# Completeness of R^m

1. Jul 13, 2012

### mateomy

I'm reading through Pugh's Real Mathematical Analysis and one of the theorems he puts down states that "R^m is complete". I assume the superscript refers to it being metric as that's what he's talking about right before this theorem. Anyway, the proof seems -to me- circular. I was hoping someone could show me why it isn't or perhaps that I'm expecting too much.

It says:

let p be a Cauchy sequence in R^m. express p in component form p={p1,p2,p3,...pmn}. Because p is Cauchy, each component sequence is Cauchy. Completeness of R implies that the component sequences converge, and therefore the vector sequence converges.

I dunno, but for some reason, and I've read it over 20+ times, that seems like circular reasoning.

(Sorry for being to lazy to post in Latex format)

2. Jul 13, 2012

### Muphrid

I think $\mathbb R^m$ refers to an m-dimensional real vector space. If so then he's proving the m-dimensional space to be complete using the fact that the 1-dimensional real line is complete.

3. Jul 13, 2012

### mateomy

Ah, you pointing that out and me re-reading the proof makes it more clear. Thank you.