An attempt at proving that $A=(0,1)$ is not compact on the real line with the usual topology. I am supposed to show that the open interval $(0,1)$ on the real line $\mathbb{R}$ (with the usual topology) is not compact. I know that one of the examples of an open cover without a finite subcover, in this case, would be $A=\left\{\left(\dfrac{1}{n},1\right)\bigg\rvert n\in\mathbb{N}\right\}$, thus proving that $A$ is not compact.
However, I am trying to prove the same using a different example of an open subcover, $$\mathscr{F}=\left\{\left(\dfrac{1}{2^n},\dfrac{3}{2^n}\right)\bigg\rvert n\in\mathbb{N}\right\}.$$ If it can be shown that the family of sets $\mathscr{F}$ forms an open cover for $(0,1)$, i.e., $\bigcup\limits_{n\in\mathbb{N}}G_n$, where $G_n=\left(\dfrac{1}{2^n},\dfrac{3}{2^n}\right)$, it is straightforward to check that for any $x=\dfrac{1}{2^n}$, where $n\in\mathbb{N}$, it only lies in one of these $G_n$'s. So, removing any one of these open sets from our open cover, in an attempt to find a finite subcover, the family of sets won't form a finite subcover of $(0,1)$. Thus, $(0,1)$ is not compact.
Where I am getting stuck is in actually formally proving that any $x\in(0,1)$ lies in one of these $G_n$'s. My intuition here is that by the use of Archimedean principle, for any $x\in(0,1)$, there exists some $k\in\mathbb{N}$ such that $x>\dfrac{1}{k}$. So, let us choose $k$, for a given $x$, such that $\dfrac{1}{k}<x<\dfrac{1}{k-1}$. Again, since this $k>1$, there exists some $n_0\in\mathbb{N}$ such that $2^{n_0-1}<k<2^{n_0}\implies \dfrac{1}{2^{n_0}}<\dfrac{1}{k}<\dfrac{1}{2^{n_0-1}}$. My claim is that $x\in \left(\dfrac{1}{2^{n_0}},\dfrac{3}{2^{n_0}}\right)$, and approach was to try and prove that $\left(\dfrac{1}{k},\dfrac{1}{k-1}\right)\subseteq \left(\dfrac{1}{2^{n_0}},\dfrac{3}{2^{n_0}}\right)$, but I am unable to proceed ahead from this point.
Any help would be highly appreciated, in either proving or disproving the claim that I make in the last sentence.
 A: I would note that $G_1=(1/2,3/2)$ which covers $(1/2,1)$, and then noting that
$$\frac{3}{2^{n+1}} \ge \frac{1}{2^n}  \iff \frac{3}{2} \ge 1  \mbox{, for all }n\in \mathbb{N}$$
which is true.
This gives us that two consecutive intervals have non-empty intersection so that
$$\bigcup_{n\le N}G_{n}=\left(\frac{1}{2^N},\frac{3}{2}\right) \mbox{, for all } N\in \mathbb{N}$$
Now if we have $x\in (0,1)$ this means $x >0$ and $\frac{1}{2^n}$ converges to $0$ so by definition if we get $\epsilon =x$, there exists $N_0$ such that $\forall n\ge N_0$, $\frac{1}{2^n}<x$.
So we have that $x\in \bigcup_{n\le N_0}G_{n}$ (finite union) $\Longrightarrow$ there exists an interval $G_k$ with $k\le N_0$.
Hope I made no mistakes.
A: Let $0<x<1$. Consider the interval $(\frac {\ln (\frac 1  x)} {\ln 2},\frac {\ln (\frac 3  x)} {\ln 2})$. The length of this interval is easily seen to be $\frac {\ln 3} {\ln 2}$. Since any interval of length greater than $1$ contains an integer there exist a positive integer $n$ in this interval. You can now verify that $x \in (\frac 1 {2^{n}},\frac 3 {2^{n}})$. Note that since $0 <x<1$ and $x \in (\frac 1 {2^{n}},\frac 3 {2^{n}})$ the integer $n$ is necessarily a positive integer.
A: You need to show those intervals "continuously overlap".  That is for any $k \in \mathbb N$ we have  $\frac 1{2^{k+1}} < \frac 1{2^k} < \frac 3{2^{k+1}} <  \frac 3{2^k}$.  That follows as $\frac 12 < 1 < \frac 32 < 3$.
This means $\cup (\frac 1{2^n},\frac 3{2^n}) = (\lim_{n\to \infty} \frac 1{2^n}, \frac 3{2^1}) = (0, \frac 32)$ and so any $x \in (0,1) \in \cup (\frac 1{2^n},\frac 3{2^n})$
However if you can't use limits in that manner, consider:
Let $x \in (0,1)$.  Let $K = \{n\in \mathbb N| \frac 1{2^n} < x\}$.  $K$ is non-empty. (Proof:  Let $n > \log_2 \frac 1x$, then $n \in K$.)
Now the well ordering principle of natural numbers says $K$, being a subset of $\mathbb{N}$, must have a least element $k = \min K; k \in K$ and $k-1 \not \in K$.  If $k= 1$ are done $\frac 12 < x < 1 < \frac 32$ so $x \in (\frac 12, \frac 32)$.  If $k > 1$ then $k\in K\implies \frac 1{2^k} < x$ and $k-1 \not \in K\implies x \le \frac 1{2^{k-1}} < \frac 3{2^k}$.  So $x \in (\frac 1{2^k}, \frac 3{2^k})$.
