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Since the the space is a normal space, there exist two disjoints closed subsets A and B.

What I don't understand is how can you associate some abstract space with a real interval and is continous as well?

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Since the the space is a normal space, there exist two disjoints closed subsets A and B.

What I don't understand is how can you associate some abstract space with a real interval and is continous as well?

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HallsofIvy

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matt grime

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However, this doesn't really bare much relation to the Urysohn lemma which staes that in a normal space, S, given two disjoint (open) sets A and B there is continuous map f from S to [0,1] with f(A)=0 f(B)=1.

How you should think of this is as follows: we'll define f(A)=0, now, we'll put A inside lots of nested subsets that you should think of as filling out to occupy all of S and in such a way that f(B)=1

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HallsofIvy

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Thanks for the insights.

Just a have another quick question, Is such a map a homemorphism?

Just a have another quick question, Is such a map a homemorphism?

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matt grime

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Further more you're defining a map to the interval [0,1]. Are you saying you think all normal spaces are homeomorphic to the unit interval?

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mathwonk

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what about a converse? i.e. given closed sets A, B in X, is there a continuous function f on X such that they are the inverse images of {a} and {b}? urysohn says what you need to assume abut X for that to be true.

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That makes alot more sense now. I always like to think that everything is homemorphic, but I fail see that.

I'm a visual learner, so I'm trying to visual this as blobs of areas mapping to an interval. But I'm not sure if thats a correct interpratation either.

I'm studying the proof now, it's a killer, heh.

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mathwonk

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so just set it equal to 0 on A and equal to 1 on B. then where should it equal 1/2? on some set that separates A from B. lets see, i guess you just need to construct a sequence of closed sets.

hmmm what is a nokormal space anyway? you can separate any two clsoed sets by open sets? or soemthing?

anyway given A and B choose two more clsoed sets, disjoint and havcing A,B respectivel;y in their interiors. These will be where the function is <= 1/4 and >= 3/4 or something. then do it again,...

its a littl complicated but just work backwards, i.e. think of having sucha fucntion and ask what the sets look, olike where it is <= k/2^n for all k,n. then the definition fo normals epace is amde so you can actually have such sets nesated in the right way, and them you just define the fucntion to behave as it should.

so tyuo should try it first then read it then try to understand it this way, or any way you like. but it is trivial. just complicated.

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matt grime

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I tend to think of Urysohn in terms of balloons. This is just a visual way of relating mathwonk's idea.

Normal means (I tihnk) that closed sets are separated, so imagine an open set around A as a balloon and you inflate it until it touches B. Think of time parametrizing the inflation as your function from [0,1].

Normal means (I tihnk) that closed sets are separated, so imagine an open set around A as a balloon and you inflate it until it touches B. Think of time parametrizing the inflation as your function from [0,1].

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mathwonk

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i like that makes it seem simpler and more imaginable.

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matt grime

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It won't skip the irrationals at all. (If it skipped any value, t, in [0,1] then the orginal set must be disconnected into at least two open/closed components, the inverse images of [0,t) and (t,1], and [0,t], [t,1], which are open/closed in the subspace topology. Thus we have at least two components one around each of A and B making the construction of such a function trivial indeed: make it 0 on the one containing A and zero on the one containing B, and do whatever you want on the other components.)

The point about continuiuty is that loosely you can infer values from surrounding points, can't you?

The point about continuiuty is that loosely you can infer values from surrounding points, can't you?

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mathwonk

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