Proof of Strict Separation of a Closed Convex Set from a Compact Convex Set in a Separable Real Normed Space

We fatten $A$ and $B$ by a third of their distance into disjoint open convex sets and separate those. Throughout, $x-a=(x-y)+(y-a)$ (Elementary Identities in a Vector Space (x)), $a+z=b+z'$ gives $b-a=z-z'$ (Elementary Identities in a Vector Space (i), Elementary Identities in a Vector Space (viii)), and $\lVert z-z'\rVert\le\lVert z\rVert+\lVert z'\rVert$ (Norms and Real Normed Spaces (iv), and Norms and Real Normed Spaces (iii) with $\lambda=-1$ via Elementary Identities in a Vector Space (iv)).

Distance to $A$. For $x\in X$ the set $\{\lVert x-a\rVert:a\in A\}$ is nonempty and bounded below by $0$ (Norms and Real Normed Spaces (i)), so

$$h(x)=\inf_{a\in A}\lVert x-a\rVert$$

exists by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness (viii); this defines a function $h:X\to\mathbb{R}$. For $x,y\in X$ and $a\in A$ we have $h(x)\le\lVert x-a\rVert$, an infimum being a lower bound (Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order (iv)), and $\lVert x-a\rVert\le\lVert x-y\rVert+\lVert y-a\rVert$ (Norms and Real Normed Spaces (iv)). So $h(x)-\lVert x-y\rVert$ is a lower bound of $\{\lVert y-a\rVert:a\in A\}$, and $h(x)\le h(y)+\lVert x-y\rVert$ by Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order (iv). Exchanging $x$ and $y$,

$$|h(x)-h(y)|\le\lVert x-y\rVert\qquad\text{for all }x,y\in X.$$

$h$ is positive on $B$. By Closed Subset of a Topological Space (i) and Metric Spaces: Balls, the Metric Topology, and the Real and Complex Numbers as Metric Spaces (iii), $X\setminus A$ is open. Let $b\in B$. Then $b\notin A$, so Open Subset of a Metric Space (i) gives $\rho>0$ with $B_{d_{X}}(b,\rho)\cap A=\emptyset$. Hence $\lVert b-a\rVert\ge\rho$ for every $a\in A$ (Open Ball in a Metric Space (ii)), so $\rho$ is a lower bound and $h(b)\ge\rho>0$ by Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order (iv).

The minimum distance. By the estimate above, the restriction of $h$ to $B$ is continuous on $B$: given $\eta>0$, take $\delta=\eta$ in Continuous Map Between Metric Spaces (i), values being compared in the distance $d_{\mathbb{R}}(h(x),h(y))=|h(x)-h(y)|$. Since $B$ is nonempty and compact, Extreme Value Theorem on a Compact Subset of a Metric Space (i) gives $b_{\min}\in B$ with $h(b_{\min})\le h(b)$ for every $b\in B$. Put

$$\gamma=h(b_{\min})>0,\qquad r=\gamma/3 .$$

Open convex neighbourhoods. Let

$$A_{r}=\{a+z:a\in A,\ \lVert z\rVert<r\},\qquad B_{r}=\{b+z:b\in B,\ \lVert z\rVert<r\}.$$

Both are nonempty. Both are convex (Convex Subset of a Real Vector Space (i)): a convex combination of $a+z$ and $a'+z'$ is the same convex combination of $a$ and $a'$ plus $tz+(1-t)z'$, and $\lVert tz+(1-t)z'\rVert<r$ by Norms and Real Normed Spaces (iv) and Norms and Real Normed Spaces (iii). Both are open (Open Subset of a Metric Space (i)): if $\lVert z\rVert<r$, the ball of radius $r-\lVert z\rVert$ about $a+z$ lies in $A_{r}$, and likewise for $B_{r}$. They are disjoint: if $a+z=b+z'$, then

$$\lVert b-a\rVert=\lVert z-z'\rVert<2r<\gamma\le h(b)\le\lVert b-a\rVert,$$

the last step because an infimum is a lower bound (Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order (iv)), which is absurd.

Separation. By Separation of an Open Convex Set from a Disjoint Convex Set in a Separable Real Normed Space (i), applied to the open set $A_{r}$ and to $B_{r}$, there are $F\in X^{*}$, not identically zero, and a real number $\alpha$ with

$$F(a+z)<\alpha\le F(b+z')\qquad\text{whenever }a\in A,\ b\in B,\ \lVert z\rVert<r,\ \lVert z'\rVert<r .$$

A strict gap. Fix $z_{0}\in X$ with $F(z_{0})\neq0$, replaced by $-z_{0}$ if necessary so that $F(z_{0})>0$; then $z_{0}\neq0_{X}$. The vector $w=\bigl(r/(2\lVert z_{0}\rVert)\bigr)z_{0}$ satisfies $\lVert w\rVert=r/2<r$ and $F(w)>0$ (Linear Functional on a Linear Subspace (ii), Norms and Real Normed Spaces (iii)); put $\varepsilon=F(w)$. Taking $z=w$ and $z'=-w$ above and using Linear Functional on a Linear Subspace (i) and Linear Functional on a Linear Subspace (ii),

$$F(a)+\varepsilon<\alpha\le F(b)-\varepsilon\qquad\text{for all }a\in A,\ b\in B .$$

So $F(a)\le\alpha-\varepsilon$ and $F(b)\ge\alpha+\varepsilon$, which proves (i).

Half-spaces. Every $x\in A$ lies in the right-hand side of (ii) by its definition. Conversely, let $x_{0}\in X\setminus A$; we find $F\in X^{*}$ and $\alpha$ with $F\le\alpha$ on $A$ and $F(x_{0})>\alpha$. If $A=\emptyset$, the zero functional with $\alpha=-1$ will do. Otherwise, the set $\{x_{0}\}$ is nonempty and disjoint from $A$; it is convex (Convex Subset of a Real Vector Space (i)), since $t\,x_{0}+(1-t)\,x_{0}=1x_{0}=x_{0}$ for $0\le t\le1$ by Vector Space over a Field (viii) and Vector Space over a Field (vi); and it is compact (Compact Topological Space and Compact Subset (ii), Compact Topological Space and Compact Subset (i)), since in any cover by open sets of the subspace some member contains $x_{0}$ and is by itself a finite subcover. By (i), applied to $A$ and $\{x_{0}\}$, there are $F\in X^{*}$ and real numbers $\alpha$ and $\varepsilon>0$ with $F(a)\le\alpha-\varepsilon\le\alpha$ for every $a\in A$ and $F(x_{0})\ge\alpha+\varepsilon>\alpha$. So $x_{0}$ is not in the right-hand side.