boundedness theorem for continuous functions question
The theorem statement is "if $f$ is continuous on $[a,b]$, $f$ is bounded on $[a,b]$". This is proven in the textbook Calculus by the author Apostol by the "method of successive bisection", which I'm sure many are familiar with. The proof is done by contradiction.
Here is my concern with this proof: we take the supremum of the leftmost endpoints of the subintervals we generate. But this assumes we can generate an infinite number of subintervals. We generated these subintervals by answering a question: "in which half is $f$ unbounded?". This is a question a person has to answer, right? So why can we do it an infinite number of times?
If let's say we stopped after a finite number of bisections $n$, we have the subintervals: $[a_n, b_n] \subset ... \subset [a_1, b_1] \subset [a,b]$. Then sup$\{a, a_1, ..., a_n\} = a_n$. $f$ has to be continuous at $a_n$, so we have a neighborhood $N(a_n;\delta)$ such that if $x\in N(a_n;\delta)$, then $|f(a_n) - f(x)| < \epsilon$. But we have no guarantee that $b_n - a_n < \delta$. So $f$ can be continuous at $a_n$ and still unbounded in $[a_n, b_n]$ (this doesn't result in a contradiction, which was the direction of the proof given). We can't take more bisections because the supremum, where we applied continuity, is not necessarily going to be in the smaller bisections.
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$\begingroup$No, you don't have to "answer" the question of in which subinterval the function is unbounded! If a function is unbounded in $A\cup B$ then it is unbounded in $A$ or unbounded in $B$ (or both). So you let $C=A$ or $C=B$, in such a way that $f$ is unbounded in $C$. Which is it, $A$ or $B$? We don't know and we don't care.
I don't follow your more detailed objection. In any case, the continuity of $f$ has nothing to do with that part of the proof; whether $f$ is continuous or not if it's unbounded on $[0,1]$ then it's unbounded on $[0,1/2]$ or unbounded on $[1/2,1]$. Continuity only comes in later, after all the bisection is done.
$\endgroup$ 4 $\begingroup$Take a look at it from another perspective. $f$ being bounded means it is either bounded above or below. Suppose $f$ is not bounded above on $[a,b]$. Then $\exists x_1,\cdots,x_k$ such that $f(x_1) > 1, \cdots, f(x_k) > k, \forall k \in \mathbb{N}$.
Your sequence $(x_n)_{n \in \mathbb{N}}$ is unbounded, but it has a subsequence $(x_{n_i})$ that converges, say to a point $c \in [a,b]$.
Since $f$ is continuous, $\lim_{i\rightarrow \infty}(x_{n_i}) = c \Rightarrow \lim_{i \rightarrow \infty} f(x_{n_i})= f(c)$. This contradicts our assumption that $f$ was didn't converge. You can do the case for boundedness below basically the same way.
$\endgroup$ $\begingroup$Your question is interesting and perceptive - and hard to answer. I think you are really asking about how a proof can make "infinitely many statements". Once you are satisfied with that, you'll manage the rest of the argument just fine.
Well, that is in fact something mathematicians have struggled with for centuries, so you can't hope for a quick answer.
What this proof says is that after you've made a bisection, you can make another. That gets around the idea that you are making them all at once.
There are lots of mathematical proofs that rely on this kind of logic. It's the heart of mathematical induction. (If you haven't learned about that yet, don't worry. You will when the time comes.)
From another perspective, this kind of proof in calculus or analysis is the modern way to deal with limits. Intuitively, limits seem to be talking about numbers being infinitely close together. The way we deal with that is to replace "infinitely close together" by infinitely many statements each of which says the numbers are a small distance apart.
I hope this helps. Good luck.
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