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Reversal via Hahn-Banach

Let X be a separable Banach space. Consider the following statements:

SEP1
(First Separation) Let A be an open convex set in X, let B be a separably closed convex set in X, and assume $A\cap B=\emptyset$. Then A and B can be separated.
SEP2
(Second Separation) Let A and B be open convex sets in X such that $A\cap B=\emptyset$. Then A and B can be strictly separated.
SEP3
(Third Separation) Let A and B be separably closed, convex sets in X such that $A\cap B=\emptyset$. Assume also that A is compact. Then A and B can be strictly separated.
HB
(Hahn-Banach) Let S be a subspace of X, and let $f:S\to{\mathbf{R}}$ be a bounded linear functional with $\Vert f\Vert\le\alpha$ on S. Then there exists a bounded linear functional $F:X\to{\mathbf{R}}$ such that F extends f and $\Vert F\Vert\le\alpha$ on X.
EHB
(Extended Hahn-Banach) Let $p:X\to{\mathbf{R}}$ be a continuous sublinear functional. Let S be a subspace of X, and let $f:S\to{\mathbf{R}}$ be a bounded linear functional such that $f(x)\le p(x)$ for all $x\in S$. Then there exists a bounded linear functional $F:X\to{\mathbf{R}}$ such that F extends f and $F(x)\le p(x)$ for all $x\in X$.
It is known [3,12] that EHB and HB are equivalent to $\mathsf{WKL}_0$ over $\mathsf{RCA}_0$. We are now going to prove that SEP1 and SEP2 are also equivalent to $\mathsf{WKL}_0$ over $\mathsf{RCA}_0$; see Theorem 4.4 below. In the next section we shall prove that SEP3 is equivalent to $\mathsf{ACA}_0$, hence properly stronger than $\mathsf{WKL}_0$, over $\mathsf{RCA}_0$; see Theorem 5.1 below.

Lemma 4.1   It is provable in $\mathsf{RCA}_0$ that SEP1 implies SEP2.

Proof. Reasoning in $\mathsf{RCA}_0$, assume SEP1 and let A and B be disjoint, open, convex sets. Let B' be the separable closure of B. Clearly B' is convex and $A\cap B'=\emptyset$. By SEP1, let F and $\alpha$ be such that $F<\alpha$ on A and $F\ge\alpha$ on B'. It follows that $F>\alpha$ on B. Thus we have SEP2. This completes the proof.

Lemma 4.2   It is provable in $\mathsf{RCA}_0$ that SEP2 implies EHB.

Proof. Reasoning in $\mathsf{RCA}_0$, assume SEP2 and let p, S, and f be as in the hypotheses of EHB. Let A be the convex hull of

\begin{displaymath}\{x\in S\mid f(x)<1\} \,\cup\, \{y\in X\mid p(y)<1\} \,\,, \end{displaymath}

and let B be the convex hull of

\begin{displaymath}\{x\in S\mid f(x)>1\} \,\cup\, \{y\in X\mid -p(-y)>1\} \,\,. \end{displaymath}

Clearly A and B are open.

We claim that A and B are disjoint. If not, then for some $0\le\alpha\le1$, $0\le\beta\le1$, $x_1\in S$, $y_1\in X$, $x_2\in S$, $y_2\in X$ we have f(x1)<1, p(y1)<1, f(x2)>1, -p(-y2)>1, and

\begin{displaymath}(1-\alpha)x_1 + \alpha y_1\quad=\quad (1-\beta)x_2 + \beta y_2 \,\,
. \end{displaymath}

Note that $\alpha y_1-\beta y_2\in S$. Hence

\begin{eqnarray*}f(\alpha y_1-\beta y_2) &\le& p(\alpha y_1-\beta y_2) \\
&\le& \alpha p(y_1)+\beta p(-y_2) \\
&\le& \alpha-\beta \,,
\end{eqnarray*}


yet on the other hand we have

\begin{eqnarray*}f(\alpha y_1-\beta y_2) &=& f((1-\beta)x_2-(1-\alpha)x_1) \\
...
...(x_1) \\
&\ge& (1-\beta)-(1-\alpha) \\
&=& \alpha-\beta \,,
\end{eqnarray*}


hence $f(\alpha y_1-\beta y_2)=\alpha-\beta$. Since at least one of the above inequalities must be strict, we obtain a contradiction. This proves our claim.

By SEP2, there exists a bounded linear functional $F:X\to{\mathbf{R}}$ such that F(x)<1 for all $x\in A$, and F(x)>1 for all $x\in B$. Clearly F extends f. It remains to show that $F(y)\le p(y)$ for all $y\in X$. Suppose not, say p(y)<F(y). If F(y)>0, then for a suitably chosen r>0 we have p(ry)<1<F(ry), a contradiction. If $F(y)\le0$, then for a suitably chosen r>0 we have p(ry)<-1<F(ry). Putting z=-ry we get -p(-z)>1>F(z), again a contradiction. This completes the proof.

Lemma 4.3   It is provable in $\mathsf{RCA}_0$ that EHB implies HB.

Proof. HB is a special case of EHB with $p(x)=\alpha\Vert x\Vert$.

Theorem 4.4   The following statements are pairwise equivalent over $\mathsf{RCA}_0$.
1.
SEP1, the first separation theorem.
2.
SEP2, the second separation theorem.
3.
EHB, the extended Hahn-Banach theorem.
4.
HB, the Hahn-Banach theorem.
5.
$\mathsf{WKL}_0$.

Proof. Lemmas 4.1, 4.2, 4.3 give the implications SEP1 $\Rightarrow$ SEP2 and SEP2 $\Rightarrow$ EHB and EHB $\Rightarrow$ HB. The equivalence HB $\Leftrightarrow$ $\mathsf{WKL}_0$ is the main result of [3]; see also [14] and [16] and Chapter IV of [20]. Theorem 3.1 gives the implication $\mathsf{WKL}_0$ $\Rightarrow$ SEP1. This completes the proof.

Corollary 4.5   The extended Hahn-Banach theorem, EHB, is provable in $\mathsf{WKL}_0$.

Remark 4.6   Corollary 4.5 has been stated in the literature; see Theorem 4.9 of [12]. However, the proof given above is new. In addition, the proof given above contains full details, while the proof in [12] was presented in a very sketchy way.

Corollary 4.7   The Hahn-Banach theorem, HB, is provable in $\mathsf{WKL}_0$.

Remark 4.8   Corollary 4.7 has been proved several times in the literature; see [3] and [16] and Chapter IV of [20]. The proof given here is new and, from some points of view, more perspicuous.

Remark 4.9   Hatzikiriakou [10] has shown that that an algebraic separation theorem for countable vector spaces over Q is equivalent to $\mathsf{WKL}_0$ over $\mathsf{RCA}_0$. This result may be compared to our Theorem 4.4. We do not see any easy way of deducing our result from that of [10] or vice versa, but the comparison is interesting.


next up previous
Next: Separation and Up: Separation and Weak König's Previous: Separation in
Stephen G Simpson
1998-10-25