Lemma 108.73.1. There exists a system $G_1 \to G_2 \to G_3 \to \ldots $ of (abelian) topological groups such that $\mathop{\mathrm{colim}}\nolimits G_ n$ taken in the category of topological spaces is different from $\mathop{\mathrm{colim}}\nolimits G_ n$ taken in the category of topological groups.

## 108.73 Different colimit topologies

This example is [Example 1.2, page 553, TSH]. Let $G_ n = \mathbf{Q} \times \mathbf{R}^ n$, $n \geq 1$ seen as a topological group for addition endowed with the usual (Euclidean) topology. Consider the closed embeddings $G_ n \to G_{n + 1}$ mapping $(x_0, \ldots , x_ n)$ to $(x_0, \ldots , x_ n, 0)$. We claim that $G = \mathop{\mathrm{colim}}\nolimits G_ n$ endowed with the topology

is not a topological group.

To see this we consider the set

Using that $jx_0$ is never an integral multiple of $\pi /2$ as $\pi $ is not rational it is easy to show that $U \cap G_ n$ is open. Since $0 \in U$, if the topology above made $G$ into a topological group, then there would be an open neighbourhood $V \subset G$ of $0$ such that $V + V \subset U$. Then, for every $j \geq 0$ there would exist $\epsilon _ j > 0$ such that $(0, \ldots , 0, x_ j, 0, \ldots ) \in V$ for $|x_ j| < \epsilon _ j$. Since $V + V \subset U$ we would have

for $|x_0| < \epsilon _0$ and $|x_ j| < \epsilon _ j$. However, if we take $j$ large enough such that $j \epsilon _0 > \pi /2$, then we can choose $x_0 \in \mathbf{Q}$ such that $|\text{cos}(jx_0)|$ is smaller than $\epsilon _ j$, hence there exists an $x_ j$ with $|\text{cos}(jx_0)| < |x_ j| < \epsilon _ j$. This contradiction proves the claim.

**Proof.**
See discussion above.
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)