So I've been trying to work through Spivak's Calculus and I keep getting stuck.
I am terribly confused on Spivak's proof on chapter 8 theorem 7-2 that a continuous function $f$ on an interval $[a,b]$ is bounded above the proof goes as follows:
They are quite a few things that confuse me in this proof for example why can we say that the set $A$ is not empty because it is bounded above by b i know that if a function is continuous at a point $a$ then there is a number $\delta > 0$ such that $f$ is bounded on the interval $(a-\delta,a+\delta)$ however how can we know that the delta interval is contained in the interval $[a,b]$ and say that then b must be an upper bound?.
My 2nd confusion has to do with why he assumes the supremum $\alpha=b$ this really confuses me like what reason is there for this to be true and in his proof by contradiction he assumes $\alpha < b$ and then states that it must be the case that $\alpha$=$b$ however why is it not true that $\alpha \leq b$?.
Thanks in advance.
Best Answer
No, that's the justification that $A$ is bounded. The justification that $A$ is non-empty is written immediately after that is claimed: $a \in A$. This isn't hard to see: $f$ is clearly bounded on $[a,a]$ since that is just one point.
It doesn't need to be contained in $[a,b]$ but if say, the interval extends outside $[a,b]$ then you can always replace it by a smaller interval which is contained inside $[a,b]$. For instance. Suppose $[a,b] = [0,1]$ and you tell me that $f$ is bounded on $(0.5, 1.1)$, well then $f$ must also be bounded on $(0.6, 1)$.
However, this has nothing to do with $b$ being an upper bound. $b$ is an upper bound because by definition, $A = \{x : a \le x \le b \text{ and ...}\}$ so automatically, if $x \in A$ then $a \le x \le b$: every element of $A$ is less than or equal to $b$.
$b$ is an upper bound of $A$. Therefore $\sup A \le b$ meaning either $\sup A < b$ or $\sup A = b$. If we can rule out that $\sup A < b$ then it must be the case that $\sup A = b$. This is what happens in the first paragraph.