Open Sets in Standard Topology of Locally Convex Space

Theorem
Let $\GF \in \set {\R, \C}$.

Let $\struct {X, \PP}$ be a locally convex space over $\GF$.

Let $\tau$ be the standard topology on $\struct {X, \PP}$.

Let $U \subseteq X$.

Then $U \in \tau$ for each $x \in U$ there exists $n \in \N$, $p_1, \ldots, p_n \in \PP$ and $\epsilon > 0$ such that:


 * $\set {y \in X : \map {p_k} {y - x} < \epsilon \text { for each } 1 \le k \le n} \subseteq U$

Sufficient Condition
Let $U \subseteq X$ and suppose that:


 * for each $x \in U$ there exists $n \in \N$, $p_1, \ldots, p_n \in \PP$ and $\epsilon > 0$ such that:


 * $U_x = \set {y \in X : \map {p_k} {y - x} < \epsilon \text { for each } 1 \le k \le n} \subseteq U$

For each $p \in \PP$, $\epsilon > 0$ and $y \in X$, define:


 * $\map {B_p} {\epsilon, x} = \set {y \in X : \map p {y - x} < \epsilon}$

so that:


 * $\ds U_x = \set {y \in X : \map {p_k} {y - x} < \epsilon \text { for each } 1 \le k \le n} = \bigcap_{k \mathop = 1}^n \map {B_{p_k} } {\epsilon, x}$

From the definition of the standard topology:


 * $\SS = \set {\map {B_p} {\epsilon, x} : p \in \PP, \, \epsilon > 0, \, x \in X}$

generates $\tau$.

Since topologies are closed under finite intersection, we have that $U_x \in \tau$.

Further, since $\tau$ is closed under set union, we have:


 * $\ds \bigcup_{x \mathop \in U} U_x \in \tau$

Since $x \in U_x$ for each $x \in U$, we have:


 * $\ds U \subseteq \bigcup_{x \mathop \in U} U_x$

On the other hand, since $U_x \subseteq U$ for each $x \in U$, we have:


 * $\ds \bigcup_{x \mathop \in U} U_x \subseteq U$

from Union of Subsets is Subset.

So we obtain:


 * $\ds U = \bigcup_{x \mathop \in U} U_x$

That is, $U \in \tau$.

Necessary Condition
Suppose that $U \in \tau$ and $x \in U$.

From the definition of the generated topology:


 * there exists an indexing set $A$ such that:


 * for each $\alpha \in A$ there exists $p_{\alpha, 1}, \ldots, p_{\alpha, n_\alpha} \in \PP$, $\epsilon_{\alpha, 1}, \ldots, \epsilon_{\alpha, n_\alpha} > 0$ and $x_{\alpha, 1}, \ldots, x_{\alpha, n_\alpha} \in X$ such that:


 * $\ds U = \bigcup_{\alpha \mathop \in A} \bigcap_{k \mathop = 1}^{n_\alpha} \map {B_{p_{\alpha, k} } } {\epsilon_{\alpha, k}, x_{\alpha, k} }$

Since $x \in U$, there therefore exists $\alpha \in A$ such that:


 * $\ds x \in \bigcap_{k \mathop = 1}^{n_\alpha} \map {B_{p_{\alpha, k} } } {\epsilon_{\alpha, k}, x_{\alpha, k} }$

That is:


 * $\map {p_{\alpha, k} } {x_{\alpha, k} - x} < \epsilon_{\alpha, k}$ for each $1 \le k \le n$.

Now note that if $y \in X$ has:


 * $\map {p_{\alpha, k} } {y - x} < \epsilon_{\alpha, k} - \map {p_{\alpha, k} } {x_{\alpha, k} - x}$

we have:


 * $\map {p_{\alpha, k} } {y - x_{\alpha, k} } \le \map {p_{\alpha, k} } {y - x} + \map {p_{\alpha, k} } {x - x_{\alpha, k} } < \epsilon_{\alpha, k}$

So, take:


 * $\ds \epsilon = \min_{1 \le k \le n_\alpha} \paren {\epsilon_{\alpha, k} - \map {p_{\alpha, k} } {x_{\alpha, k} - x} }$

Then for $y \in X$ and each $1 \le k \le n$, we have that:


 * $\map {p_{\alpha, k} } {y - x} < \epsilon$

implies:


 * $\map {p_{\alpha, k} } {y - x_{\alpha, k} } < \epsilon \le \epsilon_{\alpha, k}$

so that:


 * $\ds y \in \bigcap_{k \mathop = 1}^{n_\alpha} \map {B_{p_{\alpha, k} } } {\epsilon_{\alpha, k}, x_{\alpha, k} }$

So, we have:


 * $\ds \bigcap_{k \mathop = 1}^{n_\alpha} \map {B_{p_{\alpha, k} } } {\epsilon, x} \subseteq \bigcap_{k \mathop = 1}^{n_\alpha} \map {B_{p_{\alpha, k} } } {\epsilon_{\alpha, k}, x_{\alpha, k} } \subseteq U$

That is, setting $p_k = p_{\alpha, k}$ for each $1 \le k \le n$, and setting $n = n_\alpha$, we obtain:


 * $\set {y \in X : \map {p_k} {y - x} < \epsilon \text { for each } 1 \le k \le n} \subseteq U$

as was desired.