Local basis criterion for topological group / vector space?

Question

Let $X$ be a topological vector space (TVS), and let $\mathcal B$ be nonempty family of subsets of $X$ that each contain $0$. Are there simple conditions that guarantee $\mathcal B$ is a local basis for a topology that turns $X$ into a TVS? How does the answer change if we relax $X$ to being an abelian group or just a group?

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My first idea is as follows. If $\mathcal B$ is the local basis for a topology $\mathcal T$ that makes $X$ a TVS, then the set of all translates of elements in $\mathcal B$ must be an ordinary basis for $\mathcal T$. These sets must satisfy the "ordinary basis criterion," which here is equivalent to a local basis criterion: For any $V_1, V_2 \in \mathcal B$ and $x_1, x_2 \in X$ such that $0 \in (V_1 + x_1) \cap (V_2 + x_2),$ there is some $V \in \mathcal B$ such that $V \subseteq (V_1 + x_1) \cap (V_2 + x_2)$. This guarantees that the translates of $\mathcal B$ are the basis for a topology on $X$. However, this doesn't seem to be enough to show that $\mathcal T$ turns $X$ into a TVS. But it seems there are many natural candidate properties to assume for $\mathcal B$ that might help the situation, e.g., every $V \in \mathcal B$ is absorbing, star-shaped, balanced, etc.

Answer 1

You need singletons to be closed and you need the vector space operations to be continuous.

Singletons being closed is equivalent to saying that for any $x\neq 0$, there exists $V\in\mathcal{B}$ such that $x\notin V$. This will show that for any $x\neq 0$, $x\notin \overline{\{0\}}$, and $\overline{\{0\}}=\{0\}$. The rest follows from translation invariance of the topology. So the condition implies that singletons are closed. On the other hand, if there exists $x\in X$ such that $x\in V$ for all $V\in \mathcal{B}$, then $x\in \overline{\{0\}}$, and $\{0\}$ is not closed. So every singleton is closed iff this condition is satisfied.

Continuity of addition is equivalent to the condition that for every $V\in \mathcal{B}$, there exists $U\in \mathcal{B}$ such that the Minkowski sum $U+U\subset V$. Indeed, since $0+0=0\in V$, there exist $W_1,W_2\in \mathcal{B}$ such that $W_1+W_2\subset V$ by continuity at $(0,0)$. Take $U=W_1\cap W_2$. This shows that continuity implies the condition. On the other hand, suppose we have the condition. Fix $x,y\in X$ and $V\in \mathcal{B}$. Since the condition is satisfied, there exists $U\in \mathcal{B}$ such that $U+U\subset V$. Then $(x+U)+(y+U)\subset (x+y)+V$, and we have continuity of addition.

Last, we need continuity of scalar multiplication. First assume scalar multiplication is continuous. Fix a scalar $\alpha$, $x\in X$, and $U\in \mathcal{B}$. There exist $\varepsilon>0$ and $W_1\in \mathcal{B}$ such that for any $\beta$ with $|\alpha+\beta|<\varepsilon$, $\beta(x+W_1)\subset \alpha x+U$. Taking the particular choice $x=0$, we can see that this requires that for any scalar $\alpha$, there exists $\varepsilon>0$ such that for any $\beta$ with $|\beta-\alpha|<\varepsilon$, $\beta W_1\subset U$. For any $R>0$, let $RD$ denote the set of scalars $\alpha$ with $|\alpha|\leqslant R$, which is a compact set (assuming you're working in $\mathbb{R}$ or $\mathbb{C}$). For each $\alpha\in RD$, there exist $(\varepsilon_\alpha, W_\alpha)\in (0,\infty)\times \mathcal{B}$ such that $\beta W_\alpha\subset U$ for all $\beta\in \alpha+\text{interior}(\varepsilon_\alpha D)$. Choose a finite subset $F$ of $RD$ such that $(\alpha +\text{interior}(\varepsilon_\alpha D)$ is a cover of $RD$, and let $W\in \mathcal{B}$ be a subset of $\cap_{\alpha\in F}W_\alpha$. Then $$\{\beta y:|\beta|\leqslant R, y\in W\}\subset U.$$ Indeed, for any $\beta y$ in this set, there exists $\alpha \in F$ such that $|\beta-\alpha|<\varepsilon_\alpha$, and $\beta y\in \beta W\subset \beta W_\alpha\subset U$. Therefore we have shown that for any $U\in \mathcal{B}$ and $R>0$, there exists $W\in \mathcal{B}$ such that $RW\subset U$.

