Checking and Proving the following by the Archimedean Property of Reals. First part I believe is done. Unsure about second part. 
For each $n=1,\,2,\,3,\,\dots\;$ let
  $C_n = \displaystyle{\left[\frac{1}{n},2+\frac{1}{n}\right]}$.  Prove that
  
  
*
  
*(i) $\quad\displaystyle{\bigcup_{n=1}^\infty C_n}=(0,3]$ and
  
*(ii) $\quad\displaystyle{\bigcap_{n=1}^\infty C_n}=[1,2]$
using Archimedean Property of Reals,

This is what I've done thus far for (i). 
For all $\in[1,3]$
are inside $_1$. If $\in(0,1)$, then there exists an n such that $1/<<1$. Therefore, $\in C_n$ $\subset\bigcup_{k=1}^\infty$ $C_k$=(0,1). Since also $(0,3]⊃_$ for all , (i) follows.  
 A: $f(n) = \frac{1}{n}$ is monotonically decreasing for $n \in N$.
Therefore the largest that $1/n$ can be is 1 and the smallest it can be is 0 (tends to 0). Therefore, the smallest that 2+$1/n$ can be is 2. This should prove (ii)
Assume there exists a point outside intersection from $(0,1)$ then clearly it will not be in $C_1$, therefore the assumption is wrong. Assume that a point exists in the intersection from $(2, \infty)$. Say it is $2+\epsilon$ where $\epsilon \in (0, \infty)$. Now for the assumption to hold true this point must lie in interval $C_n$ for all $n \in N$,
$$
2 + \epsilon < 2 + \frac{1}{n}, \forall n \in N
$$ 
$$
\epsilon < \frac{1}{n}, \forall n \in N
$$
But as,
$$
\lim_{n \to \infty} \frac{1}{n} = 0 
$$
Therefore there is some $n$ such that,
$$
\frac{1}{n} < \epsilon
$$ 
Again the assumption is false. So proved by contradiction.
A: Hint: The second part is similar.  If $\ 0\le x\le 2\ $, can you show that $\ x\in C_n\ $ for all $\ n\ $?  If so, then you have $\ \left[1, 2\right]\ \subseteq\displaystyle{\bigcap_{n=1}^\infty C_n}\ $.
Likewise, if $\ x\in C_n\ $ for all $\ n\ $, can you show that $\ 0\le x\ \le 2\ $?  You would then have $\ \displaystyle{\bigcap_{n=1}^\infty C_n}\subseteq\left[1, 2\right]\  $.
A: First note that $\forall n\in\Bbb N, 1/n\le 1\wedge2+1/n>2\implies\forall C_n,[1,2]\subset C_n$. Hence, $[1,2]$ lies in their intersection. Next you need to show that $[1,2]$ is the intersection, i.e. for $x>2\vee x<1,\exists C_k|x\notin C_k$. If $x<1$, then $x\notin C_1$; can you use the Archimidean property for the other?
A: Clearly $[1,2]\subseteq C_n$ for all $n$, hence $\supseteq$ holds in (ii).
Now we prove the reverse inclusion. Suppose $x$ in the intersection. Then $x\geq\frac1n$ for all $n$, in particular $x\geq1$. Also, $x\leq2+\frac1n$ for all $n$, which implies $x\leq2$. So $x\in[1,2]$.
Note that, if we took $[\frac1n,2+\frac1n)$, nothing would change, because having $x<2+\frac1n$ for all $n$ would still allow 2. Taking $(\frac1n,2+\frac2n]$ would, instead, turn the intersection to $(1,2]$, because we would have $x>1$.
