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. $\square$

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