Next, pick an arbitrary $x\in X$ and note that, by continuity of scalar multiplication at $(0,x)$, there exist $\varepsilon>0$ and $W_2\in \mathcal{B}$ such that for any $\beta$ with $|\beta|<\varepsilon$, $\beta (x+W_2)\subset U$. Since $0\in W_2$, this means that $\{\beta x:|\beta|<\varepsilon\}\subset U$. So for each $x\in X$ and $U\in \mathcal{V}$, there exists $\varepsilon>0$ such that for all $\beta$ with $|\beta|<\varepsilon$, $\beta x\in U$.

I claim that these last two conditions, when combined with the condition from the continuity of addition, yields continuity of scalar multiplication. Therefore we have listed four conditions which are necessary and sufficient on top of your initial conditions to get a TVS. Fix a scalar $\alpha$ and $x\in X$. Fix $V\in \mathcal{B}$ and $U\in \mathcal{B}$ such that $U+U\subset V$. There exists $\varepsilon>0$ such that for all $\beta$ with $|\beta|<\varepsilon$, $\beta x\in U$. Let $R=|\alpha|+\varepsilon$. There exists $W\in \mathcal{B}$ such that $\{\beta y:|\beta|\leqslant R, y\in W\}\subset U$. Fix a scalar $\gamma$ such that $|\alpha-\gamma|<\varepsilon$. We will show that $\gamma(x+W)\subset \alpha x+V$. Indeed, write $\gamma = \alpha+\beta$ where $|\beta|<\varepsilon$. Then $$\gamma(x+W)=(\alpha+\beta)(x+W) = \alpha x + \beta x + (\alpha+\beta)W \subset \alpha x+U+U\subset \alpha x+V.$$

Answer 2

I'd like to mention a variation on bangs's answer. Let $\mathcal B$ be a family of subsets of $X$ containing $0$, and let $\mathcal T$ be the set of all unions of translates of elements in $\mathcal B$. First of all, bangs's answer can be restated as follows.

Theorem. $\mathcal T$ turns $X$ into a (not necessarily Hausdorff) TVS with local basis $\mathcal B$ iff the following four conditions are satisfied.

1. (local basis criterion) $0 \in (V_1 + x_1) \cap (V_2 + x_2) \implies 0 \in V \subseteq (V_1 + x_1) \cap (V_2 + x_2)$ for some $V \in \mathcal B$.

2. (addition rule) $V \in \mathcal B \implies U + U \subseteq V$ for some $U \in \mathcal B$.

3. (multiplication rule I) $V \in \mathcal B, r>0 \implies B_r U\subseteq V$ for some $U \in \mathcal B$, where $B_r = \{ \lambda \in \mathbb K : |\lambda| < r\}$.

4. (multiplication rule II) $V \in \mathcal B, x \in X \implies B_r x \subseteq V$ for some $r > 0$.

Now for the variation: We can replace 3 and 4 with the conditions that every element of $\mathcal B$ be absorbing and contain a balanced element of $\mathcal B$. In other words, the four conditions above are equivalent to 1 and 2 along with:

$\quad 3'$. Every element of $\mathcal B$ contains a balanced element of $\mathcal B$.

$\quad 4'$. Every element of $\mathcal B$ is absorbing.

It is well known that $3'$ and $4'$ are satisfied if $\mathcal T$ makes $X$ a TVS with local basis $\mathcal B$ (see, e.g., Rudin's Functional Analysis). On the other hand, suppose $1$, $2$, $3'$, and $4'$ hold.

Proof of 3. Let $V \in \mathcal B$ and $r > 0$. Choose $n \in \mathbb N$ such that $n > r$. By induction on 2, there is some $U' \in \mathcal B$ such that $U' + U' + \cdots + U' \subseteq V$, with $U'$ appearing $n$ times in the sum. Let $U \in \mathcal B$ be a balanced subset of $U'$. Thus $B_r U \subseteq nU \subseteq U + U + \cdots + U \subseteq V$.

Proof of 4. Let $V \in \mathcal B$ and $x \in X$. Let $U\in \mathcal B$ be a balanced subset of $V$. Since $U$ is absorbing, let $x \in t U$ for some $t > 0$. Thus $t^{-1}x \in U$, hence $B_{t^{-1}}x \subseteq U \subseteq V$, using the fact that $U$ is balanced.