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

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Unformatted text preview: 9th June 2005 Munkres Β§ 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 βŠ‚ U q . Furthermore, recall that U q = βˆ… 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 β†’ [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). ” βŠ‚ ”: 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 βŠ‚ U s βŠ‚ U t , contradiction. It follows that x ∈ T p>r U p- S q<r U q . ” βŠƒ ”: 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 βŠ‚ S q<r U q , contradiction. It follows that, contradiction....
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S33 - 9th June 2005 Munkres Β 33 Ex 33.1(Morten Poulsen Let...

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