The Topological Group That Must Be Discrete

Let $G$ be a topological group (a group that is also a topological space, where multiplication and inversion are continuous). Suppose $G$ has a subset $U$ that is open, and that $U$ contains no element other than the identity $e$ that has finite order (i.e., no non-identity element $g \in U$ satisfies $g^n = e$ for some $n \geq 1$ ).

Now consider a different scenario: suppose $G$ is a topological group and $\{e\}$ is an open set (the trivial subgroup is open).

Prove that $G$ must be discrete.

Then use this to answer: why is $\mathbb{Q}$ (with the subspace topology from $\mathbb{R}$ ) not a discrete topological group under addition, even though $(\mathbb{Q}, +)$ is a perfectly good group?

Answer: Open Identity Forces Discrete Topological Group

Key Idea / Intuition

In a topological group, the topology is "homogeneous" — it looks the same at every point, because left-translation $g \mapsto ag$ is a homeomorphism. So if one point (the identity) has an open neighborhood, you can translate that neighborhood to make every singleton open. Discreteness then follows immediately. The $\mathbb{Q}$ example grounds the abstraction: singletons in $\mathbb{Q}$ are never open in $\mathbb{R}$ 's topology, so the hypothesis fails, and indeed $\mathbb{Q}$ is not discrete.

Formal Proof / Solution

Setup. Recall that in a topological group $G$ , for any fixed $a \in G$ , the left-translation map $L_a : G \to G, \quad L_a(g) = ag$ is a homeomorphism (it is continuous with continuous inverse $L_{a^{-1}}$ ).

Step 1: Every singleton is open.

Suppose $\{e\}$ is open in $G$ . Let $g \in G$ be any element. We want to show $\{g\}$ is open.

Apply the homeomorphism $L_g$ : $L_g(\{e\}) = \{g \cdot e\} = \{g\}.$

Since $L_g$ is a homeomorphism and $\{e\}$ is open, its image $\{g\}$ is open.

Step 2: Conclude $G$ is discrete.

Since every singleton $\{g\}$ is open, every subset $S \subseteq G$ is a union of open sets: $S = \bigcup_{g \in S} \{g\},$ so $S$ is open. Thus every subset of $G$ is open, which is exactly the discrete topology.

$\blacksquare$

Why $\mathbb{Q}$ is not discrete.

In $\mathbb{Q}$ with the subspace topology from $\mathbb{R}$ , the open sets are intersections of open intervals with $\mathbb{Q}$ . Every open set in $\mathbb{Q}$ is infinite — no singleton $\{q\}$ is open (since any open interval around $q$ contains other rationals). In particular, $\{0\}$ is not open in $\mathbb{Q}$ .

By the theorem above (contrapositively), $\mathbb{Q}$ is not discrete. And indeed it isn't: every neighborhood of $0$ contains infinitely many other rationals, so we cannot isolate any point.

Conceptual takeaway. In a topological group, the topology is completely determined by the neighborhoods of the identity (you can translate any neighborhood of $e$ to a neighborhood of any $g$ ). So "the identity is isolated" is equivalent to "every point is isolated" — a beautiful rigidity that has no analogue in general topological spaces.