Can the following set exist? Say $a$ and $b$ are two numbers. Can the set $\{a,x_{1},x_{2},\dots b\}$ exist, where there are infinite $x_{i}$s?
Please note that one can't say $[0,1]$ is an example. This is because all the elements of this interval cannot be listed in the form above, as the interval is not denumerable. 
Thank you for your time. 
 A: Sure.  Consider for example $R = \left\{1-\frac1n\mid \text{$n$ is a positive integer}\right\} = \left\{0,\frac12,\frac23,\frac34,\ldots\right\}$ and $Z = \{1\}$.  Then $R\cup Z =\left\{0,\frac12,\frac23,\frac34,\ldots, 1\right\}$.
Such as set is said to have order type $\omega+1$; the possible order types of sets of rationals is an interesting question.
You then asked if this set was countable.  It is, because it is the union of a countable set and a finite set. Consider the following bijection with $\Bbb N$:  $$f(n) = \left\{\begin{aligned}
1,&\text{if $n=0$} \\
1-\frac1n,&\text{if $n>0$}
\end{aligned}\right.$$
This bijection does not preserve the order of the elements, since $0<6$ but $f(0) = 1\not\lt \frac56 = f(6)$.  No bijection of $R\cup Z$ with $\Bbb N$ preserves order, and that's exactly why its order type is not the same as that of $\Bbb N$, which has order type $\omega$.
A: My meager understanding goes that "exist" in mathematics usually means "to be a member of a set".  So, the question then means "can such a set S be a subset of another set R"?  Consider the powerset of rational numbers P(Q).  The set {0, [(1/2), (1/3), ...,] 1} is a subset of P(Q) given the natural numbers as infinite.  The part within brackets has general form (1/n) where n is a positive natural number.  So, yes such sets exist.  
If you're working with a formal axiomatic theory for sets where the existential quantifier need not in principle have an interpretation, then your question might require us looking at the axiom set to tell if this holds or not.
If you differentiate between a class and a set, and for any set X, P(X) is a class, this answer might not work. 
