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Hausdorff Spaces

A topology on a set X is a piece of data that tells us what points of X are close together. More precisely, for a given point x of X, a topology can tell us whether a given set U containing x is a neighbourhood of x. This usually means there is a small region around x and that small region is contained in U, though it could be more abstract. For example, a neighbourhood of a point x on the real line must contain a small interval (x-ε,x+ε) around x (here, ε is a positive real number), so the closed interval [-2,2] is a neighbourhood of 1 (we can set ε to be 1, giving us the interval (0,2), which is contained in [-2,2]), but the closed interval [0,6] is not a neighbourhood of 6, as for any ε > 0, we'd have that 6 + ε/2 is in (6-ε,6+ε) but not in [0,6].

A topology doesn't tell us directly what sets are neighbourhoods of what points. Instead, it tells us which sets are open. Roughly, an open set is a set that does not contain the points on its boundary, so (0,2) is open (it does not contain 0 or 2) but [0,2) is not (it contains 0, which is on the boundary). The two pieces of data, which sets are neighbourhoods of which points or which sets are open, are equivalent: a neighbourhood of a point x is a set containing an open set containing x, and an open set is a set that contains a neighbourhood of each of its points.

A topology can satisfy axioms stating how topologically distinct, or separated, its points are. These axioms are called separation axioms. The weakest separation axiom states that, for any two points x and y (where x ≠ y), there is an open set containing one point but not the other, i.e. the points have any topological distinction at all. A topology satisfying this axiom is called a Kolmogorov topology or a T₀ topology. A stronger separation axiom might state that, for any two given points x and y, we can view the space as a disjoint union of two subspaces, one containing x, the other containing y. A space satisfying this axiom is called a totally disconnected space. The latter axiom is usually not called a separation axiom, though I still think it counts.

The most important separation axiom, at least in my opinion, is the Hausdorff, or T₂, separation axiom. In this blog post, I aim to explain what a Hausdorff space is and why I think it's an important separation axiom.

Topology

Definition. A topology, topological space, or simply space, is a set X along with a family τ of subsets of X satisfying:

X ∈ τ

If A,B ∈ τ, then A ∩ B ∈ τ

If F ⊂ τ, then ⋃F ∈ τ

Elements of τ are called open sets of the topology. ❀

A lot of the terminology and notation in the definition above might be new to some people so I'll take some time to explain what it all means.

A set is any collection of elements. For example, we have a set of all natural numbers ℕ and a set of all real numbers ℝ. Given a set X and an object x, we may ask whether x is an element of the set X. We write x ∈ X to denote x is an element of X. For example, 6 ∈ ℕ as the number 6 is an element of the set of all natural numbers. We write a vertical or diagonal bar through a relation to denote its negation. For example, 3.5 ∉ ℕ means that 3.5 is not an element of the set of natural numbers.

We write {a₁, …, aₙ} for the finite set containing the elements a₁ up to aₙ, so {1,2,3} is the set containing 1, 2 and 3. We write ∅ for the empty set, i.e. the set with no elements.

A subset S of a set X is a set all whose elements are elements of X. For example, the subsets of {1,2,3} are {}, {1}, {2}, {3}, {1,2}, {1,3}, {2,3} and {1,2,3}. We write S ⊂ X to denote S is a subset of X. The set of all subsets of a set X is called its powerset, denoted P(X). So, for example, P({1,2}) = {{}, {1}, {2}, {1,2}}.

A family is a set of sets. The family τ of subsets of X in the definition of a topology is thus a set of a subsets, i.e. τ is a subset of the set P(X) of all subsets of X.

