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I need help on the proof of Theorem 7-2 in Spivak:

If $f$ is continous on $[a,b]$, then $f$ is bounded above on $[a,b]$.

So, the proof starts with this:

Let

$$A= \{x:a\le x \le b \text{ and } f \text{ is bounded above on } [a,x]\}$$

The author then went on to prove that $A$ has a least upper bound $\alpha$ and that $\alpha=b$

Here's the part I don't understand:

... we only know that $f$ is bounded on $[a,b]$ for every $x<b$, not necessarily that $f$ is bounded on $[a,b]$.

Since we already know that $A$ has a least upper bound $\alpha$ and that $\alpha=b$, doesn't this mean that $b$ is in $A$ hence $f$ is bounded on every $x$ in $[a,b]$ including $x=b$?

Then, to conclude the proof:

There is a $\delta>0$ such that $f$ is bounded on $\{x:b-\delta<x\le b\}$. There is $x_0$ in $A$ such that $b-\delta < x_0 < b$.Thus $f$ is bounded on $[a,x_0]$ and also on $[x_0,b]$, so $f$ is bounded on $[a,b]$.

Here, since $x_0$ is not equal to $b$, how can we conclude that $f$ is bounded on $[a,b]$? Furthermore, this does not appear to me to make use of $\alpha=b$, so couldn't we just have skipped to this part?

Thank you in advance for any help provided.

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    $\begingroup$ Consider the function $\frac{1}{b-x}$. All except the last step $b \in A$ works for that function. $\endgroup$ – Daniel Fischer Aug 11 '13 at 18:47
  • $\begingroup$ It doesn't mean that $\alpha=\sup A$ does not mean $\alpha\in A$. $\endgroup$ – Thomas Andrews Aug 11 '13 at 18:48
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There's an important step that goes by so quickly, you might have missed it. Consider this sentence:

There is a $\delta>0$ such that $f$ is bounded on $\{x:b-\delta<x\le b\}$.

This claim is not a consequence of the earlier steps! It's a new ingredient, and an important one. Do you see how the claim follows from the continuity of $f$? If so, do you see what the author is doing with $x_0$?

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  • $\begingroup$ yes, I see how the claim follows from the continuity of $f$ but I don't see why need to introduce $x_0$ into the proof? Why can't we just conclude that $f$ is bounded on $[a,b]$ when we know the claim is true? $\endgroup$ – mauna Aug 11 '13 at 20:00
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    $\begingroup$ @mauna Intuitively, the business with $A$ tells us that $f$ is bounded on a set that gets close to $b$ but doesn't necessarily include $b$. The business with $\delta$ tells us that $f$ is also bounded on a set that includes $b$ and extends to a neighborhood of $b$. The author then combines those two facts. The purpose of $x_0$ is to prove that the two sets where we know $f$ to be bounded actually cover all of $[a,b]$. $\endgroup$ – Chris Culter Aug 11 '13 at 20:19
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    $\begingroup$ @mauna To your second question, let me echo Daniel and Thomas: There is no guarantee that $b$ is in $A$! You cannot conclude, merely from looking at $A$, that $f$ is bounded on all of $[a,b]$. $\endgroup$ – Chris Culter Aug 11 '13 at 20:22

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