Are totally positive units of $L$ closed (in $L$) with respect to the finite-idelic topology? My question is closely related to this one, but not clearly the same in my opinion.
Let $L$ be a number field, with ring of integers $\mathcal{O}_L$, and set $L^{\times}_+\subset L^{\times}$ to be the subgroup of totally positive elements (with respect to the real embeddings $L\hookrightarrow \mathbb{R}$). Set $\mathcal{O}^{\times}_{L,+}:=\mathcal{O}_L^{\times}\cap L^{\times}_+$ to be the subgroup of totally positive units, and denote by $\overline{\mathcal{O}_{L,+}^{\times}}\subset \mathbb{A}_{L,f}^{\times}$ its closure inside the group of finite idèles (with respect to the usual restricted product topology).
I wonder whether one always has an equality
$$L^{\times} \cap \overline{\mathcal{O}_{L,+}^{\times}}=\mathcal{O}_{L,+}^{\times},$$
or whether one can find some counter-example, i.e., a sequence
$(z_n)$ of totally positive units which converges to some $z\in L^{\times}$ (hence, in $\mathcal{O}_L^{\times}$) in $\mathbb{A}_{L,f}^{\times}$ but such that $z\notin \mathcal{O}_{L,+}^{\times}$, which could eventually be the case if $\mathcal{O}_{L,+}^{\times}$ is not discrete in $\mathbb{A}_{L,f}^{\times}$ (and if I'm not missing some clear reason why it should not exist!).
Thank you in advance for hint or reference on the topic !
 A: I believe this is difficult in general, and I don't think it has been studied much yet.
A partial answer, showing where not to look for simple counterexamples:
Unwinding the definitions, everything is taking place inside a compact subquotient of the idèles where all components at finite places have absolute value 1, the infinite places have been thrown away, the distinction between idèle and adèle topology disappears and the restricted direct product reduces to an ordinary direct product, inheriting its topology from all the projections to finite residue class rings.
A global unit will thus be in the closure of a proper subgroup (e.g. of the totally positive units) when, for all ideals $\mathfrak{m}$ of $\mathcal{O}_L$, its image in $(\mathcal{O}_L/\mathfrak{m})^\times$ lies in the image of the subgroup.
So the question becomes:  When the totally positive units $\mathcal{O}_{L,+}^\times$ are a proper subgroup of the global units $\mathcal{O}_L^\times$, can we always find an ideal $\mathfrak{m}$ which can "see" this fact, i.e. where the image of the totally positive units in $(\mathcal{O}_L/\mathfrak{m})^\times$ is strictly smaller than the image of all global units?
This is vacuously true when $L$ has no real places: all global units are then totally positive.
It is true for $L=\mathbb{Q}$: Take e.g. $\mathfrak{m}=(3)$ (or any other odd prime ideal).
For number fields of unit rank $1$ and with real places, we can try to find an $\mathfrak{m}$ such that the totally positive units all map to $1$, or to a small subgroup, preferably of odd order, of the residue class ring.
In $L=\mathbb{Q}(\sqrt{5})$, with fundamental unit $\varepsilon$ a root of $x^2-x-1$, the prime ideal above $5$ separates the image $-2$ of $\varepsilon$ from the image $\{\pm 1\}$ of the powers of $\varepsilon^2$. It can't see that $-1$ isn't a totally positive unit; but each of the two prime ideals above $11$ can.
For other real quadratic fields with fundamental unit $\varepsilon$ of negative norm, $\varepsilon+1$ is a non-unit. If it has a prime ideal factor of odd norm, we can take that as $\mathfrak{m}$, and all the totally positive units will map to $1$ modulo $\mathfrak{m}$, while both $\varepsilon$ and $-1$ will map to $-1$. If $|N(\varepsilon+1)|$ is a power of $2$, as happens e.g. in $\mathbb{Q}(\sqrt{17})$, we can try to reason with $\varepsilon^3+1$ instead, and arrange for the image of $\varepsilon$  to generate a multiplicative group of order $6$ in the residue class ring, with all totally positive units landing in the subgroup of order $3$. If the cube doesn't work either, we can use a higher odd power. (One such power will always give us a suitable prime ideal of odd norm. Since $S$-unit equations in a fixed field have only finitely many solutions, only finitely many $\varepsilon^k+1$ can have all their prime ideal divisors above $(2)$.)
For real quadratic fields with fundamental unit $\varepsilon$ of positive norm, we consider $\varepsilon-1$ instead, or $\varepsilon^k-1$ with odd exponent $k$ if necessary, sending all powers of $\varepsilon$ to $1$ or to a subgroup of odd order in a finite residue class field modulo some prime ideal of odd norm, disjoint from the image of $-1$.
The same argument works for cubic fields of negative discriminant. These have only one real place, so we can always arrange for the fundamental unit to be totally positive, and only need to separate its powers from $-1$. (The fields of discriminants $-23$ and $-31$ contain exceptional units and need a little extra care.)
But I do not see how to extend this line of reasoning to any number field of unit rank $2$ or larger. It can deal with one generator at a time, but we'd need to deal with a full set of fundamental units at once. Highly composite $\mathfrak{m}$ may or may not help here.
