Disjoint Compact Set and Closed Set in Topological Vector Space separated by Open Neighborhood

Theorem
Let $F$ be a topological field.

Let $X$ be a topological vector space over $F$.

Let $K$ be a compact subspace of $X$.

Let $C \subseteq X$ be a closed set such that:


 * $K \cap C = \O$

Then there exists an open neighborhood $V$ of ${\mathbf 0}_X$ such that:


 * $\paren {K + V} \cap \paren {C + V} = \O$

Proof
If $K = \O$, we have $K + V = \O$, so we have:


 * $\paren {K + V} \cap \paren {C + V} = \O$

from Intersection with Empty Set.

Now take $K \ne \O$.

Let $x \in K$.

Since $K$ and $C$ are disjoint, it follows that $x \in X \setminus C$, so ${\mathbf 0}_X \in \paren {X \setminus C} - x$.

Since $C$ is closed, $X \setminus C$ is open.

Then from Translation of Open Set in Topological Vector Space is Open, $\paren {X \setminus C} - x$ is an open neighborhood of ${\mathbf 0}_X$.

From Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods: Corollary, there exists a symmetric open neighborhood $V'_x$ of ${\mathbf 0}_X$ such that:


 * $V'_x + V'_x \subseteq \paren {X \setminus C} - x$

Applying Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods: Corollary again, there exists a symmetric open neighborhood $V_x$ of ${\mathbf 0}_X$ such that:


 * $V_x + V_x \subseteq V'_x$

so that:


 * $V_x + V_x + V_x + V_x \subseteq \paren {X \setminus C} - x$

so that:


 * $x + V_x + V_x + V_x + V_x \subseteq X \setminus C$

Since ${\mathbf 0}_X \in V_x$, it follows that:


 * $x + V_x + V_x + V_x \subseteq X \setminus C$

We argue that:


 * $\paren {x + V_x + V_x} \cap \paren {C + V_x} = \O$

suppose that:


 * $\paren {x + V_x + V_x} \cap \paren {C + V_x} \ne \O$

Then there exists $u_1, u_2, u_3 \in V_x$ and $c \in C$ such that:


 * $x + u_1 + u_2 = c + u_3$

Then:


 * $c = x + u_1 + u_2 - u_3$

Since $V_x$ is symmetric, we have $-u_3 \in V_x$.

So:


 * $c \in x + V_x + V_x + V_x$

so:


 * $\paren {x + V_x + V_x + V_x} \cap C \ne \O$

contradicting that:


 * $x + V_x + V_x + V_x + V_x \subseteq X \setminus C$

We therefore have:


 * $\paren {x + V_x + V_x} \cap \paren {C + V_x} \ne \O$

for each $x \in K$.

Since ${\mathbf 0}_X \in V_x$, we have:


 * $x \in x + V_x$ for each $x \in K$.

So, we have:


 * $\ds K \subseteq \bigcup_{x \in K} \paren {x + V_x}$

Since $K$ is compact, there exists $x_1, x_2, \ldots, x_n \in K$ such that:


 * $\ds K \subseteq \bigcup_{i \mathop = 1}^n \paren {x_i + V_{x_i} }$

Let:


 * $\ds V = \bigcap_{i \mathop = 1}^n V_{x_i}$

Since this is the finite intersection of open sets, $V$ is open.

From Sum of Union of Subsets of Vector Space and Subset, we then have:


 * $\ds K + V \subseteq \bigcup_{i \mathop = 1}^n \paren {x_i + V_{x_i} } + V \subseteq \bigcup_{i \mathop = 1}^n \paren {x_i + V_{x_i} + V}$

From Intersection is Subset, we then have:


 * $x_i + V_{x_i} + V \subseteq x_i + V_{x_i} + V_{x_i}$

for each $i \in \set {1, 2, \ldots, n}$.

From Set Union Preserves Subsets: Families of Sets, we have:


 * $\ds \bigcup_{i \mathop = 1}^n \paren {x_i + V_{x_i} + V} \subseteq \bigcup_{i \mathop = 1}^n \paren {x_i + V_{x_i} + V_{x_i} }$

Recall that $x_i + V_{x_i} + V_{x_i}$ does not intersect $C + V_{x_i}$.

Since $C + V \subseteq C + V_{x_i}$, we have that $x_i + V_{x_i} + V_{x_i}$ does not intersect $C + V$ for any $i \in \set {1, 2, \ldots, n}$.

So, we have:


 * $\ds K + V \subseteq \bigcup_{i \mathop = 1}^n \paren {x_i + V_{x_i} + V_{x_i} } \subseteq X \setminus \paren {C + V}$

so that:


 * $\paren {K + V} \cap \paren {C + V} = \O$

So $V$ is the desired open neighborhood of ${\mathbf 0}_X$.