On the definition of radius of convergence; a small supremum technicality

[schniefen]

Homework Statement

I have a question about a definition of the radius of convergence, in particular about a small detail regarding a property of the supremum.

Relevant Equations

If $M=\sup (A)$, then for every $M'<M$, there exists an $x\in A$ such that $x>M'$.

I am reading the following passage in these lecture notes (chapter 10, in the proof of theorem 10.3) on power series (and have seen similar statements in other texts):

Let $$R=\sup \left\{\left|x\right|\ge 0:\sum a_nx^n \text{ converges}\right\}.$$ If $R=0$, then the series converges only for $x=0$. If $R>0$, then the series converges absolutely for every $x\in \mathbb R$ with $|x|<R$, since it converges for some $x_0\in\mathbb R$ with $|x|<|x_0|<R$.

I'm confused about $|x_0|<R$.

If $M=\sup (A)$, then for every $M'<M$, there exists an $x\in A$ such that $x>M'$. Then, using the definition of an upper bound (in e.g. Spivak's Calculus), it follows that $M'<x\color{red}{\leq} M$. Suppose $A=\{1,2,3\}$, then $2.8<3$, but stating $2.8<3<3$ would be incorrect.

Why is it correct then to state $|x_0|<R$ in this case?

 

[wrobel]

You are right, it should be $|x|<|x_0|\le R$.

 

[wrobel]

Your definition of sup is not complete:
$M=\sup A\Longleftrightarrow$
1) $x\in A\Rightarrow x\le M$
2) for any $M'<M$ it follows that there exists $x\in A$ such that $x>M'$

 

[schniefen]

I thought maybe the statement $|x|<|x_0|< R$ is not incorrect after all if the set $$\left\{\left|x\right|\ge 0:\sum a_nx^n \text{ converges}\right\}$$ is an interval, but this is not something we know a priori. Besides I am really not sure how to show the set is an interval -- probably requires a definition of what an interval is. Anyway, this seems far-fetched and I lean towards a typo, though I have seen this stated in other texts as well (see for instance A First Course in Mathematical Analysis by Brannan, chapter 8, section 8.3).

 

[FactChecker]

Why is it correct then to state $|x_0|<R$ in this case?

You can go directly from the statement for $|x_0| \le R$ to $\exists x_1 : |x| \lt |x_1| \lt |x_0| \le R$ to $|x| \lt |x_1| \lt R$.

 

[wrobel]

You can go directly from the statement for $|x_0| \le R$ to $\exists x_1 : |x| \lt |x_1| \lt |x_0| \le R$ to $|x| \lt |x_1| \lt R$.

Not directly, but by proving that the series is convergent at $x_1$. That is not a problem, just a redundant step.

upd

Besides I am really not sure how to show the set is an interval

Theorem. If the series is convergent at $x'$ then it is convergent in the interval $\{|x|<|x'|\}$. (It is an interval in $\mathbb{R}$ and a circle if we consider $\mathbb{C}$)

This theorem is proved in any textbook where power series are discussed. Find it.

 

[schniefen]

Ok, I think this clarified it. Thanks a lot!

To summarize; by the definition of the supremum, we have $|x| \lt |x_0| \leq R$. Now, $\sum a_nx^n$ converges for $x=x_0$ and so for all $x\in\mathbb R$ with $|x|<|x_0|$. As @FactChecker pointed out, we can find an $|x_1|$ such that $|x|<|x_1|<|x_0|$ and we are done.

 

[FactChecker]

Ok, I think this clarified it. Thanks a lot!

To summarize; by the definition of the supremum, we have $|x| \lt |x_0| \leq R$. Now, $\sum a_nx^n$ converges for $x=x_0$ and so for all $x\in\mathbb R$ with $|x|<|x_0|$. As @FactChecker pointed out, we can find an $|x_1|$ such that $|x|<|x_1|<|x_0|$ and we are done.

Yes. But as @wrobel pointed out above, there is no point in doing this. Using $\le$ as you originally suggested is more directly derived from the definitions.