Lemma 9.5.4. Let $E$ be a locally convex space over $K$, $M \subset E$ be a subspace, $U \in \cn_{M}(0)$ be convex and circled, then
There exists $V \in \cn_{E}(0)$ circled and radial such that $U = M \cap V$.
For any $x \in E \setminus \ol M$, there exists $V \in \cn_{E}(0)$ circled and radial such that $U = M \cap V$ and $x \not\in U$.
Proof. (1): Let $W \in \cn_{E}(0)$ be circled and radial with $W \cap M \subset U$, and $V = \text{Conv}(W \cup U)$.
For any $u \in U$, $w \in W$, and $t \in [0, 1]$, there exists $\alpha \in (0, 1)$ such that $x = \alpha^{-1}w \in W$. In which case,
\[(1 - \alpha)U + w = (1 - \alpha) U + \alpha x \subset V\]so $V \in \cn(0)$.
For any $\lambda \in K$ with $\abs{\lambda}\le 1$, $u \in U$, $w \in W$, and $t \in [0, 1]$,
\[\lambda (1 - t)u + \lambda tw = (1 - t)\lambda u + t \lambda w \in V\]as $U$ and $W$ are both circled.
so $V \in \cn_{E}(0)$ is convex and circled.
For any $u \in U$, $w \in W$, and $t \in [0, 1]$, if $(1 - t)u + tw \in M$, then $u \in M \cap U \subset M \cap W$, so $(1 - t)u + tw \in W$.
(2): Since $x \not\in \ol M$, there exists $W \in \cn_{E}(0)$ circled and radial with $x + W \cap M = \emptyset$. Let $V$ as in (1), then $x \not\in W + M \supset W + U = V$.$\square$