Proving that a rational and irrational sequence converge to the same value

Question

In order to prove that the Dirichlet function is not continuous for all real numbers, x, I first want to prove that there is a rational and irrational sequence that converge to the same value, $a$.

Let $a$ $\in$ $\mathbb{R} $. For any $n \in \mathbb{N}$, $a$ - $1\over n$< $a$ + $1\over n$. By a Theorem from class (If $x$ and $y$ are arbitrary, distinct real numbers, then there exists a rational number between $x$ and $y$), there exists a rational, $a_n$ between $a$ - $1\over n$ and $a$ + $1\over n$, i.e. $a$ - $1\over n$ < $a_n$ < $a$ + $1\over n$. Thus |$a_n - a$| < $1\over n$ and $a_n$ converges to $a$.

I think this takes care of the rational sequence, but I am having trouble coming up with an irrational sequence that converges to $a$. Is there a similar approach to proving that an irrational sequence exists that converges to the same value? Any hints would be greatly appreciated.

Answer 1

Here's an example of this. Let $a$ be irrational. Then the sequence $a_n = a$ for all $a$ is an irrational sequence converging to $a$. Now take the decimal expansion for $a$ (every real number has one). A rational sequence converging to $a$ is given by the sequence of partial expansions, i.e $b_n$ is the first $n$ decimal places. So if $a$ is $\pi$, then $b_n$ is the sequence $3, 3.1, 3.14, 3.141, 3.1415, ...$ etc.