If $f$ is continuous on $[a,b]$, then $f$ is bounded on $[a,b]$.
First off, I have seen a proof of this and I sort of get it but wanted to go through the proof and justify my arguments correctly to make sure I in fact get it. This proof can usually be found in calculus books which is what I am aiming for. I bring this up as I know there is a topological way to do this, but that is not what I am looking for. Okay here we go.
Lets first define the set $A = \{x : x \in [a,b]$ and $f$ is bounded at $[a,x]\}$
Here we are simply saying that $\forall x$ in this interval $[a,x]$ will be bounded. We don't know yet that $[a,b]$ is bounded but we can go with this for now.
Next it should be clear that $A$ is not the empty set since $a$ and $b$ are a part of this set. If it were empty we could bound the set but not get a least upper bound.
Since $A$ is bounded above (by some $x$), it has a least upper bound call it $\alpha$. We want to show that $\alpha = b$. To prove this, suppose $\alpha < b$. Well, since $f$ is continuous, I can construct a bounded interval around $\alpha$, call it $[\alpha – \epsilon,\alpha+\epsilon]$ for $\epsilon > 0$. But we defined our set $A$ to be bounded at $[a,x]$. Here we can have $x = \alpha – \epsilon$ since $\alpha -\epsilon \in [a,b]$. So $[a, \alpha – \epsilon]$ is bounded.
We have constructed two bounded intervals, namely $[a,\alpha – \epsilon]$ and $[\alpha – \epsilon, \alpha + \epsilon]$. Putting these together gives us the bounded interval $[a, \alpha + \epsilon]$. This creates a contradiction because $\alpha$ would not be the least upper bound of $A$. $\alpha +\epsilon$ would be the least upper bound which is not equal to $\alpha$ since $\epsilon > 0$. So it must be that $\alpha = b$, and hence $b$ is the least upper bound.
This proves that our set $A$ is bounded on any interval of the form $[a,x]$. To show that $[a,b]$ is bounded, take the interval $[b-\epsilon, b]$. This interval will be bounded since $f$ is continuous. We also know that $[a, b-\epsilon]$ is bounded because $b – \epsilon \in [a,b]$ (Here $x = b – \epsilon$). Putting them both together we see that $[a,b]$ is bounded.
Here is the proof in my own words to try and convince myself of the argument. One idea that does not make sense is showing that $\alpha = b$. Since $b$ is the least upper bound then why are we not done with the proof? I mean $b$ being the least upper bound means its bounded right?
Best Answer
I am pretty sure this proof is a valid proof. However, you seem to have confusing about the part after we prove $b$ is the least upper bound of the set and why the proof does not end there, so I will pick up the argument from there.
We have that the least upper bound of the set $A=\{ x : x \in [a, b] \wedge f \ \text{is bounded on} \ [a, b]\}$ is $b$. This does not, however, automatically mean, $b \in A$. The least upper bound of the set does not necessarily lie inside the set. Take the following example:
$$S=\{3, 3.1, 3.14, 3.141, 3.1415, 3.14159, ...\}$$
The above set slowly approaches $\pi$'s decimal expansion with rational numbers. Therefore, $\pi$ is the least upper bound of the set. However, $\pi \notin S$ because this is a set of rational numbers and $\pi$ is not rational. This might take a while to understand, so you might want to pause and think here if you are confused.
Therefore, since $b$ being the least upper bound does not imply $b \in A$, we need to prove $b \in A$ using the fact that the $f$ is continuous. The proof does this well, but it's sort of ambiguous, so I'll rewrite it here to clarify: