A is non-empty since $$a$$ is in A. The set is bounded above, since b is an upper bound. By P13, there exists a least upper bound, $$x_0$$ = sup(A).

We have $$x_0 \in [a,b]$$, since elements less than a cannot be upper bounds of A, while elements larger than b cannot be least upper bounds. We will show that f($$x_0$$) = 0. The proof is by contradiction.

Suppose f($$x_0$$) is not equal to 0. By the lemma, we may pick a $$\delta > 0$$ such that $$f$$ does not change sign on the set of all $$x \ in [a,b]$$ with $$|x-x_0| < \delta$$, and so it is either positive everywhere on this set, or negative everywhere on this set. In fact, it must be everywhere negative, since the set has non-empty intersection with A: Since $$x_0$$ = sup(A), there exists $$x_1 \in A$$ with $$x_1 \leq x_0$$ and $$|x_0 - x_1| < \delta$$. We conclude that $$f < 0$$ on [a, $$x_1$$] and also on [$$x_1, x_0$$], hence on all of [a, $$x_0$$]. That is, $$x_0 \in A$$; in particular, $$x_0 < b$$. Choose $$x_2$$ with $$x_0 < x_2 < b$$ and $$|x_2 - x_0| < \delta$$. Since $$f$$ is negative on [a, $$x_0$$] and also on $$[x_0, x_2]$$, it is negative on all of $$[a, x_2$$]. Thus, $$x_2 \in A$$, which is impossible since $$x_0$$ is an upper bound for A, and $$x_0 < x_2$$. This contradiction shows that the assumption is wrong and $$f(x_0) = 0$$.

• You can choose any $\epsilon$ such that $0<\epsilon\leq f(a)$. For example $\epsilon=f(a) /2$ would work fine. – Paramanand Singh Nov 6 '19 at 2:27

If a function $$f$$ is continuous at $$a$$ then the values of $$f(x)$$ can be made to lie near $$f(a)$$ by choosing $$x$$ near $$a$$. Now suppose $$f(a) >0$$ Then it is obvious that all the numbers which are very near to $$f(a)$$ must be positive. And by continuity we can restrict values of $$f(x)$$ to these numbers by choosing $$x$$ near $$a$$. If the argument does not seem obvious to you then you need to figure out what's not obvious here and let me know so that I can provide more explanation. This is what your lemma says in symbols.