# Confusion on Spivak's proof chapter 7-12

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 $$\alpha \leq b$$?.

• 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. Commented Feb 16, 2021 at 0:30
• 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 $A$. Commented Feb 16, 2021 at 0:43

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.

• 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 Commented Feb 16, 2021 at 21:42
• @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 $\le$. Commented Feb 16, 2021 at 23:45
• funny thank you very much I understand know. Commented Feb 17, 2021 at 7:28