April 25th, 2017, 10:45 PM  #1 
Newbie Joined: Apr 2017 From: Neither here nor there Posts: 3 Thanks: 0  Urysohn's Lemma
Hello. This did not come from a topology book, but we were asked to prove Urysohn's Lemma. We are familiar with standard proofs for this, which are all likely simpler to exhibit than our attempt here, but we were just curious about where our method here went wrong. At a glance, the lemma essentially states that all normal spaces bear a topology at least as strong as the metric space R1 ∩ [0,1]. Since preimages preserve many set operations, it seems a good deal of the problem can be handled just by thinking in terms of [0,1]. Let T be the mother set, and let F and G be the closed sets in question. The idea was to inductively tackle the problem. Stage 1: If F and G are our closed sets, then we can let F be a subset of the preimage of 0 and G be a subset of the preimage of 1. By normality, we can arbitrate that the closed set constituting the complement of their disjoint open covers is exactly the preimage of ½, while the open cover of F is the preimage of [0,½) and the open cover of G is the preimage of (½,0]. So at this point, we have identified the open sets S = { ƒ1( [0,½) ) , ƒ1( (½,1] ) }, and the closed sets S = { F ⊆ ƒ1(0) , G ⊆ ƒ1(1) , ƒ1( [0,½] ) , ƒ1( [½,1] ) , ƒ1(½) }. It is important to ensure that all set operations between the preimages make sense for the sake of consistency, and they intuitively seem to. Stage 2: We can then take the complement of the open cover of F to induce another closed set, which should be the preimage of (½,1]. We can call this set X. By normality, F and X bear disjoint open covers, so we can let the closed set constituting their complement to be the preimage of ¼. Similarly, we can do the same thing with G and the complement of its open cover to attain the preimage of ¾. So at this point, we should have been able to identify the open sets, S = { ƒ1( [0,½) ) , ƒ1( (½,1] ) , ƒ1( [0,¼) ) , ƒ1( (¼, ¾) ) , ƒ1( (¾,1] ) }, and S = { F , G , ƒ1(¼) , ƒ1(½) , ƒ1(¾) , ƒ1( [0,½] ) , ƒ1( [½,1] ) , ƒ1( [0,¼] ) , ƒ1( [¼,¾] ) , ƒ1( [¾,1] ) }. And again, it is important to ensure that all set operations between the preimages make sense...and they seem to. So the general idea is that we can continue identifying new open and closed sets by progressively inducting on our current set of open and closed sets via basic set operations and a property of normal spaces. The main points are that 1. At every nth stage, we achieve the (closed) preimages of the dyadic rationals in [0,1] that are a multiple of 1/2n. 2. We progressively achieve smaller open covers around each dyadic rational as the stages progress. The radii of the open covers decrease by ½ with each stage ever since the stage of their inception. e.g. The closed set ƒ1(1/2) has open cover ƒ1( (¼,¾) ) at stage 2, and at stage 3, it also gains the open cover ƒ1( (3/8 , 5/8) ). So what we altogether accomplish, if this is right, is an open preimage of each open sphere in a local neighborhood base surrounding each dyadic rational in the interval [0,1]. Each such local neighborhood base is a countable set of open spheres of radii 1/2n. Since the set of dyadic rationals are dense in [0,1], a countable union across this countable set of local neighborhood bases returns a countable base for the topology in R1 ∩ [0,1]. By theorem, given ƒ : U → V, if for every open set A ⊆ V, its preimage ƒ1(A) is open, then ƒ is continuous. So the mapping we have constructed should be continuous, since the preimage of any (open) set in our justconstructed base is open. The odd part about this is that it seems we could have let F = ƒ1(0) and G = ƒ1(1) right at the beginning. Then if all of our arguments are consistent, our function should precisely separate the two closed sets, which is hallmark of a perfectly normal space and not normal spaces in general. The added perk is that as our open covers converge around each dyadic rational, their preimages should converge to the closed preimage of the dyadic rational. This matches the idea that a space is perfectly normal iff every closed set is a Gδ set. In any case, we came up with this on the spot without doublechecking. We may be blunt in asking, but where exactly is our argument wrong? Last edited by skipjack; April 27th, 2017 at 12:48 PM. 
April 27th, 2017, 12:14 PM  #2  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100  Quote:
 Background: Oddly enough, the Lemma is intelligible; you can piece it together from straightforward definitions. Ref: https://en.wikipedia.org/wiki/Urysohn%27s_lemma Definition: X is a normal space if any two closed subsets have disjoint open neighborhoods. R1,R2,R3 for example. [1,0] and [1/n,2] have disjoint open neighborhoods no matter what n is. Definition: Two disjoint closed subsets A and B of X are said to be separated by functions if there is a continuous function from X to [0,1] such that f(A) = 0 and f(B) =1. Urysohn's lemma states that a topological space X is normal if and only if any two disjoint closed sets can be separated by a continuous function. Definition: Preimage Preimage  from Wolfram MathWorld "Given f:X>Y, the image of x is f(x). The preimage of y is then f$\displaystyle ^{1}$(y)={xf(x)=y}, or all x whose image is y. Images are elements of the range, while preimages are subsets (possibly empty) of the domain."  OP: Presumably, T is a normal space and you are trying to prove that if F and G are disjoint closed sets in T then there exists a continuous function f:T> [0,1] such that $\displaystyle \forall$ a $\displaystyle \epsilon$ F, f(a) =0 and $\displaystyle \forall$ b $\displaystyle \epsilon$ G, f(b) = 1: http://homepage.divms.uiowa.edu/~jsi...hnLemma_v5.pdf Last edited by skipjack; April 27th, 2017 at 12:47 PM.  
April 27th, 2017, 07:24 PM  #3 
Newbie Joined: Apr 2017 From: Neither here nor there Posts: 3 Thanks: 0 
Urysohn's lemma should apply to any normal space X. The nifty thing about having [0,1] as the codomain is that for a continuous function f: X > [0,1], the topology that the mapping induces on X is only as strong as the topology in [0,1], regardless of what the original topology in X is. The thing is that if the proposed scheme is right, then it seems that we can arbitrate right from the start that F = ƒ1(0) and G = ƒ1(1), which is not supposed to be general to normal spaces but is instead a particular feature of perfectly normal spaces. So there must be something we missed in our proof. 
April 27th, 2017, 07:34 PM  #4  
Senior Member Joined: Aug 2012 Posts: 1,922 Thanks: 534  Quote:
For my part I could use some more clarity in the basic exposition of the problem. You say, "At a glance, the lemma essentially states ..." Well is that what the lemma says or are you paraphrasing for intuition? Can you make a precise statement of exactly what it is you are trying to prove, including all terms that we may have seen a long time ago but no longer recall the exact definition of. Like normal space for example. Secondly, "Since preimages preserve many set operations ..." is handwavy. The question is whether preimage happen to preserve any particular set operations we care about in this specfic context. Perhaps you can make a precise statement here. I can't say this will solve your problem (though writing down the exact statement of what is to be proved is often helpful to working out the proof) but I can say that you might get more interested readers. As it stands, one has to Google around just to figure out what you're asking.  
April 28th, 2017, 05:06 AM  #5  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100  Quote:
Looks like you missed it in post 2, where all definitions are also referenced. EDIT: Actually, the whole premise of your proof is wrong. f$\displaystyle ^{1}$ determines f only if it is surjective. Last edited by zylo; April 28th, 2017 at 05:54 AM.  

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