Lemma 15.36.3 (Baire category theorem). Let $M$ be a topological abelian group. Assume $M$ is linearly topologized, complete, and has a countable fundamental system of neighbourhoods of $0$. If $U_ n \subset M$, $n \geq 1$ are open dense subsets, then $\bigcap _{n \geq 1} U_ n$ is dense.
Proof. Let $U_ n$ be as in the statement of the lemma. After replacing $U_ n$ by $U_1 \cap \ldots \cap U_ n$, we may assume that $U_1 \supset U_2 \supset \ldots $. Let $M_ n$, $n \in \mathbf{N}$ be a fundamental system of neighbourhoods of $0$. We may assume that $M_{n + 1} \subset M_ n$. Pick $x \in M$. We will show that for every $k \geq 1$ there exists a $y \in \bigcap _{n \geq 1} U_ n$ with $x - y \in M_ k$.
To construct $y$ we argue as follows. First, we pick a $y_1 \in U_1$ with $y_1 \in x + M_ k$. This is possible because $U_1$ is dense and $x + M_ k$ is open. Then we pick a $k_1 > k$ such that $y_1 + M_{k_1} \subset U_1$. This is possible because $U_1$ is open. Next, we pick a $y_2 \in U_2$ with $y_2 \in y_1 + M_{k_1}$. This is possible because $U_2$ is dense and $y_2 + M_{k_1}$ is open. Then we pick a $k_2 > k_1$ such that $y_2 + M_{k_2} \subset U_2$. This is possible because $U_2$ is open.
Continuing in this fashion we get a converging sequence $y_ i$ of elements of $M$ with limit $y$. By construction $x - y \in M_ k$. Since
is in $M_{k_ i}$ we see that $y \in y_ i + M_{k_ i} \subset U_ i$ for all $i$ as desired. $\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (2)
Comment #3537 by Dario Weißmann on
Comment #3669 by Johan on