Fdtxjr

For each $n=1,,2,,3,,dots;$ let
$C_n = displaystyleleft[frac1n,2+frac1nright]$. Prove that

• (i) $quaddisplaystylebigcup_n=1^infty C_n=(0,3]$ and


• (ii) $quaddisplaystylebigcap_n=1^infty C_n=[1,2]$
  

using Archimedean Property of Reals,

This is what I've done thus far for (i).
For all $𝑥in[1,3]$
are inside $𝐶_1$. If $𝑥in(0,1)$, then there exists an n such that $1/𝑛<𝑥<1$. Therefore, $𝑥in C_n$ $subsetbigcup_k=1^infty$ $C_k$=(0,1). Since also $(0,3]⊃𝐶_𝑛$ for all 𝑛, (i) follows.

$f(n) = frac1n$ 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 $n in N$,
$$
2 + epsilon < 2 + frac1n, forall n in N
$$

$$
epsilon < frac1n, forall n in N
$$

But as,
$$
lim_n to infty frac1n = 0
$$

Therefore there is some $n$ such that,
$$
frac1n < epsilon
$$

Again the assumption is false. So proved by contradiction.

Hint: The second part is similar. If $ 0le xle 2 $, can you show that $ xin C_n $ for all $ n $? If so, then you have $ left[1, 2right] subseteqdisplaystylebigcap_n=1^infty C_n $.

Likewise, if $ xin C_n $ for all $ n $, can you show that $ 0le x le 2 $? You would then have $ displaystylebigcap_n=1^infty C_nsubseteqleft[1, 2right] $.

First note that $forall ninBbb N, 1/nle 1wedge2+1/n>2impliesforall C_n,[1,2]subset C_n$. Hence, $[1,2]$ lies in their intersection. Next you need to show that $[1,2]$ is the intersection, i.e. for $x>2vee x<1,exists C_k|xnotin C_k$. If $x<1$, then $xnotin C_1$; can you use the Archimidean property for the other?

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 $xgeqfrac1n$ for all $n$, in particular $xgeq1$. Also, $xleq2+frac1n$ for all $n$, which implies $xleq2$. So $xin[1,2]$.

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 $x>1$.