Exercise TG.3

Proof. Suppose that $X$ is a subspace of $Y$ and consider two distinct points $x,y\in X$. Then there is a neighborhood $U$ of $x$ in $Y$ that does not contain $y$ by Exercise 17.15. It then follows that $U\cap X$ is a neighborhood of $x$ in $X$ that does not contain $y$ since $y\notin U$. A similar argument shows that there is a neighborhood of $y$ in $X$ that does not contain $x$. Hence $X$ also satisfies the $T_1$ axiom, again by Exercise 17.15. □

Proof. Presumably here $G$ is a topological group. Hence $G$ satisfies the $T_1$ axiom so that $H$ and $\overline{H}$ do as well by Lemma 1 since they are subspaces of $G$.

We first show that $\overline{H}$ is a subgroup of $G$, noting that of course $\overline{H}$ is nonempty since $H$ must be (as it is a subgroup) and $H\subset \overline{H}$. Let $f:G\times G\to G$ be the operation defined by $f(x_{}\times y_{})=x\cdot y^{-1}$ for $x,y\in G$. It is a well-known theorem of group theory that $\overline{H}$ is a subgroup of $G$ if and only if $x\cdot y^{-1}\in \overline{H}$ for any $x,y\in \overline{H}$, which is to say that $f(\overline{H}\times \overline{H})\subset \overline{H}$. This is exactly what we intend to show. We have the following deductions:

• As subsets of $G\times G$, $\overline{H}\times \overline{H}=\overline{H\times H}$ by Exercise 17.9 since $H\subset G$.
• Since $f$ is continuous on $G\times G$, it follows that $f(\overline{H\times H})\subset \overline{f(H\times H)}$ by Theorem 18.1.
• Since $H$ is a subgroup we have that $x\cdot y^{-1}\in H$ for every $x,y\in H$, which is to say that $f(H\times H)\subset H$. It then follows from Exercise 17.6 part (a) that $\overline{f(H\times H)}\subset \overline{H}$.

Putting these all together, we can conclude that

which shows the desired result that $\overline{H}$ is a subgroup of $G$.

Next, obviously $H$ and $\overline{H}$ are both groups by the definition of a subgroup. The operation of $H$ (or $\overline{H}$) is of course the operation of $G$ with its domain restricted to $H\times H$ (or $\overline{H}\times \overline{H}$). The continuity of this operation on $H$ (or $\overline{H}$) follows from Theorem 18.2 part (d) since it is continuous on $G$ since $G$ is a topological group. Likewise, the inversion function on $H$ (or $\overline{H}$) is a restriction of the inversion function on $G$, and so is also continuous for the same reason. Hence $H$ and $\overline{H}$ are topological groups by definition. □