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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 . Then A and B can be separated.
SEP2
(Second Separation) Let A and B be open convex sets in X such that . Then A and B can be strictly separated.
SEP3
(Third Separation) Let A and B be separably closed, convex sets in X such that . 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 be a bounded linear functional with on S. Then there exists a bounded linear functional such that F extends f and on X.
EHB
(Extended Hahn-Banach) Let be a continuous sublinear functional. Let S be a subspace of X, and let be a bounded linear functional such that for all . Then there exists a bounded linear functional such that F extends f and for all .
It is known [3,12] that EHB and HB are equivalent to over . We are now going to prove that SEP1 and SEP2 are also equivalent to over ; see Theorem 4.4 below. In the next section we shall prove that SEP3 is equivalent to , hence properly stronger than , over ; see Theorem 5.1 below.

Lemma 4.1   It is provable in that SEP1 implies SEP2.

Proof. Reasoning in , 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 . By SEP1, let F and be such that on A and on B'. It follows that on B. Thus we have SEP2. This completes the proof.

Lemma 4.2   It is provable in that SEP2 implies EHB.

Proof. Reasoning in , assume SEP2 and let p, S, and f be as in the hypotheses of EHB. Let A be the convex hull of

and let B be the convex hull of

Clearly A and B are open.

We claim that A and B are disjoint. If not, then for some , , , , , we have f(x1)<1, p(y1)<1, f(x2)>1, -p(-y2)>1, and

Note that . Hence

yet on the other hand we have

hence . 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 such that F(x)<1 for all , and F(x)>1 for all . Clearly F extends f. It remains to show that for all . 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 , 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 that EHB implies HB.

Proof. HB is a special case of EHB with .

Theorem 4.4   The following statements are pairwise equivalent over .
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.
.

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

Corollary 4.5   The extended Hahn-Banach theorem, EHB, is provable in .

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 .

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 over . 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: Separation and Up: Separation and Weak König's Previous: Separation in
Stephen G Simpson
1998-10-25