Relative geometric Hahn–Banach with the exact open, closed, and compact hypotheses

For (i), put $D=A-B=\{x-y:x\in A,y\in B\}$. It is nonempty. For $x_j-y_j\in D$ and $0\le t\le 1$, their convex combination equals $((1-t)x_1+t{x}_2)-((1-t)y_1+t{y}_2)\in D$. If $x-y\in D$, choose $r>0$ with $B(x,r)\subseteq A$; then $B(x-y,r)\subseteq D$ by keeping $y$ fixed. Thus $D$ is open and convex. If $0\in D$, then some $x\in A$ equals some $y\in B$, contrary to disjointness; hence $0\mathrm{\notin }D$.

For (ii), now suppose $A$ is closed and $B$ compact. The distance-gap lemma applied with $K=B,C=A$ gives $\delta >0$ with $\mathrm{\parallel }x-y\mathrm{\parallel }\ge \delta$ for all $x\in A,y\in B$. Put $\rho =\delta \mathrm{/}2$ and $O=A+B(0,\rho )$. The ball is convex by the triangle inequality, so for $x_j+h_j\in O$ each convex combination has its $A$ component in $A$ and its ball component of norm less than $\rho$, including the weights zero and one. Thus $O$ is convex. A ball about $x+h$ of radius $\rho -\mathrm{\parallel }h\mathrm{\parallel }$ stays in $O$, so $O$ is open; it contains nonempty $A$. If $x+h=y\in B$, then $\mathrm{\parallel }x-y\mathrm{\parallel }=\mathrm{\parallel }h\mathrm{\parallel }<\rho <\delta$, impossible. Therefore $O$ and $B$ are disjoint.

For (i), apply point separation to $D$ and $0$. It gives $0\mathrm{\ne }f\in X^{\ast }$ with $u(d)<u(0)=0$ for every $d\in D$, where $u=\mathrm{Re}f$ is nonzero bounded real-linear. Consequently $u(x)<u(y)$ for every $x\in A,y\in B$. Fix $y_0\in B$. The nonempty real set $u(A)$ is bounded above by $u(y_0)$, so it has a real supremum $a_0$. Explicitly, $a_0=-\inf (-u(A))$; reflection of lower bounds makes this the least upper bound. Since every $u(y)$ bounds $u(A)$, $a_0\le u(y)$ for all $y\in B$.

For (i), fix a vector $v$ with $u(v)>0$: nonzero $u$ has a nonzero value, and negation makes that value positive. For each fixed $x\in A$, choose $r_x>0$ with $B(x,r_x)\subseteq A$ and set $t=r_x\mathrm{/}(2\mathrm{\parallel }v\mathrm{\parallel })>0$. Then $x+tv\in A$ and $u(x)<u(x)+tu(v)=u(x+tv)\le a_0$. This proves the strict left inequality. If $B$ is open, for each $y\in B$ choose $s_y>0$ with $B(y,s_y)\subseteq B$ and set $t=s_y\mathrm{/}(2\mathrm{\parallel }v\mathrm{\parallel })$. Then $y-tv\in B$, and step 2.1 gives $a_0\le u(y-tv)<u(y)$. Thus both inequalities are strict in that case.

If only $B$ is open, apply the result just proved to $(B,A)$ to get a functional $g$ and level $b$ with $\mathrm{Re}g(y)<b\le \mathrm{Re}g(x)$ for $y\in B,x\in A$. Taking $f=-g$ and $a=-b$ gives $\mathrm{Re}f(x)\le a<\mathrm{Re}f(y)$. This completes (i) in each orientation.

For (ii), apply the already proved open-side assertion to $(O,B)$. It gives $0\mathrm{\ne }f\in X^{\ast }$ and $a_0\in \mathbb{R}$ with $u(x)+u(h)<a_0\le u(y)$ whenever $x\in A$, $\mathrm{\parallel }h\mathrm{\parallel }<\rho$, and $y\in B$, where $u=\mathrm{Re}f\mathrm{\ne }0$. Regard $u$ as a member of the real dual of the underlying normed space, and let $N=\mathrm{\parallel }u\mathrm{\parallel }={\sup }_{\mathrm{\parallel }v\mathrm{\parallel }\le 1}\mathrm{\mid }u(v)\mathrm{\mid }$. The bound $\mathrm{\mid }u(v)\mathrm{\mid }\le \mathrm{\parallel }f\mathrm{\parallel }\mathrm{\parallel }v\mathrm{\parallel }$ makes this supremum finite, and a normalized vector with nonzero value shows $N>0$.

Continuing (ii), we show ${\sup }_{\mathrm{\parallel }h\mathrm{\parallel }<\rho }u(h)=\rho N$. Normalization gives $u(h)\le \mathrm{\mid }u(h)\mathrm{\mid }\le N\mathrm{\parallel }h\mathrm{\parallel }\le \rho N$ (including $h=0$). For any $q<\rho N$ with $q<0$, $h=0$ has $u(h)>q$. For $0\le q<\rho N$, use the supremum criterion for $N$ with error $N-q\mathrm{/}\rho >0$ to obtain $v$ with $\mathrm{\parallel }v\mathrm{\parallel }\le 1$ and $\mathrm{\mid }u(v)\mathrm{\mid }>q\mathrm{/}\rho$. Let $w=v$ if $u(v)>0$ and $w=-v$ otherwise; then $m=u(w)=\mathrm{\mid }u(v)\mathrm{\mid }>0$. Set $t=(q\mathrm{/}m+\rho )\mathrm{/}2$, which satisfies $0<t<\rho$ and $tm>q$. Thus $h=tw$ has $\mathrm{\parallel }h\mathrm{\parallel }<\rho$ and $u(h)>q$. No number smaller than $\rho N$ is an upper bound, proving the identity.

For (ii), for each fixed $x\in A$, step 4.2 says $a_0-u(x)$ is an upper bound for all $u(h)$ with $\mathrm{\parallel }h\mathrm{\parallel }<\rho$. The identity just proved yields $u(x)+\rho N\le a_0$. Put $\epsilon =\rho N\mathrm{/}2>0$ and $a=a_0-\epsilon$. Then $u(x)\le a_0-\rho N=a-\epsilon$, while $a+\epsilon =a_0\le u(y)$ for every $y\in B$. Since $\epsilon >0$, these are precisely the required uniform strict separation inequalities.

Finally, if $z$ lies outside a nonempty closed convex $A$, the set $B=\{z\}$ is nonempty, disjoint from $A$, and convex since $(1-t)z+tz=z$. It is intrinsically compact: any open cover of its one-point metric space has a member containing $z$, and that one member is a finite subcover. Thus the hypotheses of (ii) hold and steps 1.2, 4.2, 5.1 and 6.1 give the final specialization.