Can you give me an example of a totally disconnected subgroup of a topological group that is not a topological group? [closed]

Question

In the book The topology of fiber bundles, Steenrod characterize bundle over a base $X$ and totally disconnected structural group $G$ as follows.

Theorem: Let $X$ be arcwise connected, arcwise locally connected, and semi locally 1-connected. Let $G$ be totally disconnected topological group. Then the equivalence classes of principal bundles over $X$ with group $G$ are in 1-1 correspondence with the equivalence classes (under inner automorphisms of $G$) of homomorphisms of $\pi_1(X)$ into $G$.

In the differentiable context (every manifold here is closed and connected), examples of Ehresmann-Feldbbau fiber bundles (fiber bundles where the group be not, necessarily, a topological group) with totally disconnected groups occur in foliated bundles:

Proof: A foliated bundle is characterizazed by a homomorphisms $\varphi:\pi_1(M)\longrightarrow \text{Diff}(F)$, where $F$ is the fiber. With the $C^\infty$-topology, the group $\text{Diff}(F)$ is a metrizable topological group that act continuously in $F$. Since that the fundamental group of a compact manifold is finitely genereted (a discussion can be found here), the group $G=\text{Image}(\varphi) $ is an enumareable set. Now, given $f,g\in G$, $f\neq g$, denoting by $d$ a metric that generates the topology of $\text{Diff}(F)$, the function $F(h)=\frac{d(f,h)}{d(f,h)+d(f,g)}$, $F:\text{Diff}(F)\rightarrow \mathbb{R}$ , is a continuous function satisying $F(f)=0$ and $F(g)=1$. Since G is an enumerable set, there exists $\alpha$, $0<\alpha<1$, such that $F(h)\neq \alpha $ for all $h\in G$. Thus, $G$ is equal the disjoint union $(G\cap F^{-1}(-\infty,\alpha))\cup (G\cap F^{-1}(\alpha,\infty))$, $f\in \cap F^{-1}(-\infty,\alpha)$ and $g\in \cap F^{-1}(\alpha,\infty)$, therefore $G$ is totally disconnected.

I want to show that a foliated fiber bundle is also a fiber bundle with a group $G$. This motives the question below:

Question: The group $G$ is a topological group? More generally, every totally disconnected subgroup of a topological group is, itself, a topological group?

Answer 1

If $H$ is a subgroup of a topological group $G$ and is equipped with the subspace topology, then $H$ is a topological group.

It's mostly a matter of seeing that the product topology $Top_{prod}$ on $H \times H$ matches the subspace topology $Top_{sub}$ on $H \times H$ coming from $G \times G$. The product topology on $H \times H$ is the smallest topology such that the two projections $\pi_1, \pi_2: H \times H \to H$ are continuous. An open in $H$ is of the form $U \cap H$ for some open $U$ of $G$, and $\pi_1^{-1}(U \cap H) = (U \cap H) \times H = (U \times G) \cap (H \times H)$ is open in the subspace topology on $H \times H$. A similar statement holds for $\pi_2$ in place of $\pi_1$. Hence $Top_{prod} \subseteq Top_{sub}$. But also, the subspace topology on $H \times H$ is the smallest topology such that the inclusion $i: H \times H \to G \times G$ is continuous. For opens $U, V$ of $G$ we have $i^{-1}(U \times V) = (U \times V) \cap (H \times H) = (U \cap H) \times (V \cap H)$; the latter is open in the product topology. So $Top_{sub} \subseteq Top_{prod}$.

If $m_H: H \times H \to H$ is the restriction of multiplication $m_G: G \times G \to G$, and $U \cap H$ is a typical open in $H$, then, $m_H^{-1}(U \cap H) = m_G^{-1}(U) \cap (H \times H)$ is open in $H \times H$ under the subspace topology, hence under the product topology by the above paragraph.

If $i_G: G \to G$ denotes inversion and $i_H: H \to H$ is its restriction, then $i_H^{-1}(U \cap H) = i_G^{-1}(U) \cap H$ is open in $H$ for a typical open $U \cap H$.