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- 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):

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

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

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?