Topology of multiplicative subgroups of real line without zero, is this subgroup topologically closed or not?

Question

I've been thinking about a problem recently and stumbled upon this question that I couldn't seem to answer easily. Let's consider some finite set of primes $p_1, \dots, p_n$ and their multiplicative subgroup of $(\mathbb{R} \setminus {0}, \cdot) $ given by numbers of the form $ p_1^{a_1} \cdot p_2^{a_2} \cdots p_n^{a_n} $. This is obviously a subgroup of $(\mathbb{Q} \setminus {0}, \cdot) $. But can you help me figure out whether it is topologically closed or not? If we just take one prime number it is obviously closed, as we excluded 0. But for 2 primes already I find it quite difficult to tell. I tried to look at sequences like $a_0 = 1$ where $a_n = 2a_{n-1}$ if $a_{n-1}<5$ and $a_n=\frac{a_{n-1}}{5}$ if $a_{n-1} \geq 5$. This sequence is contained within the group I defined (for $p_1 = 2$ and $p_2 = 5$ )and it also is bounded in $\mathbb{R}$ thus it has a convergent subsequence. Intuitively I feel like the limit of a subsequence of this is not in the subgroup I constructed cause the number of decimal places of this sequence increases with every number in the sequence but I might totally wrong. This was just my idea of trying to prove whether this subgroup is closed or not. Can any of you help me?

Answer 1

Good question!

A useful fact to know is that $(\Bbb R_{> 0}, \cdot)$ and $(\Bbb R, +)$ are isomorphic as groups, and as topological spaces, via the map $\log$. So we can "transform" the problem to $(\Bbb R, +)$. There, in the case of two distinct primes $p$ and $q$, the question is "is the subgroup of $(\Bbb R, +)$ generated by $\log p$ and $\log q$ closed?" We can scale everything down by $\log p$, so then the question is "is the subgroup of $(\Bbb R, +)$ generated by $1$ and $\frac{\log q}{\log p} = \log_p q$ closed?" (This answers if it's closed in the subspace topology in $\Bbb R \setminus 0$, since $\Bbb R_{>0}$ is closed in this space. In fact it will be fairly clear from the discussion that it's also not closed in $\Bbb R$.)

Finitely generated subgroups of $(\Bbb R, +)$ are fairly well understood. In particular, if $\alpha$ is irrational then the subgroup generated by $1$ and $\alpha$ is dense in $\Bbb R$. This is proved here, but hopefully it sounds believable!

In this case, $\log_p q$ is indeed irrational. This follows from the fundamental theorem of arithmetic, as in this question.

So the generated subgroup is dense. If it was closed, it would be all of $\Bbb R$. There are many ways to show this is impossible. For example $\Bbb R$ is not a finitely generated group, eg because it is a divisible group. (see here). Or, for example, back in $(\Bbb R_{>0}, \cdot)$, if you take any prime not in the list $p_1, \dotsc, p_n$, then clearly that prime is not in the generated subgroup, by the fundamental theorem of arithmetic. Or, as you observed, all elements are rational, so it's not all of $\Bbb R_{>0}$.

Therefore, this subgroup is not closed for any finite $n$. In fact, it's still not closed if you take all the primes. For example this generated subgroup is still countable, so it can't be all of $\Bbb R$. In this case the generated subgroup in $(\Bbb R_{>0}, \cdot)$ is just $(\Bbb Q_{>0}, \cdot)$, so arguing that it doesn't contain an irrational also works.

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In general any nontrivial subgroup of $(\Bbb R, +)$ is either discrete or dense, and it's discrete if and only if it's isomorphic to $\Bbb Z$, if and only if it has a smallest positive element. If it's finitely generated, this happens if and only if the generators are all rational multiples of each other. It's dense if and only if it has arbitrarily small nonzero elements.

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The sequence you came up with is a clever idea! I suspect that making the strategy of picking an arbitrary convergent subsequence work is difficult. For example, it has a subsequence converging to $5$. In fact, it has a subsequence converging to any any number in $[1, 10]$. In other words, $\{a_n : n \in \Bbb N\}$ is dense in $[1, 10]$. We can see this by using our understanding of $(\Bbb R, +)$ again. It will follow from this:

Let $\alpha \in \Bbb R$ be irrational. Consider the sequence $x_n = n\alpha - \lfloor n\alpha \rfloor$. This sequence is dense in $[0, 1]$. This fact is closely related to the density of $\{m + n \alpha\}$ in $\Bbb R$. In fact the sequences version is exactly what the first answer at the linked question shows.

This certainly shows that $\{a_n : n \in \Bbb N\}$ is dense in $[1, 5]$. This is because when you divide by $5$ every time you go past $5$, this is essentially the same as continually subtracting $\log 5$ to stay in the interval $[0, \log 5]$ - which is analagous to taking the fractional part. Then density in $[5, 10]$ follows, because if $a_{n_k} \to a \in [1, 5)$, then $a_{n_k + 1} \to 2a$.

Informally, I think this problem is pretty tricky to come up with a solution for without basically doing what my answer does and classifying all subgroups of $\Bbb R$. This is because showing that the subgroup generated by $1$ and $\alpha$ is not closed essentially immediately implies it's dense, which suggests there's not a particularly "clever" sequence or limit you could have considered, and you might as well just prove it's dense right away.