Description: Form [0 AM], a form of the Hahn-Banach Theorem for separable vector spaces, is proved to be provable in ZF\(^{0}\)
Content:
Form [0 AM] is a form of the Hahn-Banach Theorem for separable vector spaces. We shall prove by induction that it is provable in ZF\(^0\).
Assume \(V\) is a separable vector space, \(D = \{d_0, d_1, \cdots\}\) is a countable dense subset of \(V\), and \(p :V \to \Bbb R\) satisfies \(p(x+y) \le p(x) + p(y)\), and \((\forall t\ge 0)(\forall x\in V)(p(tx) = tp(x))\). Also assume that \(p\) satisfies the following condition: \[(*) (\forall \epsilon > 0)(\forall x\in V-S)(\exists y\in D) \mathrm{\; such \; that \;} (p(x - y) < \epsilon \mathrm{ \; and \; }p(y - x) < \epsilon. \] Let \(S\) be a subspace of \(V\) and let \(f\) be a linear function from \(S\) to \(\Bbb R\) such that for all \(x\in S\), \(f(x)\le p(x)\). We shall define by induction a countable set of linear functions \(G=\{g_n: n\in\omega\}\) such that \(g_n\subseteq g_{n+1}\) and such that the linear extension of \(f\), \(f^*=\bigcup G\), can be extended to all of \(V\) by using the condition (\(*\)).
We define \(g_0 = f\). Suppose \(g_n\) has been defined and \(S_n\) is the domain of \(G_n\). In what follows, we shall let \(d=d_{n+1}\) and \(g=g_{n+1}\). If \(d\) is in the domain of \(g_n\), then define \(g=g_n\). Suppose that \(d\) is not in the domain of \(g_n\). We would like to define \(g\) so that for any \(t\in\Bbb R\) and any \(x\in S_n\), \[g(td + x) = tg(d) + g(x) \le p(td + x).\] Then \[tg(d) \le p(td + x) - g(x).\] If \(t = 1\), \(g(d) \le p(d + x) - g(x)\) and if \(t = -1\), \(g(d) \ge g(x) - p(x - d)\). Let \(s=\mathrm{sup}_{x\in S}(g(x) - p(x - d))\) and let \(i=\mathrm{inf}_{x\in S}(p(d + x) - g(x))\). Then \(s\le g(d)\le i\). Define \(g(d)\) to be the mid-point of the interval \([s,i]\). For any \(x\in S_n\), We define \(g(x) = g_n(x)\) and, for each \(t\in\Bbb R\), we define \(g(td + x)=tg(d)+g(x)\). Let \(f^* = \bigcup\{g_n: n\in\omega\}\). It is clear that \(f^*\) is a linear function, \(f^*(x)\le p(x)\) for all \(x\) in the domain of \(f^*\), and \(D\) is a subset of the domain of \(f^*\).
Let \(\epsilon > 0\) and \(x\in V\). Then, by (\(*\)), there is a \(y\in D\) such that \(p(x - y) < \epsilon\) and \(p(y - x) < \epsilon\). If \(f^*(x - y)\ge 0\), then \(0\le f^*(x - y)\le p(x - y) < \epsilon\). If \(f^*(x - y) < 0\), then \(0 < -f^*(x - y) = f^*(y - x)\le p(y - x)<\epsilon\). Consequently, \(|f^*(x - y)| < \epsilon\) and \(f^*\) can be extended to all of \(V\).
Howard-Rubin number: 143
Type: Proof
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