Can the ratio of the two smallest element of an iid sample converge to 1 if the support of $X$ is positive? We have: $\mathbb P(X \leq 0)=0$ and $\mathbb P (X \leq a)>0$ for any $a>0$.
 A: The answer is no. Here is a counterexample. Lets $X_1, X_2, \dots, X_n$ random samples from an exponential distribution with parameter $\lambda$.the order statistics $X_{(1)}, X_{(2)}$ each have distribution 
\begin{align*} 
X_{(1)} &= \frac{Z_1}{\lambda n} \\ 
X_{(2)} &= \frac{1}{\lambda}\left(\frac{Z_1}{n}+\frac{Z_2}{n-1}\right)
\end{align*}
where the $Z_1, Z_2$ are iid standard exponential random variables (see Wikipedia).
Therefore 
$$
\frac{X_{(2)}}{X_{(1)}} = \frac{Z_1}{Z_1+Z_2n/(n-1)} \rightarrow \frac{Z_1}{Z_1+Z_2} \neq0
$$
A: You're right for most cases, however this depends on the distribution as well.
If $X_1, \dots, X_n \sim \Bbb{P}_\theta$ where $Var_\theta(X_i) = 0$, this doesn't hold true, because
$$\forall i,j \in \{1,\dots,n\}\quad\Bbb{P}(X_i = X_j) = 1 \rightarrow \Bbb{P}\left(\lim_{n \to \infty}\frac{X_i}{X_j}=1\right)=1$$
But for most cases, where $Var_\theta(X_i) \ne 0$, you can prove this indirectly:
Let's assume that
$$\exists \lim_{n \to \infty}\frac{X^{(2)}}{X^{(1)}} := A \in \Bbb{R}^+_0 \cup \{+\infty\}$$
We can assume that $A \ge 0$, because $X^{(2)} \ge X^{(1)}$.
If 
$$\lim_{n \to \infty}\frac{X^{(2)}}{X^{(1)}} = A \stackrel{\text{then }}{\longrightarrow}\lim_{n \to \infty}\Bbb{E}\left(\frac{X^{(3)}}{X^{(2)}}\right) = A$$
Since $\forall \varepsilon, N_1 \in \Bbb{R}^+ \exists, N_2 \in \Bbb{N}^+$ such that from a sample of $X_1,\dots,X_{N_1}$
$$A - \varepsilon < \frac{X^{(2)}}{X^{(1)}} < A + \varepsilon,$$
We can assure that from a sample of $X_1,\dots,X_{N_2}$, where $N_2$ is sufficiently large:
$$A - \varepsilon < \frac{X^{(3)}}{X^{(2)}} < A + \varepsilon$$
Since as $n \to \infty$,
$$\lim_{n \to \infty}\left(\frac{X^{(3)}}{X^{(2)}} - \frac{X^{(2)}}{X^{(1)}}\right) = 0$$
Now we'll have to prove $2$ cases separately:
Case $1$: $A = 1$
From our previous argument it follows that
$$\lim_{n \to \infty}\frac{X^{(2)}}{X^{(1)}} = \lim_{n \to \infty}\frac{X^{(3)}}{X^{(2)}} = \dots = \lim_{n \to \infty}\frac{X^{(n)}}{X^{(n-1)}} = 1$$
$$\Rightarrow \lim_{n \to \infty}\frac{X^{(n)}}{X^{(1)}} = 1$$
$$\Rightarrow \forall i \in \{1,\dots, n\} \quad \sup {Ran}_{\theta} X_i = \inf {Ran}_{\theta} X_i$$
$$\Rightarrow \forall i \in \{1,\dots, n\}\quad Var(X_i) = 0$$
Which is a contradiction, becuase in this case we assumed that $Var(X_i) \ne 0$.
Case $2$: $A \in \Bbb{R}^+_0 \backslash \{1\}$
Again, we get
$$\lim_{n \to \infty}\frac{X^{(2)}}{X^{(1)}} = \lim_{n \to \infty}\frac{X^{(3)}}{X^{(2)}} = \dots = \lim_{n \to \infty}\frac{X^{(n)}}{X^{(n-1)}} = A$$
Which is this case means:
$$\lim_{n \to \infty}\frac{X^{(2)}}{X^{(1)}} = \lim_{n \to \infty}\frac{1}{A}\frac{X^{(3)}}{X^{(1)}} = \dots =\lim_{n \to \infty}\frac{1}{A^n}\frac{X^{(n)}}{X^{(1)}}$$
$$\Rightarrow \lim_{n \to \infty}\frac{X^{(n)}}{X^{(1)}} = \lim_{n \to \infty}A^n\frac{X^{(2)}}{X^{(1)}} = \\ = 0, +\infty \text{ or does not exist (depending on the value of } A)$$
It can't be $0$, since then $$\Bbb{P}(X_n = 0) = 1 \rightarrow \forall i \in \{1\dots n\}\quad \Bbb{P}(X_i = 0) = 1 \rightarrow Var(X_i) = 0$$
Which is a contradiction again.
It can' be $+ \infty$ either, because that can only occur if
$X_n \ne 0$ and $X_{n-1} = 0$, but as we've already discussed,
$$\lim_{n \to \infty}\frac{X^{(n)}}{X^{(n-1)}} = \lim_{n \to \infty}\frac{X^{(n-1)}}{X^{(n-2)}} (= + \infty)$$
Which also means that $X_{n-1} \ne 0$ and $X_{n-2} = 0$, and we get a contradiction from $X_{n-1} = 0$ and $X_{n-1} \ne 0$.
The only option in this case is for the limit not to exist, which isn't technically a contradiction, but it further proves our point.
Result:
Both cases lead to contradiction, which means that the given limit does not exist.