The intersection of two sets A and B, denoted A ∩ B, is the set of all elements that are in both A and B. For example, {1,2,3} ∩ {2,3,4} = {2,3}, as 2 and 3 are the only elements contained in both the left and right set. We say two sets meet if their intersection is non-empty, otherwise we say they're disjoint. The union of sets A and B, denoted A ∪ B, is the set of all elements that are in A or B (or both). For example, {1,2,3} ∪ {2,3,4} = {1,2,3,4}. The intersection of a family of sets F, denoted ⋂F, is the set of all elements that are in all sets in F and the union of a family of sets F, denoted ⋃F, is the set of all elements that are in at least one set in F. For example, ⋂{{1,2,3},{2,3,4},{3,4,5}} = {3} and ⋃{{1,2,3},{2,3,4},{3,4,5}} = {1,2,3,4,5}.

So, a topology (X,τ) consists of a set X and a family τ of subsets of X called the open sets. The entire space X is open, the intersection of two open sets is open and an arbitrary union (possibly infinite) of open sets is open.

Given three open sets A, B and C, we have that the intersection A ∩ B of A and B is open and the intersection (A ∩ B) ∩ C of A ∩ B with C is also open. In general, by repeatedly taking the intersection of two sets, we can show that the intersection of a finite number of open sets is open.

Note that, in the last point, the family F of open sets can be empty. We have ⋃{} = {}, so the empty set is always open.

Example. Let ℝ be the set of real numbers. A set U ⊂ ℝ is open iff, for all x ∈ U, there is some ε > 0 for which (x-ε,x+ε) ⊂ U. This topology is called the usual topology on ℝ. ⌟

Example. Let X be a set. Then, (X,P(X)) is a topology where every subset of X is open. This topology is called the discrete topology on X. ⌟

Example. Let X be a set. Then, (X,{∅,X}) is a topology where only the empty set and X are open. This topology is called the indiscrete topology on X. ⌟

Example. Let X be a topology. A set U ⊂ X is open iff U is empty or there are only a finite number of elements of X that U does not contain. This topology is called the cofinite topology on X. ⌟

Exercise 1. Verify the examples above are topologies. ▢

Exercise 2. List all four topologies on X = {0,1}. ▢

Operations and Kinds of Sets

Throughout this chapter, we fix a topology (X,τ).

Definition. Let x be a point in X. A neighbourhood (abbreviated: nbhd) of x is a set U ⊂ X for which there is an open set O ⊂ X for which x ∈ O and O ⊂ U. The neighbourhood filter on the point x, denoted N(x), is the family of all neighbourhoods of x. ❀

For example, in ℝ with the usual topology, [0,1] is a neighbourhood of ½.

Definition. Let A ⊂ X be a set. A point x ∈ X lies in the interior of A iff one of the following equivalent statements holds:

There is a neighbourhood U of x for which U ⊂ A.

There is an open set O for which x ∈ O and O ⊂ A.

The interior of A is denoted int(A) or Å. ❀

For example, the interior of [0,1] is (0,1). The interior of a set A can be defined as the union of all open sets contained in A. As the interior of a set is a union of open sets, it is itself open. It is, in fact, the unique open set that is contained in A and contains all other open sets contained in A.

Definition. Let A ⊂ X be a set. A point x ∈ X lies in the closure of A iff for every neighbourhood U of x, there is a point y in A ∩ U (it is possible that y is x itself). The closure of A is denoted cl(A) or A̅. ❀

For example, consider ℝ with the usual topology and let A be the set of all rational numbers between 0 and 1. Then, A̅ = [0,1].

Definition. Let C ⊂ X be a set. We say C is closed iff one of the following equivalent statements holds:

C is the complement X\O of an open set O.

C̅ = C. ❀

Here, the complement X\O of O in X is the set of all points x in X that are not in O. You'll prove in the exercises that the two definitions of closed set above are equivalent.

As closed sets are the complements of open sets, closed sets satisfy properties that are dual to the properties open sets must satisfy:

The empty set ∅ is closed.

The union of two closed sets is closed.

An arbitrary (possibly infinite) intersection of closed sets is closed.

The closure A̅ of a set A is the intersection of all closed sets that contain A.

Definition. Let A ⊂ X be a set. A point x ∈ X lies on the boundary of A iff one of the following equivalent statements holds:

x ∈ A̅\Å.

