Balanced Core: $U\text{ open }\implies U^*\text{ open}$

Question

I need one last lemma for the proof of finite dimensional subspaces are closed:

Is it true that if a subset is open so is its balanced core??

Answer 1

Yes, the balanced core of an open set is open.

Let $x$ be a point in the balanced core $U^\ast$ of $U$.

Then $K = \overline{\mathbb{D}}\cdot x = \{\lambda\cdot x : \lvert\lambda\rvert \leqslant 1\}$ is a subset of $U$, and $K$ is balanced, so $K\subset U^\ast$.

Furthermore, as the image of the compact set $\overline{\mathbb{D}}$ under the continuous map $\lambda \mapsto \lambda\cdot x$, $K$ is quasicompact.

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Interlude: Some facts about topological vector spaces we need

In a topological vector space, Hausdorff or not, the balanced open neighbourhoods of $0$ form a neighbourhood basis at $0$: Let $N$ be a neighbourhood of $0$. By the continuity of scalar multiplication in $(0,0)$, there is an $\varepsilon > 0$ and a neighbourhood $N_1$ of $0$ such that $N_2 := D_\varepsilon(0)\cdot N_1 \subset N$. $N_2$ is balanced, since $D_\varepsilon(0)$ is balanced.

The interior of a balanced set $B$ is balanced if and only if it is empty or $0\in\overset{\large\circ}{B}$, for $A = \bigcup\limits_{0 < \lvert\lambda\rvert\leqslant 1}\lambda\cdot \overset{\large\circ}{B}$ is open as the union of open sets, and contained in $B$ since $B$ is balanced, hence $A\subset \overset{\large\circ}{B}$, and, if the interior of $B$ is not empty, $\overline{\mathbb{D}}\cdot \overset{\large\circ}{B} = A\cup \{0\}$ is therefore a subset of $\overset{\large\circ}{B}$ if and only if $0\in \overset{\large\circ}{B}$.

Thus $\overset{\large\circ}{N}_2$ is an open balanced neighbourhood of $0$ contained in $N$. Since $N$ was arbitrary, the claim about the neighbourhood basis is proved.

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Back to the main proof:

Since $U$ is open, for every $k\in K$, there is a neighbourhood $V_k$ of $0$ such that $k + V_k \subset U$. By the continuity of addition, there is a neighbourhood $W_k$ of $0$ with $W_k + W_k \subset V_k$. By the interlude, we may assume $W_k$ as open and balanced.

Then $\mathfrak{W} = \{ k + W_k : k \in K\}$ is an open cover of $K$. Since $K$ is quasicompact, there is a finite subset $F = \{k_1,\dotsc,k_n\}$ of $K$ such that $\mathfrak{F} = \{ k + W_k : k\in F\}$ is also an open cover of $K$. Define

$$W := \bigcap_{k\in F} W_k.$$

As a finite intersection of balanced open sets, $W$ is also a balanced open set. And we have

$$K + W \subset U.$$

Namely, for every $y \in K$, there is a $k\in F$ such that $y \in k + W_k$, since $\mathfrak{F}$ is a cover of $K$, and therefore

$$y + W \subset (k+ W_k) + W \subset k + W_k + W_k \subset k + V_k \subset U.$$

Since $K$ and $W$ are both balanced, $K+W$ is balanced, therefore $K+W\subset U^\ast$.

Since $W$ is open, $K+W$ is open, and hence $K+W \subset \left(U^\ast\right)^{\Large\circ}$, thus, finally,

$$x \in K \subset K+W \subset \left(U^\ast\right)^{\Large\circ}$$

shows that $U^\ast$ is open, since $x\in U^\ast$ was arbitrary.