Locally Convex Spaces and Continuous Separation

Constant maps are continuous because the preimage of an open set is empty or the whole domain; identity maps are continuous by their preimages. Thus $x\mapsto (a,x)$ and $x\mapsto (x,a)$ into the appropriate products are continuous. Composing with the structure maps proves continuity of translations $x\mapsto x+a$, fixed dilations $x\mapsto bx$, and the orbit maps $s\mapsto sx$ for fixed $x$.

Translation by $-a$ inverts translation by $a$; when $b\mathrm{\ne }0$, dilation by $b^{-1}$ inverts dilation by $b$. The vector axioms and zero/negative identities verify these inverse formulas. The inverses are continuous by step 1.1, so these maps are homeomorphisms. Consequently translates of open sets and nonzero dilates of open sets are open.

Fix an open $U_0$ with $0\in U_0\subseteq U$. Continuity of $s\mapsto sx$ at $0$, where $0x=0$, gives $\delta >0$ such that $\mathrm{\mid }s\mathrm{\mid }<\delta$ implies $sx\in U_0$. For every positive real $t>1\mathrm{/}\delta$, $(1\mathrm{/}t)x\in U$ and hence $x\in tU$. For $x=0$ every positive $t$ works.

Joint addition continuity at $(0,0)$ gives open zero-neighborhoods $U_1,U_2$ with $U_1+U_2\subseteq U_0$. Put $W=U_1\cap U_2\cap (-U_1)\cap (-U_2)$. This is open, contains zero, satisfies $-W=W$, and has $W+W\subseteq U_1+U_2\subseteq U$.

For $\epsilon >0$ put $c=\epsilon \mathrm{/}(M+1)>0$. The set $cN$ is a zero-neighborhood by step 2.1, and $h=cn$ in it satisfies $\mathrm{\mid }f(h)\mathrm{\mid }=c\mathrm{\mid }f(n)\mathrm{\mid }\le cM<\epsilon$. Thus $f$ is continuous at zero. At $x$, the neighborhood $x+cN$ maps into the $\epsilon$-ball about $f(x)$ because $f(x+h)-f(x)=f(h)$. This includes $M=0$ and the zero functional.

For scalar addition at $(a,b)$, errors $\mathrm{\mid }s-a\mathrm{\mid },\mathrm{\mid }t-b\mathrm{\mid }<\epsilon \mathrm{/}2$ give $\mathrm{\mid }(s+t)-(a+b)\mathrm{\mid }<\epsilon$. For multiplication, if $\mathrm{\mid }t-b\mathrm{\mid }<1$ then $\mathrm{\mid }st-ab\mathrm{\mid }\le \mathrm{\mid }s-a\mathrm{\mid }\mathrm{\mid }t\mathrm{\mid }+\mathrm{\mid }a\mathrm{\mid }\mathrm{\mid }t-b\mathrm{\mid }\le (\mathrm{\mid }b\mathrm{\mid }+1)\mathrm{\mid }s-a\mathrm{\mid }+\mathrm{\mid }a\mathrm{\mid }\mathrm{\mid }t-b\mathrm{\mid }\mathrm{.}$ Taking both errors below $\min (1,\epsilon \mathrm{/}(2(\mathrm{\mid }a\mathrm{\mid }+\mathrm{\mid }b\mathrm{\mid }+1)))$ makes this less than $\epsilon$. These are product-open neighborhoods, so they prove joint topological continuity, including $a=0$ or $b=0$. Together with the preceding steps this proves all assertions, without a choice principle or a separation axiom.

Let $x,y\in \bar{C}$ for convex $C$, and $0<t<1$. The affine map $(a,b)\mapsto (1-t)a+tb$ is continuous by the vector operations. For an open neighborhood $O$ of $(1-t)x+ty$, its preimage contains a product neighborhood $P\times Q$ of $(x,y)$. There exist $a\in P\cap C$ and $b\in Q\cap C$ by the closure test. Their convex combination is in $O\cap C$. Thus $(1-t)x+ty\in \bar{C}$. At $t=0,1$ the assertion follows from membership of $x,y$; for $C=\mathrm{\varnothing }$ its closure is empty.

For $x,y\in \mathrm{int}C$ and $0<t<1$, the open set $(1-t)\mathrm{int}C+t\mathrm{int}C$ contains their combination and lies in $C$. It is open as a union of translates of a nonzero dilate of an open set. The endpoints $t=0,1$ are immediate, and an empty interior is convex vacuously. If $y\in \mathrm{int}C$ and $x\in C$, then for $0<s\le 1$ the open set $(1-s)x+s\mathrm{int}C$ lies in $C$. Hence $(1-s)x+sy$ belongs to its interior. Continuity of the orbit at $s=0$ shows every neighborhood of $x$ contains such a point, so $x\in \overline{\mathrm{int}C}$.