For every neighbourhood U of x, there are points y and z in U for which y is in A and z is not in A.

We write ∂A to denote the boundary of A. ❀

For example, consider ℝ with the usual topology. Let A be the set of all rational numbers between 0 and 1. Then, ∂A = [0,1] and the boundary of that set is ∂∂A = ∂[0,1] = {0,1}.

In the exercises, if multiple definitions of a concept are given, use the first one. For example, you may use the definition of a closed set as the complement of an open set, but not as a set equal to its own closure unless you prove these two definitions are equivalent. This is to avoid circular reasoning.

Definition. Let D ⊂ X be a set. We say D is dense iff one of the following equivalent statements holds:

For every non-empty open set O, we have that D ∩ O is non-empty.

D̅ = X.

Exercise 3. Let O ⊂ X. Show that the following are equivalent:

O is open.

For every point x in O, there is a neighbourhood U of x for which U ⊂ O.

O = O̊.

∂O ∩ O = ∅. ▢

Exercise 4. Let C ⊂ X. Show that the following are equivalent:

C is the complement of an open set.

C̅ = C.

∂C ⊂ C. ▢

Exercise 5. Let A ⊂ X and let x be a point in X. Show that the following are equivalent:

For every nbhd U of x, there are points y and z in U where y is in A and z is not in A.

x ∈ A̅\Å. ▢

Exercise 6. Let A,B ⊂ X. Show that:

A ⊂ A̅ = A̅̅.

int(int(A)) = int(A) ⊂ A.

∂∂∂A = ∂∂A. ▢

Continuous Functions

For sets A and B, we write f: A → B to denote f is a function from A to B. A function from A to B assigns, to each element a of A, an element f(a) of B. For example, we have a function f: {0,1,2} -> {0,1} where f(0) = 0, f(1) = 1 and f(2) = 0. The element a of A is called the argument of the function and the element f(a) of B is called the function value at a.

Given a function f: A -> B and a subset S ⊂ A, we write f(S), f[S] or f"S for the image of S under f. I.e. the set {f(a) | a ∈ S} of values of the function f at elements of S. The notation {f(a) | a ∈ S} means "the set of f(a) for all a in S" and is called set-builder notation. Sometimes a colon : is used instead of a pipe | to separate the element from the condition. We sometimes write im(f) for the image f(A) of all of A under f.

Informally, a continuous function is a function where a small change in the input results in a small change in the output. Formally:

Definition. Let (X,τ₀) and (Y,τ₁) be topologies and let f: X -> Y be a function. We say f is continuous at a point x ∈ X iff, for every neighbourhood V of f(x) in Y, there is a neighbourhood U of x in X for which f(U) ⊂ V. ❀

Intuitively, V is the small change in the output we request and U represents the small change in the input that would give that small change in the output.

Example. Let f: ℝ → ℝ be a function defined as follows: f(x) = 3 if x ≤ 0 and f(x) = 4 if x > 0. Then, f is continuous everywhere except at 0. We have f(0) = 3 and (2½,3½) is a neighbourhood of 3. Yet, for any ε > 0, no matter how small, we have that f((0-ε,0+ε)) contains 4 (it is the value of f at ε/2), which is not in the neighbourhood (2½,3½). ⌟

Though it can be interesting to study the kinds of sets you can get as the set of arguments at which a given function is continuous, we're mostly interested in functions that are continuous everywhere. We can define a function to be continuous iff it is continuous at x for all x in X, but there is a definition of continuous that uses the data of a topology more directly.

For a function f: A -> B and a subset S ⊂ B, we write f⁻¹(S) or f⁻¹[S] for the preimage of S under f. I.e. the set {a ∈ A | f(a) ∈ S} of a in A for which f(a) is in S. Note that the preimage of B itself is A and the preimage of a complement is the complement of the preimage: f⁻¹(B\S) = A\f⁻¹(S). In this sense, preimages are more well-behaved than images.

