Exercise 2.15

Proof. Assume $Y$ is a subgroup of $X$. Under our assumptions, there exists a sequence $E_{n}n$ of $X$ such that

(i)

${({\bar{E}}_n)}^{\text{∘}}=\varnothing ;$

(ii)

$X\setminus Y={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{E}_n$.

By (i), the complement $V_n$ of ${\bar{E}}_n$ is a dense open set. Since $X$ is an F-space, it follows from the Baire’s theorem that the intersection $S$ of the $V_n$’s is dense in $X$: So is $x+S$ ( $x\in X$). To see that, remark that

follows from 1.3 (b). Since $S$ and $x+S$ are both dense open subsets of $X$, the Baire’s theorem asserts that

Thus,

Moreover, it follows from (ii) that $X\setminus Y\subset {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_n{E}_n$, i.e. $Y\supset S$. Combined with (3 ), this shows that $x+Y$ cuts $Y$. Therefore, our arbitrary $x$ is an element of the subgroup $Y$. We have thus established that $X\subset Y$, which achieves the proof. □