1. ## Complete subset

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

I tried to show that any convergent sequence in $\displaystyle K$ must have its limit in $\displaystyle C^n$ which means $\displaystyle K$ is closed in a complete space, but I failed to do this.

2. ## Re: Complete subset

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

3. ## Re: Complete subset

Originally Posted by FernandoRevilla
Hint: $\displaystyle K$ 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 $\displaystyle C^n$, so it must closed.

4. ## Re: Complete subset

Presumably $\displaystyle C^n$ is $\displaystyle \mathbb{C}^n$. Note then that since $\displaystyle \mathbb{C}^n$ is itself, you have that a given subspace is complete if and only if it's closed. But, evidently $\displaystyle K$ is closed since the mapping $\displaystyle (x_1,\cdots,x_n)\mapsto x_1+\cdots+x_n$ is a continuous (linear functional) map $\displaystyle \mathbb{C}^n\to\mathbb{C}$, and $\displaystyle K$ is the preimage of the closed set $\displaystyle \{1\}\susbseteq\mathbb{C}$ under this map.

5. ## Re: Complete subset

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.