Let $C$ be balanced. For $0<\mathrm{\mid }\lambda \mathrm{\mid }\le 1$, the homeomorphism $D_{\lambda }:x\mapsto \lambda x$ carries $\bar{C}$ onto $\overline{\lambda C}$. Indeed, pull an open neighborhood back by $D_{\lambda }$ for one inclusion and use its inverse for the other. Since $\lambda C\subseteq C$, the closure test gives $\overline{\lambda C}\subseteq \bar{C}$. For $\lambda =0$ and $C\mathrm{\ne }\mathrm{\varnothing }$, balance gives $0\in C$, so $0\bar{C}=\{0\}\subseteq \bar{C}$; for empty $C$ the dilation has empty image. Thus the closure is balanced.

Suppose $n\ge 1$ and each $K_j$ is nonempty. The simplex $\mathrm{\Delta }=\{t\in {\mathbb{R}}^n:t_j\ge 0,{\sum }_j{t}_j=1\}$ is closed: coordinate maps and their finite sum are continuous, and the conditions are inverse images of closed real rays and $\{1\}$. It is bounded since $0\le t_j\le 1$ and ${\sum }_j{t}_j^2\le n$. It is compact by Heine–Borel. Finite product compactness makes $\mathrm{\Delta }\times {\prod }_j{K}_j$ compact. The map $\mathrm{\Phi }(t,x)={\sum }_j{t}_j{x}_j$ is continuous: coordinate projections and inclusions are continuous by preimages of basic opens, scalar multiplication is jointly continuous, and iterating addition preserves continuity. Therefore its image is compact.

Every value of $\mathrm{\Phi }$ is a convex combination from the union. Conversely, write a hull point as ${\sum }_{\mathrm{\ell }=1}^m{a}_{\mathrm{\ell }}{y}_{\mathrm{\ell }}$. Assign each $y_{\mathrm{\ell }}$ the least index $j$ for which $y_{\mathrm{\ell }}\in K_j$, and let $t_j$ be the sum of the weights with that label. For $t_j>0$, their normalized combination $x_j=t_j^{-1}{\sum }_{\mathrm{\ell }\text{ labelled }j}{a}_{\mathrm{\ell }}{y}_{\mathrm{\ell }}$ belongs to $K_j$ by finite convexity. For the finitely many zero $t_j$, choose any $x_j\in K_j$ using finite choice. Then $t\in \mathrm{\Delta }$ and ${\sum }_j{t}_j{x}_j={\sum }_{\mathrm{\ell }}{a}_{\mathrm{\ell }}{y}_{\mathrm{\ell }}$. This proves equality of the two sets.

The hull is compact by step 1.4 and step 2.1, and Hausdorffness gives closedness. Each singleton is compact since any cover has one member covering its sole point, and convex since its every combination is that point; hence finite point hulls are covered. Delete empty members of a finite family in their original order. If none remain, its union and hull are empty, and the empty subcover proves compactness. For one nonempty convex member the hull is that member. All choices made above are finite.

Choose a symmetric open zero-neighborhood $O$ with $O+O\subseteq U$, then an open convex zero-neighborhood $C\subseteq O$. Joint scalar continuity at $(0,0)$ supplies $\delta >0$ and an open zero-neighborhood $W$ with $\{a:\mathrm{\mid }a\mathrm{\mid }<\delta \}W\subseteq C$. Put $W_0=(\delta \mathrm{/}2)W$, which is open and contains zero.

Define $B={\bigcup }_{\mathrm{\mid }a\mathrm{\mid }\le 1}a{W}_0$. If $aw\in B$ and $\mathrm{\mid }\lambda \mathrm{\mid }\le 1$, then $\lambda (aw)=(\lambda a)w\in B$, so $B$ is balanced. It contains $W_0$. For $w=(\delta \mathrm{/}2)v\in W_0$, $aw=(a\delta \mathrm{/}2)v\in C$ since $\mathrm{\mid }a\delta \mathrm{/}2\mathrm{\mid }<\delta$, so $B\subseteq C$. Moreover $B$ is open: all nonzero dilates are open, and the zero dilate contributes only zero, which already belongs to $W_0$.

