S33 - 9th June 2005 Munkres 33 Ex. 33.1 (Morten Poulsen)....

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Unformatted text preview: 9th June 2005 Munkres 33 Ex. 33.1 (Morten Poulsen). Let r [0 , 1]. Recall from the proof of the Urysohn lemma that if p < q then U p U q . Furthermore, recall that U q = if q < 0 and U p = X if p > 1. Claim 1. f- 1 ( { r } ) = T p>r U p- S q<r U q , p, q Q . Proof. By the construction of f : X [0 , 1], p> U p- [ q< U q = p> U p = f- 1 ( { } ) and p> 1 U p- [ q< 1 U q = X- [ q< 1 U q = f- 1 ( { 1 } ) . Now assume r (0 , 1). : Let x f- 1 ( { r } ), i.e. f ( x ) = r = inf { p | x U p } . Note that x / S q<r U q , since f ( x ) = r . Suppose there exists t > r , t Q , such that x / U t . Since f ( x ) = r , there exists s Q such that r s < t and x U s . Now x U s U s U t , contradiction. It follows that x T p>r U p- S q<r U q . : Let x T p>r U p- S q<r U q . Note that f ( x ) r , since x T p>r U p . Suppose f ( x ) < r , i.e. there exists t < r such that x U t S q<r U q , contradiction. It follows that, contradiction....
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This note was uploaded on 01/12/2011 for the course MATH 110 taught by Professor Brown during the Fall '08 term at Arizona Western College.

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S33 - 9th June 2005 Munkres 33 Ex. 33.1 (Morten Poulsen)....

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