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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: enter image description here

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.

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  • $\begingroup$ For the first point, it seems you misread Spivak's argument: $f$ is bounded on the interval $(\color{red}\alpha-\delta,\color{red}\alpha+\delta)$. For the second point, he doesn't assume $\alpha=b$ – he proves it. $\endgroup$
    – Bernard
    Commented Feb 16, 2021 at 0:30
  • $\begingroup$ The fact that $b$ is an upper bound is from the definition of $A$, which confines $a\le x\le b$. The creativity of this proof is the construction of set $A$. $\endgroup$ Commented Feb 16, 2021 at 0:43

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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

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.

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?

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 $b$.

My 2nd confusion has to do with why he assumes the supremum α=b this really confuses me like what reason is there for this to be true and in his proof by contradiction he assumes α<b and then states that it must be the case that α=b however why is it not true that α≤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.

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  • $\begingroup$ just a quick question spivak states that there is an $x_0$ in $A$ satisfying $\alpha - \delta < x_0<\alpha$ however couldn't $x_0$ be equal to $\alpha$ since a closed interval contains its supremum so why don't we say $\alpha - \delta <x_0\leq\alpha$ where the possibility that $x_0$=$\alpha$ is allowed $\endgroup$ Commented Feb 16, 2021 at 21:42
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    $\begingroup$ @Thehomeschooler Right, $x_0 \le \alpha$ would be a better way to write it since in general there might not be an element of $A$ in $(\sup A - \delta, \sup A)$. Here there is because $A$ is an interval: if $f$ is bounded on $[a, x]$ then $f$ is bounded on $[a, x']$ for any $a \le x' \le x$. But it isn't important that $x_0 \neq \alpha$ so it would be better to write $\le$. $\endgroup$
    – Sera Gunn
    Commented Feb 16, 2021 at 23:45
  • $\begingroup$ funny thank you very much I understand know. $\endgroup$ Commented Feb 17, 2021 at 7:28

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