Put $V=\mathrm{co}(B)$. Then $W_0\subseteq B\subseteq V\subseteq C\subseteq O$, and $V$ is convex. For $\mathrm{\mid }\lambda \mathrm{\mid }\le 1$, distributing $\lambda$ through a finite convex sum leaves its summands in $B$, proving balance of $V$. To prove openness, represent $v={\sum }_j{t}_j{b}_j\in V$. Some $t_k>0$, and $t_{k}B+{\sum }_{j\mathrm{\ne }k}{t}_j{b}_j$ is an open subset of $V$ containing $v$. Thus $V$ is an open balanced convex zero-neighborhood. Zero coefficients cause no problem because the sum of the coefficients is one.

If $x\in \bar{V}$, then $x-O$ is an open neighborhood of $x$ and meets $V$. Write $v=x-o\in V$ with $o\in O$; hence $x=v+o\in V+O\subseteq O+O\subseteq U$. Thus $\bar{V}\subseteq U$. By closure preservation, $\bar{V}$ is convex and balanced; it is closed and contains the open zero-neighborhood $V$. Finally balance implies $-V\subseteq V$ and applying negation again gives $V=-V$; the same applies to every balanced set, including the empty set.

Write $S_x=\{t>0:x\in tU\}$. If $t\in S_x$ and $a\ge t$, then $x\mathrm{/}a=(t\mathrm{/}a)(x\mathrm{/}t)+(1-t\mathrm{/}a)0\in U$, so $a\in S_x$. If $a>p(x)$, it cannot be a lower bound of $S_x$; hence some $t\in S_x$ satisfies $t<a$, and then $a\in S_x$. This uses only the defining greatest-lower-bound property, not an assumption that the infimum is attained.

For $r>0$, $S_{rx}=r{S}_x$ by direct substitution, so $p(rx)=rp(x)$: multiplication by $r$ bijects lower bounds of $S_x$ with lower bounds of $r{S}_x$ and preserves their order. At $r=0$, both sides are zero. For $\eta >0$ put $a=p(x)+\eta$ and $b=p(y)+\eta$. These are positive admissible numbers, and $(x+y)\mathrm{/}(a+b)=\frac{a}{a+b}(x\mathrm{/}a)+\frac{b}{a+b}(y\mathrm{/}b)\in U\mathrm{.}$ Thus $p(x+y)\le p(x)+p(y)+2\eta$ for every $\eta >0$. If subadditivity failed by a positive gap $d$, take $\eta =d\mathrm{/}4$ to contradict this bound. Hence $p$ is sublinear.

If $p(x)<1$, choose $a$ with $p(x)<a<1$, for example $(p(x)+1)\mathrm{/}2$. It is admissible, and convexity with zero gives $x\in aU\subseteq U$. Conversely, if $x\in U$, continuity of $s\mapsto sx$ at $s=1$ and openness of $U$ give an $\eta >0$ with $(1+\eta )x\in U$. Thus $1\mathrm{/}(1+\eta )\in S_x$, and $p(x)\le 1\mathrm{/}(1+\eta )<1$. These prove both inclusions of the strict-sublevel identity.

Given $\epsilon >0$, the open zero-neighborhood $N=\epsilon (U\cap (-U))$ has $p(h)<\epsilon$ and $p(-h)<\epsilon$ for $h\in N$, by homogeneity and the strict-sublevel identity. Subadditivity gives $p(x+h)-p(x)\le p(h)$ and $p(x)-p(x+h)\le p(-h)$, hence $\mathrm{\mid }p(x+h)-p(x)\mathrm{\mid }<\epsilon$. Translating $N$ proves continuity at every $x$. This argument does not assert absolute domination by an asymmetric gauge.

Suppose $U$ is balanced. For $\mathrm{\mid }a\mathrm{\mid }=1$, balance gives $aU\subseteq U$ and $a^{-1}U\subseteq U$, whence $aU=U$ and $S_{ax}=S_x$. For $\lambda \mathrm{\ne }0$, write $\lambda =\mathrm{\mid }\lambda \mathrm{\mid }a$ with $\mathrm{\mid }a\mathrm{\mid }=1$, and obtain $p(\lambda x)=\mathrm{\mid }\lambda \mathrm{\mid }p(ax)=\mathrm{\mid }\lambda \mathrm{\mid }p(x)$. At $\lambda =0$ this follows from $p(0)=0$. Thus the finite nonnegative continuous sublinear $p$ is a seminorm.

