# Relation between topological closure and algebraic closure

#### huram2215

Suppose $\displaystyle ((G , \cdot), \tau_G)$ is a topological group, i.e., a group with a topology and the multiplication and inverse operators are continuous. Let H be a subgroup of G. I can prove that H is a topological group. The problem is with $\displaystyle \overline{H}$.

Denote $\displaystyle \overline{H}$ as the topological closure of H and $\displaystyle \tau_{\ \overline{H}}$ as the subspace topology on $\displaystyle \overline{H}$. Munkres claims in Ch2, exercise 3 that $\displaystyle ( ( \overline{H} , \cdot ) , \tau_{ \ \overline{H}} )$ is a topological space and I'm trying to prove it. If I can prove $\displaystyle \overline{H}$ is a subgroup of G, I can finish the proof (that the multiplication operator and inverse operators in $\displaystyle \overline{H}$ are continuous).

All we have to work with is the structure induced on $\displaystyle \overline{H}$ by the general topological group G. If this were a metric space, we could take convergent sequences and work with that. However, in a general topological space, we might work with topological convergence; if I take neighborhoods of a point of closure, I can get a sequence, but without compactness I don't see how to get a convergent sequence to the point of closure.

Ideas?

#### Opalg

Suppose $\displaystyle ((G , \cdot), \tau_G)$ is a topological group, i.e., a group with a topology and the multiplication and inverse operators are continuous. Let H be a subgroup of G. I can prove that H is a topological group. The problem is with $\displaystyle \overline{H}$.

Denote $\displaystyle \overline{H}$ as the topological closure of H and $\displaystyle \tau_{\ \overline{H}}$ as the subspace topology on $\displaystyle \overline{H}$. Munkres claims in Ch2, exercise 3 that $\displaystyle ( ( \overline{H} , \cdot ) , \tau_{ \ \overline{H}} )$ is a topological space and I'm trying to prove it. If I can prove $\displaystyle \overline{H}$ is a subgroup of G, I can finish the proof (that the multiplication operator and inverse operators in $\displaystyle \overline{H}$ are continuous).

All we have to work with is the structure induced on $\displaystyle \overline{H}$ by the general topological group G. If this were a metric space, we could take convergent sequences and work with that. However, in a general topological space, we might work with topological convergence; if I take neighborhoods of a point of closure, I can get a sequence, but without compactness I don't see how to get a convergent sequence to the point of closure.

Ideas?
As you point out, if the topology on the group is metrisable then you can prove the result by using the fact that every element of $\displaystyle \overline{H}$ is the limit of a sequence in H. For the non-metrisable case, all you have to do is to use the same argument, but replacing sequences with directed nets.

huram2215

#### huram2215

Thank you; I'm not quite familiar enough with nets to see this right off, but will dig in and try to figure it out. Nice idea.

#### Drexel28

Thank you; I'm not quite familiar enough with nets to see this right off, but will dig in and try to figure it out. Nice idea.
It looks like you're trying to prove that the closure of a subgroup in a topological group is a subgroup, right? If so, let $\displaystyle H\leqslant G$ and $\displaystyle h\in\overline{H}$. We claim that $\displaystyle h^{-1}\in\overline{H}$. To do this it suffices to show that every neighborhood of $\displaystyle h^{-1}$ intersects $\displaystyle H$, right? So, as usual, let $\displaystyle \mu$ be the inversion map and let $\displaystyle U$ be any neighborhood of $\displaystyle h^{-1}$. Then, $\displaystyle \mu^{-1}\left(U\right)$ is a neighborhood of $\displaystyle h$ and since $\displaystyle h\in\overline{H}$ it follows there exists some $\displaystyle h'\in\mu^{-1}\left(U\right)\cap H$. Thus, $\displaystyle \mu(h')=(h')^{-1}\in U\cap H$. But, since $\displaystyle U$ was arbitrary it follows that $\displaystyle h^{-1}\in\overline{H}$. The multiplicative closure is done similarly.

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