# Thread: completion of set in metrix spaces

1. ## completion of set in metrix spaces

Hi, can one help me please;

For S=(0,1)

We know the Cauchy sequences Xn=1/n converges to zero which is not in the set
and Xn=1/n+1 converges to one which is also not in the set

so S=[0,1].

For S= [0,1] n {Rational numbers}

is it just the set of all Real numbers.

For S={1/k:k=1,2,3,...}

Many thanks

2. What does completion mean?
Is that what the rest of us know as closure?

3. Hi, yes it is closure. so that the metric can be complete.

4. a) & b) $~\overline{S}=[0,1]$

c) $~\overline{S}=S\cup \{0\}$

5. Originally Posted by Plato
a) & b) $~\overline{S}=[0,1]$

c) $~\overline{S}=S\cup \{0\}$
For B) if one took the set Xn= x/2+1/x then this in the set S= [0,1] n {Rational numbers} and the sequence converges to
square root 2 but that not within the set [0,1].

SO how can For B) b) $~\overline{S}=[0,1]$

6. Originally Posted by nerdo
For B) if one took the set Xn= x/2+1/x then this in the set S= [0,1] n {Rational numbers} and the sequence converges to
square root 2 but that not within the set [0,1].

SO how can For B) b) $~\overline{S}=[0,1]$
The fact that $\overline{\mathbb{Q}\cap[0,1]}=[0,1]$ follows since $\mathbb{Q}$ is dense in $\mathbb{R}$ and the fact that $\overline{\mathbb{Q}\cap [0,1]}\supseteq\overline{\mathbb{Q}}\cap\overline{[0,1]}=\mathbb{R}\cap[0,1]=[0,1]$. But, since $\mathbb{Q}\cap[0,1]\subseteq[0,1]$ we have that $\overline{\mathbb{Q}\cap[0,1]}\subseteq\overline{[0,1]}=[0,1]$ from where the conclusion follows.