For concrete boundary calculations on the real line, $U=(-1,1)$ gives $S_x=(\mathrm{\mid }x\mathrm{\mid },\mathrm{\infty })$ and $p(x)=\mathrm{\mid }x\mathrm{\mid }$, with $S_0=(0,\mathrm{\infty })$. On ${\mathbb{R}}^2$ the open strip $U=\{(x,y):\mathrm{\mid }x\mathrm{\mid }<1\}$ gives $p_U(x,y)=\mathrm{\mid }x\mathrm{\mid }$, since admissibility is exactly $t>\mathrm{\mid }x\mathrm{\mid }$; thus $p_U(0,1)=0$ despite $(0,1)\mathrm{\ne }0$. These computations show why no positive-definiteness conclusion is available. All claimed properties are established without HB or AC.

First work over the reals and let $D$ be nonempty open convex with $z\mathrm{\notin }D$. Fix $d_0\in D$ and put $U=D-d_0$ and $v=z-d_0$. Then $U$ is an open convex zero-neighborhood, $v\mathrm{\ne }0$, and $p_U(v)\ge 1$ because $v\mathrm{\notin }U$. The set $M=\{tv:t\in \mathbb{R}\}$ is a real subspace: sums and real multiples remain on the line. If $tv=sv$ then multiplying $(t-s)v=0$ by $(t-s{)}^{-1}$ when $t\mathrm{\ne }s$ would give $v=0$, so the coefficient is unique. Thus $h(tv)=t$ is well-defined and real-linear.

For $t\ge 0$, $h(tv)=t\le t{p}_U(v)=p_U(tv)$; for $t<0$, $h(tv)=t<0\le p_U(tv)$. All hypotheses of the relative extension theorem are now met. Apply HB once to obtain real-linear $H:X\to \mathbb{R}$ extending $h$ with $-p_U(-x)\le H(x)\le p_U(x)$. In particular $H(v)=1$. This is the only non-ZF input in the proof.

On the open zero-neighborhood $U\cap (-U)$ both gauges $p_U(x),p_U(-x)$ are less than one. Hence $\mathrm{\mid }H(x)\mathrm{\mid }<1$ there, so $H$ is continuous by the modulus-bound criterion. For $d\in D$, $H(d)-H(d_0)=H(d-d_0)\le p_U(d-d_0)<1=H(z)-H(d_0)$. Therefore $H(d)<H(z)$, and $H\mathrm{\ne }0$ since $H(v)=1$.

For the given real $A,B$, take $D=A-B$ and $z=0$. This is open, since it is the union of the translates $A-b$ for $b\in B$; it is convex by distributing each real convex combination through the difference. It is nonempty and excludes zero by disjointness. The preceding construction therefore produces a nonzero continuous real-linear $H$ with $H(a-b)<0$, or $H(a)<H(b)$ for every $a,b$. It also produces a vector $v$ with $H(v)=1$.

The set $-H(A)$ is nonempty and bounded below by $-H(b_0)$ for any fixed $b_0\in B$. Let $\alpha =-\inf (-H(A))$. By reversing the defining lower-bound inequalities, $\alpha$ is the least upper bound of $H(A)$. Every $H(b)$ is an upper bound, so $\alpha \le H(b)$. For $a\in A$, openness and continuity of $s\mapsto a+sv$ give a positive $s$ with $a+sv\in A$; hence $H(a)<H(a)+s\le \alpha$. If $B$ is open and $H(b)=\alpha$, a small $s>0$ with $b-sv\in B$ would give $\alpha \le H(b-sv)=\alpha -s$, an impossibility. Thus both inequalities are strict when both sets are open.

For complex $X$, restrict scalars to $\mathbb{R}$. The scalar inclusion $\mathbb{R}\to \mathbb{C}$ is continuous since it preserves distance, so the restricted scalar action is jointly continuous. Convexity and openness are unchanged. Apply the real construction to obtain $H$ and $\alpha$. Set $f(x)=H(x)-iH(ix)$. It is additive and real-homogeneous, and $f(ix)=H(ix)-iH(-x)=H(ix)+iH(x)=if(x)$. For $\lambda =a+ib$, additivity and real homogeneity therefore give $f(\lambda x)=af(x)+bf(ix)=\lambda f(x)$. Both $H$ and $x\mapsto H(ix)$ are continuous; scalar addition and multiplication are continuous, so $f$ is continuous. Its real part is $H$, so it is nonzero and obeys the same inequalities. This proves the complex case as well, with the same single HB application and no assumption of AC.

