Complete subset

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• Dec 10th 2011, 08:54 PM
jackie
Complete subset
Could someone give me a hand on this problem? I want to show that the set $K=\{x=(x_n): \sum x_n=1\}$ in complete in $C^n$

I tried to show that any convergent sequence in $K$ must have its limit in $C^n$ which means $K$ is closed in a complete space, but I failed to do this.
• Dec 10th 2011, 09:20 PM
FernandoRevilla
Re: Complete subset
Quote:

Originally Posted by jackie
Could someone give me a hand on this problem? I want to show that the set $K=\{x=(x_n): \sum x_n=1\}$ in complete in $C^n$

Hint: $K$ is compact and so sequentially compact.
• Dec 11th 2011, 05:26 PM
jackie
Re: Complete subset
Quote:

Originally Posted by FernandoRevilla
Hint: $K$ is compact and so sequentially compact.

Thanks a lot for your help, Fernando. However, I don't really see the connection between your hint and this problem yet. Also, I don't think I learn that sequentially compact implies complete. Or you want me show that K is compact in $C^n$, so it must closed.
• Dec 11th 2011, 08:58 PM
Drexel28
Re: Complete subset
Presumably $C^n$ is $\mathbb{C}^n$. Note then that since $\mathbb{C}^n$ is itself, you have that a given subspace is complete if and only if it's closed. But, evidently $K$ is closed since the mapping $(x_1,\cdots,x_n)\mapsto x_1+\cdots+x_n$ is a continuous (linear functional) map $\mathbb{C}^n\to\mathbb{C}$, and $K$ is the preimage of the closed set $\{1\}\susbseteq\mathbb{C}$ under this map.
• Dec 11th 2011, 10:08 PM
FernandoRevilla
Re: Complete subset
Quote:

Originally Posted by jackie
Also, I don't think I learn that sequentially compact implies complete.

I immediately thought about a general result: Every compact metric space is complete. Drexel28's proposal is better.