Cardinality of the set of all numbers in [0, 1] with decimal expansions ending in infinite sequence of 5's My idea is this: 
Let $T$ denote the set in question. 
Some typical elements of $T$ will be:
$T_1 = a_{11}a_{12}...a_{1n}555...$    ,   $n$ a natural number
$T_2 = a_{21}a_{22}...a_{2n}555...$          
and so on (I'm assuming there are infinitely many elements in T, but I'm actually not certain this is correct and  I wouldn't mind a proof either way). 
Define a bijection $f$ by $f:\Bbb N \to T$ by $f$($b_{11}b_{12}...b_{1n}$) $= 0.b_{11}b_{12}...b_{1n}555...$
Now I think this is kind of sketchy, especially how I "tack on" the infinite sequence of $5$'s in the definition of my bijection. Legal? Why/why not? 
 A: Your map is almost an injection. A map like this will have a hard time hitting all of $T$ though, because of the presence of leading $0$'s.
I'm suggesting that things in $T$ are all rational, because they all have repeating decimal representations. $$$$
In particular, since $0.555\ldots = \frac59$, you can show that $$0.d_1d_2\ldots d_{n-1}d_n5555\ldots = \dfrac{d_1d_2\ldots d_{n-1}d_n}{10^n} + \dfrac{5}{9\cdot 10^n} = \dfrac{9d_1d_2\ldots d_{n-1}d_n + 5}{9\cdot 10^n}\in \Bbb Q.$$
If you know that all countably infinite sets have the same size and can show your set $T$ is indeed infinite, you're done.

Modifying your map into a bijection is annoyingly tricky. As mentioned, naturals of the form $55\ldots55$ all get sent to $0.5555\ldots \in T$. Even then, the map
\begin{align*}
f : \Bbb N &\to T\\
d_1d_2\ldots d_{n-1}d_n &\mapsto 0.d_1d_2\ldots d_{n-1}d_n555\ldots
\end{align*}
will never hit anything with two or more leading $0$s (even a single leading $0$, if your $\Bbb N$ doesn't contain $0$). 
You could find a way to send integers of the form $55\ldots55$ to decimals of the form $0.00\ldots00555\ldots$. There are still things in $T$ with leading $0$s left out by this map; $0.000123555\ldots$ for example. It's injective now, but I'm having a hard time modifying it to be surjective.
I love a good bijection as much as the next guy, but I personally would file this one under the "too much trouble" category.
