Question about open interval with a finite decimal expansion 
Let $Y$ denote the set of numbers in $(0, 1)$ with a decimal expansion that contains only $0$s and $1$s, and only finitely many $0$s. Decide if you think $Y$ is countably infinite or uncountable – I promise, $Y$ is infinite.

I understand that the open interval $(0,1)$ is uncountable and have the proof for that.


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*I'm confused what the original question is asking in terms of the "decimal expansion that contains only $0$s and $1$s". Does this mean numbers such that $0.x_1x_2x_3\dotso x_n$ such that $n$ can only be a $0$ or a $1$? So for example, $0.11$ and $0.101$ would be okay but not $0.21$ (as an example)?

*Does have this decimal expansion change the fact that the open interval is uncountable? Because if it were to be countable, then we should be able to list all its elements with no double listing. I'm just confused as to whether or not that is the case.

 A: If there are only finitely many zeros in each expansion, then each number in the list is a rational number. Hence the list will be countably infinite.
A: *

*I think the question is stating that each of the digits of the (potentially infinite) decimal expansion can only be a $0$ or a $1$. It states that there must be only finitely many $0$s, so as Arturo Magidin points out, both of your examples are actually not valid. Something more like $0.10001 \dots$, $0.01111 \dots$ or $0.010101111 \dots$ would work.

*You are being asked about a subset of $(0,1)$, so since this interval is uncountable, the subset can either be uncountable or countable. For example, $\mathbb{R}$ is countable. $(0,1) \in \mathbb{R}$ is uncountable, but $\mathbb{N} \in \mathbb{R}$ is countable. You are correct that a set is countable if we can ‘list its elements’. Formally a set $A$ is countable if and only if there exists a bijection (one-to-one function) $f: \mathbb{N} \to A$, or equivalently, $f: A \to \mathbb{N}$ between the natural numbers and the set. These two definitions are equivalent, since a listing can be thought of as just a mapping from the naturals to the elements of the set.
As for the actual question: since we can only have a finite number of $0$s in the expansion, we are forced to have an infinite trail of $1$s at the end. If we didn’t have this infinite trail then our expansion would terminate and this would be the same as having an infinite trail of $0$s, which is clearly not allowed.
If we now assume that every expansion in our set must end in an infinite sequence of $1$s, then we can just consider the decimal expansions we can ‘stick to the front’ of this trail of $1$s and try to list these. So now we are considering the set of all terminating decimal expansions in $(0,1)$ containing only $0$s and $1$s. I will conjecture that this set is countable based on the fact that every finite binary expansion is rational, and the rationals are countable. Try for yourself and see if you can list its elements in a systematic way, or argue that it is equivalent to some uncountable set.
A: Let $(x_n)$ be a sequence consisting of  $0$s,and $1$s, with finitely many $0$s.
Let $U_m:=${ $(x_n)| x_n =1$ for $n \ge m$}, $m \ge 0$.
$U_m$ is finite.
$\bigcup_{m} U_m$ as a  countable union of finite sets is countable.
