Notes

Description:

[94 O] \(\rightarrow\) Form 74.

Content:

[94 O] \(\rightarrow\) Form 74.

[94 O] is:
  For all \(A\subseteq{\Bbb R}\), \(x\in A\) and \(f: A\rightarrow{\Bbb R}\) the following are equivalent:

  1. \((\forall\epsilon>0)(\exists\delta>0)(\forall y\in A)(|y - x| <\delta\)  implies \(|f(y) - f(x)|<\epsilon)\)
  2. Whenever \(\{x_{n}\}\subseteq A\)and \(\lim_{}x_{n} = x\) then \(\lim_{}f(x_{n}) = f(x)\).
and
Form 74 is:
For every \(A\subseteq\Bbb R\) the following are equivalent:
  1. \(A\) is closed and bounded.
  2. Every sequence \(\{x_{n}\}\subseteq A\) has a convergent subsequence with limit in A.)
Assume [94 O]. The usual proof  that part (1) of Form 74 implies part (2) doesn't require any choice.  So assume \(A \subseteq\Bbb R\) and \(A\) satisfies (2) of Form 74 but not (1) of Form 74. Assume first that \(A\) is bounded.  Then \(A\) is not closed so there is an \(x_{0} \not\in  A\) such  that every neighborhood of \(x_{0}\) intersects \(A\).  Define \(f : A\cup \{x_{0}\} \rightarrow {\Bbb R}\) by \(f(x) = 1\) if \(x\in A\) and \(f(x)=0\) if \(x = x_0\). Then (1) of [94 O] fails (with \(\epsilon = 1\)) so there is a sequence \(\{x_{n}\}\subseteq A\cup \{x_{0}\}\) such that \(\lim_{}f(x_{n})\neq f(x_{0})\).  This means we can assume \(\{x_{n}\}\subseteq A\) (replacing \(\{x_{n}\}\) by a subsequence if necessary).  By (2) of Form 74, \(\lim_{} x_{n}\in A\) that is \(x_0\in A\), a contradiction. Now assume that \(A\) is unbounded. Let \(f\) be a monotone, increasing function from the interval \((-1,1)\) one to one, onto \({\Bbb R}\) which satisfies\((\forall a,b\in  (-1,1))( | a - b|  \le  | f(a) - f(b)|\)).(For example, \(\displaystyle{f(x)={2x\over {1-x^{2}}}}\) would work.)  Then \(f^{-1}(A)\)is  bounded  and satisfies (2) of Form 74 because if \(\{x_{n}\}\) is a sequence in \(f^{-1}(A)\),  then \(\{f(x_{n})\}\) is a sequence in \(A\). Since \(A\) satisfies (2) of Form 74, \(\{f(x_{n})\}\) has a convergent subsequence with limit in \(A\).  Using the  property which we required \(f\) to have we  infer  that \(f^{-1}\)  applied to  the convergent subsequence  mentioned  above   gives a convergent subsequence of\(\{x_{n}\}\) with limit in \(f^{-1}(A)\).  Since  we  have  proved (2) of Form 74 implies (1) of Form 74 for bounded  sets,  we  conclude  that \(f^{-1}(A)\) is closed.  Hence there are  numbers \(a\) and \(b\)  such  that\(-1 < a < b < 1\) and \((\forall  x \in  f^{-1}(A) )( a < x < b )\). Since \(f\) is monotone increasing \((\forall  x \in  A)( f(a) < x < f(b))\). Hence  \(A\)  is bounded.

Howard-Rubin number: 5

Type: proof of equivalencies

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