Analysis II HT23, Intermediate value theorem
Flashcards
@State the intermediate value theorem in full.
Suppose $f : [a,b] \to \mathbb{R}$ and $f$ is continuous on $[a,b]$. Let $c$ be any real number between $f(a)$ and $f(b)$. Then $\exists \xi \in [a,b]$ where $f(\xi) = c$.
@Define what it means for $f$ to have the intermediate value property.
$\forall I$ interval, $f(I)$ is also interval.
When proving the intermediate value theorem by using bisection, you construct a sequence of intervals $[a _ n, b _ n]$ such that $[a _ {n+1}, b _ {n+1}] \subseteq [a _ n, b _ n]$ and $b _ n - a _ n \to 0$ with each step satisfying $f(a _ n) \le c \le f(b _ n)$. Where does the magical value $\xi$ then come from (without proving that $f(\xi) = c$)?
Since $a _ n$ and $b _ n$ are both strictly monotonic bounded sequences, they must converge to some $\xi$ and $\xi'$. Then $b _ n - a _ n \to 0$ implies $\xi = \xi'$.
When proving the intermediate value theorem by using bisection, you construct a sequence of intervals $[a _ n, b _ n]$ such that $[a _ {n+1}, b _ {n+1}] \subseteq [a _ n, b _ n]$ and $b _ n - a _ n \to 0$ with each step satisfying $f(a _ n) \le c \le f(b _ n)$. Both $a _ n$ and $b _ n$ tend to some value $\xi$. How do you use trichotomy and continuity to verify that $f(\xi) = c$?
On one hand
\[f(\xi) = \lim _ {n \to \infty} f(a _ n) \le c\]and on the other
\[f(\xi) = \lim _ {n \to \infty} f(b _ n) \ge c\]so $f(\xi) = c$.
Proofs
@Prove the intermediate value theorem via nested intervals and bisection.
Todo.
@Prove the intermediate value theorem by considering infimums and supremums.
Todo.