There is a lemma on page 95 of W. Boothby's An Introduction to Differentiable Manifolds and Riemannian Geometry as follows:
(8.4) Lemma Let $G$ be a Lie Group and $\Gamma$ an algebraic subgroup. Then there is a neighborhood $U$ of the identity $e$ such that $U \cap \Gamma = \{e\}$ if and only if $\Gamma$ is a discrete subspace, in which case $\bar{\Gamma} = \Gamma$, that is, $\Gamma$ is closed and countable.
To prove the closedness of $\Gamma$, the author states that:
if $\Gamma$ is not a closed subset, it is easy to see that there must be a sequence $\{h_{n}\} \subset \Gamma$ with $\lim_{n \to \infty}{h_{n}} = e$. This contradicts the existence of $U$ with $U \cap \Gamma = \{e\}$.
This contradicts what? Does it preclude the existence of the sequence $\{h_{n}\}$, or does it imply that we have found a neighborhood of the point of convergence $e$ of the sequence that contains only one point? If this is the case, then isn't it true that one-point sets are open in a discrete space? Even if so, what does it have to do with closedness of $\Gamma$?!
Thank you.
The author starts with the assumption that there's a neighborhood $U$ of $e$ such that $U \cap \Gamma = \{e\}$. Moreover one can assume that $UU^{-1} \subset U$ (he explains why).
We'd like to prove that $\Gamma$ is closed, which is the same as being sequentially closed because $G$ is a manifold. So take a sequence $c_n \in \Gamma$ with limit $g \in G$, we aim to prove that $g \in \Gamma$.
The set $Ug$ is a neighborhood of $g$, so by definition of convergence, there exists $N$ such that $c_n \in Ug$ for $n \ge N$. Say $c_n = u_n g$ with $u_n \in U$.
Then since $\Gamma$ is a subgroup, $c_n c_m^{-1} \in \Gamma$; but for $n,m \ge N$, you also have $c_n c_m^{-1} = u_n u_m^{-1}$, which belongs to $UU^{-1} \subset U$. Therefore $c_n c_m^{-1}$ is in $U \cap \Gamma = \{e\}$, i.e. $c_n = c_m$. In other words the sequence is constant starting from $c_N$, and so $g = \lim c_n = c_N$ is actually in $\Gamma$.
I have to admit I don't really see how it follows easily from the non-closedness of $\Gamma$ that there is a sequence $h_n$ in $\Gamma$ with $\lim h_n = e$. But if you admit that such a sequence exists, then by definition of convergence, there exists $N$ such that $h_n \in U$ (a neighborhood of $e$) for all $n \ge N$. Presumably the sequence $h_n$ constructed also satisfies $h_n \neq e$ for all $n$, since then you get a contradiction: $h_N$ is in $U \cap \Gamma$ but isn't $e$.