Finite dimensional normed vector spaces complete ?

[geoffrey159]

Homework Statement

Show that finite dimensional normed vector spaces are complete.

Homework Equations

$E$ is a finite dimensional vector space and $N$ a norm on $E$

The Attempt at a Solution

If $\{x_n\}_n$ is a Cauchy sequence in $(E,N)$, then it is bounded and each term of the Cauchy sequence belongs to a closed ball of $E$ centered in 0 with radius $r = \text{sup}_n\{N(x_n)\}$. This ball is compact in finite dimension, so $\{x_n\}_n$ has at least an adherence value in this ball, and a Cauchy sequence that has an adherence value converges to this value. Which proves $E$ is complete. Is it correct ?

 

[S.G. Janssens]

Is it correct ?

Very nearly, I would say.

so $\{x_n\}_n$ has at least an adherence value in this ball,

In my opinion, an adherent point is something that pertains to a set, not to a sequence. I would say: $(x_n)_n$ has a subsequence that converges to a point $x$ in the ball...

and a Cauchy sequence that has an adherence value converges to this value.

...and a Cauchy sequence that admits a subsequence converging to $x$ will converge to $x$ as well.

 

[geoffrey159]

OK I get it. Thanks !

 

[HallsofIvy]

A finite dimensional vector space over a complete field is complete. You don't say anything about the underlying field of the vector space. If you intend it to be the field of all real numbers, you should say that!

 

[S.G. Janssens]

A finite dimensional vector space over a complete field is complete. You don't say anything about the underlying field of the vector space. If you intend it to be the field of all real numbers, you should say that!

Sorry, but this I find rather unnecessary nitpicking, phrased in a rather unfriendly manner. The OP has clearly made an above average effort to be precise and accurate when he asked his question. Other posters could learn from this.

In standard functional analysis, when the field is not specified, it is always understood that the field must be $\mathbb{R}$ or $\mathbb{C}$. In both cases, the statement holds. (When the problem involves spectral theory (not the case here) it is of course always assumed that the field is $\mathbb{C}$.)

 

[micromass]

I agree, it's pretty obvious that the field should be $\mathbb{R}$ or $\mathbb{C}$. And this is not the fault of the OP, since he clearly said "normed vector space", which implies it. So the nitpicking is incorrect.