# Proving Urysohn's Lemma: Normal vs Perfectly Normal Spaces

• I
• Gear300
In summary, the conversation discusses an attempt to prove Urysohn's Lemma, which states that all normal spaces have a topology at least as strong as the metric space R1 ∩ [0,1]. The approach involves using preimages and set operations to gradually identify open and closed sets, but it is unclear where the argument may have gone wrong. The conversation also touches on the difference between normal and perfectly normal spaces, and how Urysohn's Lemma applies to finite normal spaces.
Gear300
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 bet 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 = { ƒ-1( [0,½) ) , ƒ-1( (½,1] ) },

and the closed sets

S = { F ⊆ ƒ-1(0) , G ⊆ ƒ-1(1) , ƒ-1( [0,½] ) , ƒ-1( [½,1] ) , ƒ-1(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 = { ƒ-1( [0,½) ) , ƒ-1( (½,1] ) , ƒ-1( [0,¼) ) , ƒ-1( (¼, ¾) ) , ƒ-1( (¾,1] ) },

and

S = { F , G , ƒ-1(¼) , ƒ-1(½) , ƒ-1(¾) , ƒ-1( [0,½] ) , ƒ-1( [½,1] ) , ƒ-1( [0,¼] ) , ƒ-1( [¼,¾] ) , ƒ-1( [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 ƒ-1(1/2) has open cover ƒ-1( (¼,¾) ) at stage 2, and at stage 3, it also gains the open cover ƒ-1( (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 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.

The odd part about this is that it seems we could have let F = ƒ-1(0) and G = ƒ-1(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 double-checking. We may be blunt in asking, but where exactly is our argument wrong? Or at the very least, what are the various differences between normal and perfectly normal spaces?

Also, how does Urysohn's Lemma apply to finite normal spaces?

Gear300 said:
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 bet 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 = { ƒ-1( [0,½) ) , ƒ-1( (½,1] ) },

and the closed sets

S = { F ⊆ ƒ-1(0) , G ⊆ ƒ-1(1) , ƒ-1( [0,½] ) , ƒ-1( [½,1] ) , ƒ-1(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 = { ƒ-1( [0,½) ) , ƒ-1( (½,1] ) , ƒ-1( [0,¼) ) , ƒ-1( (¼, ¾) ) , ƒ-1( (¾,1] ) },

and

S = { F , G , ƒ-1(¼) , ƒ-1(½) , ƒ-1(¾) , ƒ-1( [0,½] ) , ƒ-1( [½,1] ) , ƒ-1( [0,¼] ) , ƒ-1( [¼,¾] ) , ƒ-1( [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 ƒ-1(1/2) has open cover ƒ-1( (¼,¾) ) at stage 2, and at stage 3, it also gains the open cover ƒ-1( (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 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.

The odd part about this is that it seems we could have let F = ƒ-1(0) and G = ƒ-1(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 double-checking. We may be blunt in asking, but where exactly is our argument wrong? Or at the very least, what are the various differences between normal and perfectly normal spaces?

Also, how does Urysohn's Lemma apply to finite normal spaces?

What kind of topology do you use on finite spaces?

WWGD said:
What kind of topology do you use on finite spaces?

However way I see it, the preimage of a good number of elements in [0,1] would have to be ∅ in this case. Looking at the definition of continuity closely, it looks like the notion of continuity for such cases is vacantly true. So I guess that solves that.

Then could you explain to me where in the proposed method we ended up narrowing our scope from what was supposed to be normal topologies in general?

Last edited:
Sorry, I did not have much deep to say, only that I thought a finite set would be given the discrete topology under which it is metrizable through the discrete metric. I can't see any other reasonable topology for finite spaces, though I am no expert in this area.

## 1. What is Urysohn's Lemma?

Urysohn's Lemma is a fundamental theorem in topology that states that for any two disjoint closed sets in a normal space, there exists a continuous function that separates them. In other words, it provides a way to prove the normality of a space.

## 2. What is the difference between a normal and a perfectly normal space?

In topology, a normal space is a topological space in which any two disjoint closed sets can be separated by disjoint open sets. A perfectly normal space is a stronger condition where any two disjoint closed sets can be separated by a continuous function. This means that in a perfectly normal space, there exists a continuous function that maps one closed set to 0 and the other to 1.

## 3. What is the significance of proving Urysohn's Lemma for normal and perfectly normal spaces?

Proving Urysohn's Lemma is important because it allows us to determine whether a space is normal or perfectly normal. This is crucial in many areas of mathematics, such as topology and functional analysis, as it helps us understand the properties and behavior of topological spaces.

## 4. What are some examples of normal and perfectly normal spaces?

Examples of normal spaces include the real line with the usual topology, the metric space of rational numbers with the usual topology, and any Hausdorff space. Examples of perfectly normal spaces include metric spaces, topological vector spaces, and locally compact spaces.

## 5. What are some techniques for proving Urysohn's Lemma for normal and perfectly normal spaces?

There are several techniques for proving Urysohn's Lemma, including using the Tietze Extension Theorem, the Stone-Weierstrass Theorem, and the existence of partitions of unity. Additionally, the use of compactness and connectedness properties of spaces can also be helpful in proving this lemma.

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