Fix $x\mathrm{\ne }y$ and put $v=x-y\mathrm{\ne }0$. Hausdorffness gives an open neighborhood $U$ of zero not containing $v$, by taking disjoint open neighborhoods of zero and $v$. Refine $U$ to an open convex zero-neighborhood $V\subseteq U$. The singleton $\{v\}$ is convex and nonempty, and misses $V$.

Apply open separation to $V$ and $\{v\}$. There are $f\in X^{\mathrm{\prime }}$ and $\alpha \in \mathbb{R}$ with $0=\mathrm{Re}f(0)<\alpha \le \mathrm{Re}f(v)$. Consequently $\mathrm{Re}f(x)-\mathrm{Re}f(y)=\mathrm{Re}f(v)>0$. The only non-ZF input is the HB application inside that separation theorem.

Every linear functional vanishes at zero, so zero belongs to the intersection of the kernels. For nonzero $v$, step 2.1 applied to $x=v,y=0$ gives a functional with nonzero real part at $v$, hence $v$ does not belong to that intersection. This proves the kernel identity from point separation.

Conversely, assume the kernel identity and fix $x\mathrm{\ne }y$, with $v=x-y$. Some $f\in X^{\mathrm{\prime }}$ has $z=f(v)\mathrm{\ne }0$. Over $\mathbb{R}$ this already separates real parts. Over $\mathbb{C}$, if $\mathrm{Re}z\mathrm{\ne }0$ again use $f$; otherwise $\mathrm{Im}z\mathrm{\ne }0$, and $g=-if\in X^{\mathrm{\prime }}$ has $\mathrm{Re}g(v)=\mathrm{Im}z\mathrm{\ne }0$. Thus the kernel identity implies the stated real-part separation. For $X=\{0\}$ there are no distinct points, and the kernel identity still holds. The proof selects a functional only for one fixed pair, never a simultaneous family.

Consider all pairs $(k,N)$ with $k\in K$ and $N$ an open convex zero-neighborhood such that $k+N\subseteq X\setminus C$. There is such a pair above every $k$: since $C$ is closed, $(X\setminus C)-k$ is an open zero-neighborhood, and it admits an open convex refinement. Each $G_{k,N}=K\cap (k+\frac{1}{2}N)$ is open in $K$ and contains $k$. The family of all such sets covers $K$; it is defined by a property, without choosing a neighborhood for every $k$.

Compactness gives finitely many cover members $G_1,\dots ,G_m$ with $m\ge 1$. For each $j$ its set of representing pairs $(k,N)$ is nonempty. Finite choice gives representatives $(k_j,N_j)$ for this finite list. Put $W={\bigcap }_{j=1}^m\frac{1}{2}{N}_j$. It is an open convex zero-neighborhood.

For $k\in K$, some $j$ has $k=k_j+n\mathrm{/}2$ with $n\in N_j$. If $w\in W$, write $w=n^{\mathrm{\prime }}\mathrm{/}2$ with $n^{\mathrm{\prime }}\in N_j$. Convexity gives $n\mathrm{/}2+n^{\mathrm{\prime }}\mathrm{/}2\in N_j$, so $k+w\in k_j+N_j\subseteq X\setminus C$. Therefore $(K+W)\cap C=\mathrm{\varnothing }$. The set $K+W$ is nonempty and open as a union of translates of $W$, and convex because the convex combinations of its $K$ and $W$ components stay in those respective sets.

Apply open separation to $K+W$ and $C$, using HB once. Obtain a nonzero continuous scalar-linear $f$ and $\beta \in \mathbb{R}$ such that $u(k+w)<\beta \le u(c)$, where $u=\mathrm{Re}f$. The restriction $u{\mathrm{\mid }}_K$ is continuous, since real part is continuous and restrictions are continuous. It attains a maximum $d=u(k_{\ast })$ for some $k_{\ast }\in K$. Since $0\in W$, the strict inequality at $k_{\ast }+0$ gives $d<\beta$. This attainment step turns pointwise strict separation into a uniform gap.

Put $\alpha =(d+\beta )\mathrm{/}2$ and $\epsilon =(\beta -d)\mathrm{/}2>0$. Then $\alpha -\epsilon =d$ and $\alpha +\epsilon =\beta$, so $u(k)\le d=\alpha -\epsilon <\alpha +\epsilon =\beta \le u(c)$ for all $k,c$. A singleton $K$ is allowed and simply has its sole value as the maximum. Nonemptiness of $K$ is used for attainment and of $C$ in open separation; no other separation axiom or choice principle is used.