Is there a random variable $X$ with positive support such that the ratio of the two smallest realizations of an iid sample goes to one? Imagine I have given a random variable $X$ with supp$(X)=(0,\infty)$ and $\mathbb P(X \in (0,a))>0$ for any fixed $a>0$ 
Now  given an iid sample $X_1,...,X_n$ - is it possible that
$$X^{(2)}/X^{(1)}\xrightarrow{\mathbb P}1$$ for $n \to \infty$, where $X^{(i)}$ describes the $i$-th smallest element?
 A: Yes, there are such distributions where the ratio of the second-smallest to smallest values approaches unity in probability as the sample size grows large.  They have to behave "essentially" like distributions with strictly positive support in the sense of approaching zero probability very, very rapidly at the origin.  See the first figure below for an illustration.
(I rely here on the correct intuition that when the distribution has a strictly positive minimum, eventually the two smallest values in large samples will both be  close to that minimum with high probability, whence their ratio approaches unity.  This intuition won't work when the minimum is zero.)

For convenience, let's work with independent and identically distributed random variables $X_i$ with continuous distributions.  This means they have a common density $f$ and $F(x)=F(0)+\int_0^x f(t)\mathrm{d}t$ is their common distribution function.
The question concerns the smallest two among the first $n$ of these variables, $X\lt Y,$ where $n$ will grow arbitrarily large.  The joint distribution of the two smallest values among them has density
$$f_{n;2}(x,y) = n(n-1)f(x)f(y)\left(1-F(y)\right)^{n-2}$$
for all $0\le x\le y.$  Introducing a variable $u$ defined by
$$x = uy$$
to represent the ratio $x/y\le 1,$ changing variables from $(x,y)$ to $(u,y),$ integrating from $u=0$ to $u=r$ (which can be expressed in terms of $F$), and integrating over all possible values of $y$  will give us the distribution function of the ratio $U=X/Y$ as

$$\Pr(U \le r) = n(n-1)\int_0^\infty f(y)F(ry)(1-F(y))^{n-2}\,\mathrm{d}y$$ for $0 \le r \le 1.$

This expression will be the object of our analysis.

In some cases the distribution of $U$ is easy to evaluate.  Although this next section is a diversion, it reveals a thought process that leads to an answer.
Take, for instance, $$F_p(y) = y^p$$ for $0\le y\le 1$ where $p\gt 0.$  I obtain an answer that does not vary with $n$ at all: for possible ratios $0\le r \le 1,$
$$\Pr(U \le r) = r^p.\tag{1}$$
This indicates that any distribution $F$ that behaves like $F(y)\approx y^p$ near the origin will yield something like this power distribution for the ratio; in particular, it will not converge to $1$ in probability. 

As $p$ grows large, $(1)$ does converge to the constant value $1$ in probability.  In other words, although we have not discovered any distributions where the ratio $U$ approaches $1$, we do have a family of distributions where this ratio can be made as close to $1$ as we might like by choosing a suitable member of the family (that is, by picking the power $p$ to be sufficiently large).  We might therefore consider distributions where, at the origin, $F$ is flatter than any polynomial.  The classic, and one of the simplest, such functions is
$$F(x) = \exp\left(1 - \frac{1}{x^2}\right)$$
for $0 \le x \le 1.$
Obviously its support extends down to $0,$ because the exponential is never zero.  $F$ is infinitely differentiable at $0$ but all derivatives are zero there.

In this case the integral for $\Pr(U\le r)$ still can be evaluated.  It is simpler to express the result in terms of $s \ge 1,$ where $$1/s^2 = r,$$ as
$$\Pr(U \le r) = \Pr(U \le 1/s^2) = \frac{e^{1-s}n!}{(1+s)^{(n-1)}}= s\,e^{1-s}\frac{1^{(n)}}{s^{(n)}}\tag{2}$$
where
$$s^{(n)} = s(1+s)(2+s)\cdots(n-1+s)$$
is the Pochhammer function.  Here are plots of this probability (as a function of $s$) for $n=2, 2^2, 2^{2^2}, 2^{2^3},$ and $2^{2^4}.$  The graphs drop towards a level of zero as $n$ increases:

It is easy to show that for all $z = s-1 \gt 0,$
$$\frac{1^{(n)}}{s^{(n)}} = \frac{1^{(n)}}{(1+z)^{(n)}} = \frac{1}{1+z}\frac{2}{2+z}\frac{3}{3+z}\cdots\frac{n}{n+z} \to 0$$
as $n$ grows large.  (Examine the MacLaurin series of its logarithm.)  Thus, for all $s=1+z \gt 1,$ $(2)$ goes to $0,$ demonstrating the ratio $U$ approaches the constant $1$ in probability.
