The Coarsest Topology Determined by a Set of Seminorms

The Coarsest Topology Determined by a Set of Seminorms on a Vector Space

Theorem: Let $E$ be a vector space and let $Q$ be a collection of seminorms on $E$. Then if $E$ is equipped with the coarsest topology determined by $Q$, then $E$ becomes a locally convex topological vector space, with a base of neighbourhoods of the origin given by sets of the form $\{ x : \sup_{1 \leq i \leq n} p_i(x) \leq \epsilon \}$ where $\epsilon > 0$ and where $p_1, p_2, ..., p_n \in Q$.
  • Proof: Let $\mathcal U$ be the collection of all sets of the form $\{ x : \sup_{1 \leq i \leq n} p_i(x) \leq \epsilon \}$. Then if $U, V \in \mathcal U$ with:
(1)
\begin{align} \quad U := \{ x : \sup_{1 \leq i \leq n} p_i(x) \leq \epsilon \} \quad \mathrm{and} \quad V := \{ x : \sup_{1 \leq i \leq m} p_i'(x) \leq \epsilon' \} \end{align}
  • where $p_1, p_2, ..., p_n, p_1', p_2', ..., p_m' \in Q$ and $\epsilon, \epsilon' > 0$. Then clearly:
(2)
\begin{align} \quad U \cap V = \{ x : \sup \{ p_1(x), p_2(x), ..., p_n(x), p_1'(x), p_2'(x), ..., p_m'(x) \} \leq \min \{ \epsilon, \epsilon' \} \} \in \mathcal U \end{align}
  • If $U \in \mathcal U$ with $U = \{ x : \sup_{1 \leq i \leq n} p_i(x) \leq \epsilon \}$ and if $\alpha \in \mathbf{F}$ with $\alpha \neq 0$, then:
(3)
\begin{align} \alpha U = \alpha \{ x : \sup_{1 \leq i \leq n} p_i(x) \leq \epsilon \} = \{ \alpha x : \sup_{1 \leq i \leq n} p_i(x) \leq \epsilon \} = \{ x : \sup_{1 \leq i \leq n} p_i(x) \leq |\alpha| \epsilon \} \in \mathcal U \end{align}
  • Furthermore, for each $p \in Q$ and for all $\epsilon > 0$, given the open set $(-\epsilon - 1, \epsilon + 1)$ containing $p(o) = 0$, we have that the neighbourhood $\{ x : p(x) \leq \epsilon \}$ of the origin $o$ is such that $p(\{ x : p(x) \leq \epsilon \}) \subseteq (-\epsilon - 1, \epsilon + 1)$, so that $p$ is continuous at the origin. Thus by the proposition on the Continuity of Seminorms on Vector Spaces page, each seminorm $p \in Q$ is continuous on $E$.
  • If $\tau$ is any other topology on $E$ for which every $p \in Q$ is continuous, then $\tau$ must contain sets of the form $\{ x : p(x) \leq \epsilon \}$. But $\mathcal U$ consists of all finite intersections of sets of these forms, and thus $\tau$ must contain $\mathcal U$. So the topology specified by the base of neighbourhoods $\mathcal U$ of the origin, is the coarsest such topology. $\blacksquare$

We given the topology in the theorem above a special name.

Definition: Let $E$ be a vector space and let $Q$ be a collection of seminorms on $E$. The Coarsest Topology Determined by $Q$ is the coarsest topology on $E$ for which each seminorm in $Q$ is continuous.
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