Definition. Let (X,τ₀) and (Y,τ₁) be topologies. A function f: X → Y is continuous iff the preimage of every open set is open. I.e. for all O ∈ τ₁, we have f⁻¹(O) ∈ τ₀. ❀

The two definitions of continuous, 'continuous everywhere' and 'preimages of opens are open', are equivalent. The latter definition is often preferable as it is shorter and uses the data of a topology directly.

Theorem. Let (X,τ₀) and (Y,τ₁) be topologies and let f: X → Y be a function. Then, the following are equivalent:

f is continuous everywhere;

Preimages of opens under f are open.

Proof. We want to prove an equivalence between two statements. So, we will show that the first statement implies the second, and then that the second statement implies the first, establishing the equivalence.

(1 ⟹ 2) Assume that f is continuous everywhere. I.e. for all x in X, f is continuous at x. We want to show that preimages of open sets under f are open. So, given an open O ∈ τ₁ in Y, we want to show that f⁻¹(O) is open in X. Let O be an open subset of Y. Recall the definition of preimage: f⁻¹(O) is the set of all x in X for which f(x) is in X. Recall the characterization of open sets in terms of neighbourhoods from exercise 3: a set A is open iff, for all x in X, there is some neighbourhood U of x for which U ⊂ A. As we want to show that f⁻¹(O) is open, we want to show that, for all x ∈ f⁻¹(O) there is a neighbourhood U of x for which U ⊂ f⁻¹(O). So, let x ∈ f⁻¹(O), aiming to show that there is a neighbourhood U of x for which U ⊂ f⁻¹(O). For this, we use our premise that f is continuous everywhere. In particular, f is continuous at x. So, for every neighbourhood V of f(x), there is a neighbourhood U of x for which f(U) ⊂ V. The set O is a neighbourhood of f(x) as it includes the open set O (itself) and that contains f(x). So, by the definition of 'continuous at x', there is a neighbourhood U of x for which f(U) ⊂ O. This is equivalent to U ⊂ f⁻¹(O) as they both mean that f maps all elements of U to elements of O.

(2 ⟹ 1) Exercise 7. ∎

Definition. Let (X,τ) be a topology and let S ⊂ X be a subset. The subspace topology on S is given by τ_S = {O ∩ S | O ∈ X}. ❀

Example. Let ℝ be the real line with the usual topology. Let S be the set of 1/n for natural numbers n along with the number 0. I.e. S = {0, ..., 1/4, 1/3, 1/2, 1/1}. Then, a set O ⊂ S is open in the subspace topology of S iff O does not contain 0, or O does contain 0 and for some large enough m it contains 1/n for all n ≥ m. ⌟

The subspace topology on S is the coarsest topology on S for which the function i: S → X defined by i(a) = a (called the inclusion function) is continuous. The preimage of a set A ⊂ X under this function f is the set A ∩ S. Coarsest means that it has the smallest number of open sets. Formally, a topology τ₀ on S is coarser than a topology τ₁ on S (or τ₁ is finer than τ₀) iff τ₀ ⊂ τ₁.

Exercise 7. Let (X,τ₀) and (Y,τ₁) be topologies and let f: X -> Y be a function. Assume that preimages of open sets under f are open. Show that, for all x in X, f is continuous at x. ▢

Limits

A sequence in X is an infinite sequence (x₀, x₁, x₂, ...) where, for each n, xₙ is an element of X.

Definition. Let (X,τ) be a topology and let (x₀, x₁, x₂, ...) be a sequence. We say that the point L is a limit of the sequence (x₀, x₁, x₂, ...), denoted xₙ -> L, iff, for every neighbourhood U of L, there is a large enough number m for which, for all n ≥ m, we have xₙ ∈ U. ❀

I.e. all neighbourhoods of L contain most points of the sequence.

