Closures of Subspaces of a Topological Vector Space

# Closures of Subspaces of a Topological Vector Space

 Proposition 1: Let $E$ be a topological vector space. If $M$ is a subspace of $E$, then $\overline{M}$ is a subspace of $E$.
• Proof: Let $x, y \in \overline{M}$ and let $\lambda \in \mathbf{F}$. We aim to show that $x + y \in \overline{M}$ and $\lambda x \in \overline{M}$ to conclude that $\overline{M}$ is a subspace of $E$.
• 1. Showing that $x + y \in \overline{M}$: Let $U$ be a neighbourhood of the origin. Since $E$ is a topological vector space and $U$ is a neighbourhood of the origin, by the proposition on the Bases of Neighbourhoods for a Point in a Topological Vector Space page, there exists a (balanced) neighbourhood of the origin $V$ such that $V + V \subseteq U$. Then, $x + V$ is a neighbourhood of $x$ and $y + V$ is a neighbourhood of $y$.
• Since $x$ and $y$ are in the closure of $M$, we have by definition that $(x + V) \cap M \neq \emptyset$ and $(y + V) \cap M \neq \emptyset$. So there exists $m_1, m_2 \in M$ such that $m_1 \in x + V$ and $m_2 \in y + V$. But:
(1)
\begin{align} \quad m_1 + m_2 \in (x + V) + (y + V) = (x + y) + (V + V) \subseteq (x + y) + U \end{align}
• So $(M + M) \cap ((x + y) + U) \neq \emptyset$. However, observe that $M + M = M$ since $M$ is subspace of $E$. Thus $M \cap ((x + y) + U) \neq \emptyset$. Since this holds for all neighbourhoods $U$ of the origin, and since every neighbourhood of $x + y$ is of the form $(x + y) + U$ for some neighbourhood of the origin $U$, we see that every neighbourhood of $x + y$ intersects $M$. Thus $x + y \in \overline{M}$.
• 2. Showing that $\lambda x \in \overline{M}$: On the other hand, observe that if $\lambda = 0$ then clearly $\lambda x = o \in \overline{M}$. Assume that $\lambda \neq 0$. Let $U$ be a neighbourhood of the origin. Then $\frac{1}{\lambda} U$ is also a neighbourhood of the origin. Let $V$ be a neighbourhood of the origin such that $V \subseteq \frac{1}{\lambda} U$. Then $\lambda V \subseteq U$.
• Observe that $x + V$ is a neighbourhood of $x$. Since $x$ is in the closure of $M$, we have that $(x + V) \cap M \neq \emptyset$. So there exists an $m \in M$ such that $m = x + v$ where $v \in V$. Then:
(2)
\begin{align} \quad \lambda m = \lambda x + \lambda v \in \lambda x + (\lambda V) \subseteq \lambda x + U \end{align}
• So $\mathbf{F} M \cap (\lambda x + U) \neq \emptyset$. However, observe that $\mathbf{F} M = M$ since $M$ is a subspace of $E$. Thus $M \cap (\lambda x + U) \neq \emptyset$. Since this holds for all neighbourhoods $U$ of the origin, and since every neighbourhood of $\lambda x$ is of the form $\lambda x + U$ for some neighbourhood of the origin $U$, we conclude that $\lambda x \in \overline{M}$.
• Thus $\overline{M}$ is a subspace of $E$. $\blacksquare$