Cardinality of the set of multiple-representation decimals 
The set of real numbers in the open interval $(0,1)$ which have more
  than one decimal expansion is
(A) Empty
(B)non-empty but finite
(C)Countably infinite
(D)uncountable

I know that, $$\frac{1}{10}=.10000...=.0999999...$$
              $$\frac{1}{10}=.010000...=.099999...$$
              $$ \frac{1}{100}=.0010000...=.0099999...$$
$$...$$
$$\frac{1}{10^n}$$ has also the two binary representation.
I think answer is (C). Am I correct? How do I prove it rigorously. Please help me.
When I was doing the above question, this question came in to mind.

I have question that The set of real numbers in the open interval
  $(0,1)$ which have more than one binary expansion is
(A) Empty
(B)non-empty but finite
(C)Countably infinite
(D)uncountable

I think the elements of the form $\frac{1}{2^n}$ has two binary expansion in $(0,1)$ . but I couldn't give the rigorous proof. Please help me.
 A: In any base $b$, using $h$ to denote the highest possible digit, all the numbers in $(0,1)$ that have more than one representation have the form
$$0.xxx\dots p000\ldots=0.xxx\dots qhhh\dots$$
where $p\ne0$ and $q=p-1$. In other words, they have a terminating expansion.
Thus the set of all such numbers is countably infinite by the following bijection between the terminating expansions and the positive integers:
$$(0.x_1x_2\dots x_k)_b\to(x_k\dots x_2x_1)_b$$
(C) is correct for both questions.
A: 
A number has more than one decimal expansion if and only if it has terminating decimal expansion.

To see this, if it does terminate, then if it goes like $\overline{a_1\cdots a_n . b_1\cdots b_n}$ then an alternating decimal expansion is also $\overline{a_1\cdots a_n.b_1\cdots b_{n-1} 999...}$
On the other hand, any non-terminating decimal expansion is determined uniquely, for example by repeated long division, which does not terminate and therefore does not give us room to introduce the chain of $9$s, as done in the terminating case.
Hence, the set of such real numbers actually is countable.
A similar argument can be given for the binary case, with the same conclusion.
