Closed Subsets

• Mar 1st 2012, 11:20 AM
bugatti79
Closed Subsets
Folks,

Are the following subsets closed?

1) $S=(0,1]; in X=R$

2) $S=\{x=(x_1,x_2) \in R^2 : x_1^2+x_2^2 \le 1 \} in X=R$

For 1) $S=(0,1]; in X=R$ is not closed

for if $S_n$ is a convergent sequence with $0 < S_n \le 1 \forall n \in N$

$S-N \le 1 \forall n \implies Lim S_n \le Lim 1 =1 as n \rightarrow \infty$

I dont know if this is right or how to continue.....?
• Mar 1st 2012, 12:46 PM
Plato
Re: Closed Subsets
Quote:

Originally Posted by bugatti79
Are the following subsets closed?
1) $S=(0,1]; in X=R$

For each $n\in\mathbb{Z}^+$ define $x_n=\frac{1}{n}$

If it true that $x_n\in S~?$ that is the limit of $(x_n)~?$
• Mar 1st 2012, 01:52 PM
bugatti79
Re: Closed Subsets
Quote:

Originally Posted by Plato
For each $n\in\mathbb{Z}^+$ define $x_n=\frac{1}{n}$

If it true that $x_n\in S~?$ that is the limit of $(x_n)~?$

ok

Let $\displaystyle x_n=1/n$ be a sequence $\in S \forall n \in Z^+$

ie $\displaystyle 0 < 1/n \le 1 \forall n \in Z^+$ but the $lim_{ n \rightarrow \infty } 1/n= 0 \implies x_n \notin S \implies S=(0,1]$ is not closed...

Correct?
• Mar 1st 2012, 02:16 PM
Plato
Re: Closed Subsets
Quote:

Originally Posted by bugatti79
Let $\displaystyle x_n=1/n$ be a sequence $\in S \forall n \in Z^+$
ie $\displaystyle 0 < 1/n \le 1 \forall n \in Z^+$ but the $lim_{ n \rightarrow \infty } 1/n= 0 \implies x_n \notin S \implies S=(0,1]$ is not closed...Correct?

NO indeed.
The whole point is that $\forall n,~x_n\in S$ but $\displaystyle\lim_{n\to\infty}x_n\notin S$.
• Mar 1st 2012, 02:26 PM
bugatti79
Re: Closed Subsets
Quote:

Originally Posted by Plato
NO indeed.
The whole point is that $\forall n,~x_n\in S$ but $\displaystyle\lim_{n\to\infty}x_n\notin S$.

Does this still mean the set is not closed? Thanks
• Mar 1st 2012, 02:32 PM
Plato
Re: Closed Subsets
Quote:

Originally Posted by bugatti79
Does this still mean the set is not closed? Thanks

You don't understand because you don't know what it means for a set to be closed. Do you? How do you expect to work any problem if you don't understand the underlying definitions?
• Mar 2nd 2012, 05:05 AM
bugatti79
Re: Closed Subsets
Quote:

Originally Posted by Plato
You don't understand because you don't know what it means for a set to be closed. Do you? How do you expect to work any problem if you don't understand the underlying definitions?

Definition: If S is a subset of an n.l.s, we say S is a closed set if for every

convergent series (S_n) of points $S_n \in S$, the

$\displaystyle \lim_{n\to\infty}S_n \in S$

in our case $\implies \lim_{n\to \infty} S_n \notin S$

Therefore the set is not closed...?
• Mar 9th 2012, 09:59 AM
bugatti79
Re: Closed Subsets
Quote:

Originally Posted by bugatti79
Definition: If S is a subset of an n.l.s, we say S is a closed set if for every

convergent series (S_n) of points $S_n \in S$, the

$\displaystyle \lim_{n\to\infty}S_n \in S$

in our case $\implies \lim_{n\to \infty} S_n \notin S$

Therefore the set is not closed...?

Yes, the set is not closed.

Quote:

Originally Posted by bugatti79
Folks,

Are the following subsets closed?

2) $S=\{x=(x_1,x_2) \in R^2 : x_1^2+x_2^2 \le 1 \} in X=R$

We need to show $||x|| \le 1$

By Triangle inequality we have $||x|| \le ||x-x_n||+||x_n|| \forall n \in N$
but we know $\lim_{n\to\infty}x_n=x$

so $\forall \epsilon > 0, \exists n_0 \in N s.t ||x-x_n|| < \epsilon$

In particular taking n=n_0

$||x|| \le ||x-x_n_0||+||x_n_0|| < \epsilon +1$

I dont understand the above line, why is there a '+1' on the RHS?

Thanks