Example. The sequence (1/1, 1/2, 1/3, 1/4, ...) in ℝ with the usual topology has limit 0. ⌟

Example. The sequence (0, 1, 0, 1, 0, 1, ...) in {0,1} with the discrete topology has no limit. We say the sequence diverges. ⌟

Example. The sequence (0, 0, 0, 0, ...) in {0,1} with the indiscrete topology has two limits: 0 and 1. It's important to keep in mind that limits don't need to be unique. ⌟

Of course, we do often expect the limit of a sequence to be unique. We'll later see what conditions on the space are sufficient for limits in that space to be unique.

Definition. A subsequence of a sequence x = (x₀, x₁, x₂, ...) in X is a sequence y = (y₀, y₁, y₂, ...) for which there is exists a sequence (i₀, i₁, i₂, ...) of natural numbers that is increasing, i.e. i₀ < i₁ < i₂ < ..., and for which yₙ = x_iₙ. I'll use y ⊑ x to denote y is a subsequence of x. ❀

Example. The sequence (0, 1, 0, 1, 0, 1, ...) alternating between 0 and 1 has a subsequence (0, 0, 0, 0, ...) of even entries and a subsequence (1, 1, 1, 1, ...) of odd entries, among other subsequences. ⌟

If (x₀, x₁, x₂, ...) is a sequence in a topology (X,τ) that converges to a point L, then every subsequence (y₀, y₁, y₂, ...) of it also converges to L. However, the converse doesn't need to be true. For example, consider the discrete topology on {0,1} in the example above. The sequence (0, 1, 0, 1, 0, 1, …) diverges, the subsequence (0, 0, 0, 0, ...) converges to 0 and the subsequence (1, 1, 1, 1, ...) converges to 1.

In other words, if y ⊑ x, then the set of limits of the sequence x is a subset of the set of limits of the sequence y.

There are a few ways to generalize limits of sequences. One of them is using nets, but we won't go into that today. Another generalization comes from realizing finding a limit of a sequence is equivalent to solving a certain extension problem.

We endow the set of naturals ℕ with the discrete topology. With this topology, a sequence (x₀, x₁, x₂, ...) in some topology (X,τ) is equivalent to a continuous function from ℕ to X. The function is defined by sending a natural number n to the nth entry xₙ of the sequence. This function is continuous as ℕ is discrete.

The set ℕ̅ is the set of natural numbers along with an extra element called the point at infinity, denoted ∞. The topology on ℕ̅ is defined as follows: a set O ⊂ ℕ̅ is open iff one of the following holds:

O does not contain the point at infinity.

O does contain the point at infinity and, for some natural number m, O contains all natural numbers n with n ≥ m.

The subspace topology of ℕ as a subset of ℕ̅ is the discrete topology. We have a continuous function i: ℕ -> ℕ̅ defined by i(n) = n.

Theorem. Let (X,τ) be a topology. A function f: ℕ̅ -> X is continuous iff f(∞) is a limit of the sequence (f(0), f(1), f(2), ...).

Proof. Exercise 8. ∎

"Iff" is shorthand for "if and only if".

In other words, finding a limit of a sequence (x₀, x₁, x₂, ...) is equivalent to extending the continuous function g: ℕ -> X, defined by g(n) = xₙ, to a continuous function g̃: ℕ̅ -> X. We can thus generalize limits by considering different extension problems. E.g. by considering extensions of continuous function g: ℚ -> X to continuous functions g̃: ℝ -> X (here, ℚ has the subspace topology of ℝ with the usual topology).

Exercise 8. Show that a function f: ℕ̅ -> X is continuous iff f(∞) is a limit of the sequence (f(0), f(1), f(2), ...). ▢

Hausdorff Spaces

Definition. We say a topology (X,τ) is Hausdorff (abbreviated: Hd), is a Hausdorff topology, or satisfies the T₂ axiom, iff for all points x,y in X, if x ≠ y, then there is a nbhd U of x and a nbhd V of y for which U ∩ V = ∅ (i.e. U and V are disjoint). ❀

Instead of neighbourhoods, we could have also taken U and V to simply be disjoint open sets containing x and y respectively. In other words, points of a Hausdorff topology are separated by open sets.

