Proof of Separation of an Open Convex Set from a Disjoint Convex Set in a Separable Real Normed Space

We separate the origin from the open convex set of differences $A-B$ and then move the level to the supremum of $F$ on $A$.

Let

$$C=\{a-b:a\in A,\ b\in B\}.$$

It is nonempty, as $A$ and $B$ are. It is convex: for $a,a'\in A$, $b,b'\in B$ and $0\le t\le1$,

$$t(a-b)+(1-t)(a'-b')=\bigl(ta+(1-t)a'\bigr)-\bigl(tb+(1-t)b'\bigr)$$

by Elementary Identities in a Vector Space (ix), Elementary Identities in a Vector Space (vi), Vector Space over a Field (i) and Vector Space over a Field (ii), and the two convex combinations on the right lie in $A$ and in $B$ by Convex Subset of a Real Vector Space (i). It is open (Open Subset of a Metric Space (i)). Here we use that $(u+w)-(v+w)=u-v$ and $(u+w)-w=u$ for $u,v,w\in X$ (The Zero Vector, Negatives and Differences in a Vector Space (iii), Elementary Identities in a Vector Space (vi), Elementary Identities in a Vector Space (vii), The Zero Vector, Negatives and Differences in a Vector Space (i), Vector Space over a Field (i), Vector Space over a Field (ii)). Let $a\in A$, $b\in B$ and $\rho>0$ with $B_{d_{X}}(a,\rho)\subseteq A$, and let $z\in B_{d_{X}}(a-b,\rho)$. By The Norm Metric of a Real Normed Space (i), Elementary Identities in a Vector Space (viii) and Open Ball in a Metric Space (ii),

$$d_{X}(a,z+b)=\bigl\lVert\bigl((a-b)+b\bigr)-(z+b)\bigr\rVert=\lVert(a-b)-z\rVert=d_{X}(a-b,z)<\rho ,$$

so $z+b\in A$, and therefore $z=(z+b)-b\in C$. Hence $B_{d_{X}}(a-b,\rho)\subseteq C$. And $0_{X}\notin C$, since $a-b=0_{X}$ would give $a=b+(a-b)=b$ (Elementary Identities in a Vector Space (viii)), contradicting $A\cap B=\emptyset$.

By Separating a Point from an Open Convex Set in a Separable Real Normed Space (i), applied to $C$ and the point $0_{X}\notin C$, there is $F\in X^{*}$ with $F(a-b)<F(0_{X})=0$ for all $a\in A$ and $b\in B$; by Linear Functional on a Linear Subspace (i) and Linear Functional on a Linear Subspace (ii) this says

$$F(a)<F(b)\qquad\text{for all }a\in A,\ b\in B.$$

As $C\neq\emptyset$, $F$ takes a negative value, so it is not identically zero.

Fix $b_{0}\in B$. The set $\{F(a):a\in A\}$ is nonempty and bounded above by $F(b_{0})$, so its supremum

$$\alpha=\sup_{a\in A}F(a)$$

exists by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness (vii). Every $F(b)$ with $b\in B$ is an upper bound of this set, so $\alpha\le F(b)$ by Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order (iv).

It remains to see that $F(a)<\alpha$ for $a\in A$; here the openness of $A$ is used. Fix $z_{0}\in X$ with $F(z_{0})\neq0$; replacing $z_{0}$ by $-z_{0}$ if necessary (Linear Functional on a Linear Subspace (ii)), $F(z_{0})>0$, and $z_{0}\neq0_{X}$ because $F(0_{X})=0$. Let $a\in A$, and fix $\rho>0$ with $B_{d_{X}}(a,\rho)\subseteq A$ (Open Subset of a Metric Space (i)). With $s=\rho/(2\lVert z_{0}\rVert)$ we have $\lVert s z_{0}\rVert=\rho/2<\rho$, so $a+sz_{0}\in A$, and

$$F(a)<F(a)+sF(z_{0})=F(a+sz_{0})\le\alpha .$$

Hence $F(a)<\alpha\le F(b)$ for all $a\in A$ and $b\in B$, which proves (i).