A dense set in $\mathbb{T}$

Question

Let $$\mathbb{T}=\{ z \in \mathbb{C}: |z|=1\} $$

Consider $\mathbb{T}$ as a topological group under multiplication with it's usual topology, I'm reading a proof wich states that a dense set in $T$ cannot be discrete but i don't understand why, how can I prove that?.

Answer 1

Your space is homeomorphic to $[0, 1)$ with the metric $d(x, y) = \min(|x - y|, |x - y + 1|)$. If $E \subseteq [0, 1)$ is discrete in this topology, then for every $x \in E$ exists a neighborhood $U_{x}$ such that $U_{x} \cap E = \{x\}$.

Assume for contradiction that $E$ is dense and discrete. Since $E$ is dense, there is a point $x \in E \cap (1/4, 3/4)$. Let $r = \min( x - 1/4, 3/4 - x)$. Let $\delta > 0$ be such that $(x - \delta , x +\delta) \cap E = \{x\}$. Then so does $U = (x - s, x + s)$, where $s = \min(r/2, \delta)$. Then $(U \setminus \{x\}) \cap E = \emptyset$, a contradiction since $U \setminus \{x\}$ is open and does not intersect $E$, making $E$ non-dense.

Answer 2

In such questions it's often more convenient to work with $\mathbb R$ instead of $\mathbb T$. $\mathbb R$ is a 'covering space' of $\mathbb T$ so has lots of nice properties. We have a map $p: \mathbb R \to \mathbb T; \theta \mapsto e^{2\pi i \theta}$.

Notice that locally $p$ is a homeomorphism (e.g. on $(x, x+1)$ for any $x \in \mathbb R$). This means that if $F$ is a discrete set, then so is $p^{-1}(F)$. Similarly, if $F$ is dense then $p^{-1}(F)$ is dense too.

This means it's enough to show that we can't find a discrete, dense subset of $\mathbb R$, which is easier to think about!

From here the argument is very similar to AJY's, but we can use the nicer Euclidean metric: if we pick any point $x \in F$ then since $F$ is discrete we have a little open set $U_x \subseteq \mathbb R$ with $U_x \cap F = \{x\}$, let's say $(x-\delta, x+\delta)$ with $\delta > 0$. But $F$ is also dense, so there must be an element of $F$ in $(x, x+\delta)$, contradiction.