Notes - Analysis II HT23, Intermediate value theorem


Flashcards

Can you 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 $f(\xi) = c$.

What does it mean 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.




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