If $a \in \mathbb{I}$ , how is $$\overline{\mathbb{Z}+ a\mathbb{Z}}=\mathbb{R}$$
It says in my notebook that this set in dense in $\mathbb{R}.$ How do I prove this density? With say $\mathbb{Q}$ and $\mathbb{I}$ being dense in $\mathbb{R}$ I thought it was dense because every point in $\mathbb{R}$ was either in $\mathbb{Q}(\mathbb{I})$ or a limit point of $\mathbb{Q}\text{ -being } \mathbb{I}(\mathbb{I}\text{ -being }\mathbb{Q}).$
Now here say $a\to \frac{1}{2}\in \mathbb{I}$, take example $b=\frac{1}{4}\in \mathbb{R}$ which I don't see as being a limit point of $\mathbb{Z}+ a\mathbb{Z}...$
Key points: Your set is closed under internal addition and subtraction (that is, if $x$ and $y$ are in your set then so is $x + y$ and $x - y$. Further, your set is closed under multiplication by arbitrary integers (clearly).
Suppose that $0$ were a limit point of your set. That means that given an arbitrarily small $\epsilon > 0$ I can find integers $\{m,n\}$ such that $|m+na| < \epsilon$. But then look at the set of points $\{k(m+na) \;\; k \in \mathbb Z\}$. Clearly every real number $x$ is within $\epsilon$ of one of those. In this way I see that every real number must be a limit point, so your set is dense.
Note, too, that if there are any limit points at all, then $0$ is a limit point (because if your set draws arbitrarily near some $x$ then differences between those elements, which are also in your set, draw arbitrarily near $0$).
Finally, then, we just need to consider the case in which there are no limit points at all. But then I can find integers $\{m,n\}$ such that $m + na$ is the smallest positive element of your set. It follows that every element of your set is of the form $\{k(m+na)\}$ for some integer $k$ since, otherwise, that element would fall in the gap between two multiples of your minimal element and I could construct a smaller element. The fact that $a$ is in your set means that $n$ can't be $0$ (as $a$ is not an integer) But $1$ is in your set so $1 = km + kna$ for integers $r$ and $s$, whence $a = \dfrac {1-km}{kn}$ so a would be rational, a contradiction.