Example. ℝ with the usual topology is Hausdorff. ⌟

Example. For an infinite set X, the cofinite topology on X is not Hausdorff. ⌟

Limits in Hausdorff spaces are unique. If K and L were to be two different limits of a sequence x in a Hausdorff space, then we could find disjoint nbhds U and V of K and L respectively. By definition of limit, both U and V must contain most of the sequence x, but by disjointness, they cannot.

In the previous chapter, we saw that limits are a specific case of a certain kind of extension problem. And, in fact, the unicity of limits is a weaker condition than being Hausdorff. Consider the following example:

Let X be an uncountable set and define a subset O of X to be open iff O is empty or X\O is countable.

In the space X described above, limits are unique. In fact, no sequence in X has a limit. However, X is not a Hausdorff space. If U and V are nbhds of x and y respectively, then U ∩ V always contains an uncountable number of points.

So, to accurately represent what it means for a space to be Hausdroff in terms of (generalized) limits, we must consider the more general case of continuous extension problems.

We do not expect all extension of a continuous function to a Hausdorff space to be unique. For example, consider {0,1} with the discrete topology and the subspace {0}. Define the continuous function g: {0} -> ℝ by g(0) = 69420. Then, for every real number r, we can make a continuous extension g̃: {0,1} -> ℝ of g where g̃(0) = 69420 and g̃(1) = r. Clearly, such an extension is not unique.

The informal reason why we expect extensions from ℕ to ℕ̅ to be unique but not extensions from {0} to {0,1} is because ℕ already "approximates" all of ℕ̅, while {0} does not "approximate" {0,1}. The formal reason is because ℕ is a dense subset of ℕ̅ while {0} is not a dense subset of {0,1}. Recall that a subset D of a space X is dense iff it meets all non-empty open subsets of X.

This leads to the theorem that I think is the main motivation for Hausdroff spaces:

Theorem. Let (X,τ) be a topology, let D ⊂ X be a dense subset, let (Y,τ) be a Hausdroff space, let f,g: X -> Y be continuous functions and assume f|D = g|D. I.e. for all x ∈ D, we have f(x) = g(x). Then, f = g.

Proof. Assume, towards contradiction, f ≠ g. Let x ∈ X be so that f(x) ≠ g(x). As the space Y is Hausdorff, there are disjoint open sets V₁ and V₂ containing f(x) and g(x) respectively. By continuity of f and g, the preimages f⁻¹[V₁] and g⁻¹[V₂] are open in X. We set U₁ = f⁻¹[V₁] and U₂ = f⁻¹[V₂]. The intersection of two open sets is open. Therefore, U = U₁ ∩ U₂ is open. Further, as x is in both U₁ and U₂, we have that x is in U and therefore U is non-empty. As the set D is dense, there must be some point y in D ∩ U. We then have that f(y) is in V₁ and g(y) is in V₂. By assumption, we have f(y) = g(y). Therefore, f(y) is in both V₁ and V₂, contradicting that V₁ and V₂ are disjoint. Our initial assumption that f ≠ g must therefore be wrong, and we have f = g. ∎

You might see some similarities between this proof and the proof that limits in a Hausdorff space are unique I gave in the beginning of this chapter. If you don't, try to look for them.

We saw that a space being Hausdorff is a sufficient condition for having unique limits, but it is not necessary. In other words, having unique limits is a different condition we can ask of a space than being Hausdorff. It is then natural to ask if unicity of continuous extension of functions from a dense subset to a space (as proven above for Hausdorff space) requires the space to be Hausdorff, or if (just like the unicity of limits) this is a weaker condition.

It turns out that this condition does require the space to be Hausdorff:

Theorem. Let (Y,τ) be a topology and assume, for all topologies (X,τ), every dense set D ⊂ X and any two continuous functions f,g: X -> Y, if f|D = g|D, then f = g. Then, Y is Hausdorff. ▯

That's all I had to say for now. Bye!

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