My Math Forum Urysohn's Lemma

 Topology Topology Math Forum

April 27th, 2017, 01:14 PM   #2
Senior Member

Joined: Mar 2015
From: New Jersey

Posts: 1,217
Thanks: 93

Quote:
 Originally Posted by Gear300 Hello. By theorem, given ƒ : U → V, if for every open set A ⊆ V, its preimage ƒ-1(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 just-constructed base is open.
This only applies for U and V in Rn. Not for U in Rn and V in Rm. It looked like this was where you were heading right at stage 1.

-------------------------------------------------------------------
Background:
Oddly enough, the Lemma is intelligible; you can piece it together from straight-forward 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)={x|f(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 01:47 PM.

 April 27th, 2017, 08:24 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 = ƒ-1(0) and G = ƒ-1(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, 08:34 PM   #4
Senior Member

Joined: Aug 2012

Posts: 1,701
Thanks: 448

Quote:
 Originally Posted by Gear300 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].
I haven't looked at this in detail so take this for what it's worth.

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, 06:06 AM   #5
Senior Member

Joined: Mar 2015
From: New Jersey

Posts: 1,217
Thanks: 93

Quote:
 Originally Posted by Gear300 Hello. 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 ƒ-1(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 just-constructed base is open. In any case, we came up with this on the spot without double-checking. We may be blunt in asking, but where exactly is our argument wrong?
This only applies for U and V in Rn. Not for U in Rn and V in Rm. That's where your argument is wrong.

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 06:54 AM.

 Tags lemma, urysohn

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post bonildo Calculus 3 December 17th, 2014 03:29 PM wopereis Linear Algebra 1 July 1st, 2013 04:27 AM johnr Number Theory 2 May 5th, 2013 11:34 AM bgBear Algebra 2 August 13th, 2009 10:40 AM bigli Real Analysis 0 July 26th, 2007 05:02 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top