Description:
Form 375 (Tietze-Urysohn Extension Theorem) implies Form 119 \(C(\aleph_0\),uniformly orderable with order type of the integers).
Content:
In this note we include a proof due to K. Keremedis that
Form 375 (Tietze-Urysohn Extension Theorem ) implies Form 119 ( \(C(\aleph_0\), uniformly orderable with order type of the integers).
Fix \({\cal A}=\{A_i:i\in \omega \}\) as in Form 19, where the \(A_i\) are pairwise disjoint. Let \(\bar{A}_i=\{a_i\}\cup A_i\cup\{b_i\}\), where \(a_i, b_i\) are chosen so that \(a_i, b_i\notin A_i\) and the \(\bar{A_i}\) are pairwise disjoint. Declare \(a_i\) to be less than every member of \(A_i\) and \(b_i\) to be greater than every member of \(A_i\). Put \(X=\cup \{\bar{A}_i:i\in \omega \}.\) Order \(X\) by requiring: elements of \(\bar{A}_i\) are ordered in \(X\) the way they are ordered in \(\bar{A}_i,\) otherwise if \(x\in \bar{A}_i\)and \( y\in \bar{A}_j\) then \(x\) is less than \(y\) iff \(i< j.\) Put \(\bar{X}=X\cup \{a\}, a\notin X\) and declare \(a\) to be greater than every member of \(X\). Clearly \(\bar{X}\) with the order topology is a compact T\(_2\) space and consequently a \(T_{4}\) space. Furthermore, \(G=A(=\{a_i:i\in \omega \})\cup B(=\{b_i:i\in \omega \})\cup\{a\}\) is a closed subset of \(\bar{X}\) and the function \(f:G\rightarrow[0,1]\) given by \(f(a)=0\) and for \(i\in \omega \) \(f(a_i)=1/(i+2),\) \(f(b_i)=1/(i+1)\) is clearly continuous. Hence, \(f\) has a continuous extension \(\bar{f}:\bar{X}\rightarrow [0,1].\) Put \(f_i=\bar{f}|\bar{A_i}.\) As \(\bar{f}\) is continuous it follows that \(f_i\) is also continuous. Since \(\bar{A}_i\) is compact it follows that \(f_i\) attains both its minimum and maximum values. We may assume that \(f_i\) attains its minimum at \(a_i\), its maximum at \(b_i\) and, at every other point of \(A_i\), it takes on values other than \(f_i(a_i)\) and \(f_i(b_i)\). Let \(C_i=f_i^{-1}((y_i,1/(1+i)))\) where \(y_i=\frac{2i+3}{2(i+1)(i+2)}\) is the midpoint of the interval \([1/(2+i),1/(1+i)]\). Clearly \(C_i\neq\emptyset\) and by the completeness of \(\bar{A}_i\) and the continuity of \(f_i\) it follows that \(c_i=\inf(C_i)\in A_i.\) Hence, \(c=\{c_i:i\in \omega \}\) is a choice set for \({\cal A}\), finishing the proof of the theorem.
Howard-Rubin number:
138
Type:
Theorem
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