Wikipedia defines convergence in probability as
A sequence ${X_n}$ of random variables converges in probability towards the random variable $X$ if for all ε > 0 $$\lim_{n\rightarrow\infty} P(|X_n-X|>\varepsilon) = 0$$
I wonder why the limit $X$ is a random variable. I think $X$ as a r.v. can only be a constant, and defining convergence to a number $a$ instead of a r.v. $X$
$$\lim_{n\rightarrow\infty} P(|X_n-a|>\varepsilon) = 0$$
makes the definition more clear.
That the sequence of rv's $(X_n)$ can converge to another rv $X$ is clear from the example$$X_n=X+\upsilon_n/n\qquad \upsilon_n\stackrel{\text{i.i.d.}}{\sim}\cal{N}(0,1)$$ where $X$ is an arbitrary random variable. Indeed, then$$|X_n-X|=|\upsilon_n|/n$$which converges to zero in probability:$$\mathbb{P}(|\upsilon_n|/n>\epsilon)=2(1-\Phi(n\epsilon))\stackrel{n\to\infty}{\longrightarrow}0$$ Now, if the $X_n$ are independent, then they cannot converge in probability to a random variable $X$ unless $X$ is a.s. constant.
Convergence in probability to a constant is a special case of the more general result of convergence in probability to a random variable, so it is somewhat natural to allow for the more general case. It is also worth bearing in mind that once you allow convergence to a constant, or even just convergence to zero, that is sufficient to get the essence of convergence to a random variable, even if you don't name it that. To see this, suppose we let $R_n = X_n - X$ denote the series of residuals that occurs from subtracting $X$ from each of the values $X_n$. From the definition of convergence in probability, we have the following equivalence:
$$X_n \overset{p}{\rightarrow} X \quad \quad \iff \quad \quad R_n \overset{p}{\rightarrow} 0.$$
Thus, once you define convergence in probability to zero, this also gives a corresponding condition for convergence to a random variable. It then makes sense to refer to that condition as entailing convergence to the random variable.
