Discrete Subgroup of Real Numbers is Closed

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $G$ be a subgroup of the additive group of real numbers.

Let $G$ be discrete.


Then $G$ is closed.


Proof

By Subgroup of Real Numbers is Discrete or Dense, there exists $a\in \R$ such that $G=a\Z$.

If $a=0$, then $G$ is closed.

Let $a>0$.

Then:

$\displaystyle\R\setminus G = \bigcup_{z\in \Z}(az,az+a)$

By Union of Open Sets of Metric Space is Open, $\R\setminus G$ is open.

Thus $G$ is closed.

$\blacksquare$


Also see