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**ben****Member**- Registered: 2006-07-12
- Posts: 106

Having bored you all rigid with group theory, let me now take you on a more

interesting journey, to the Lie groups. (pronounce it as "Lee")

**Definition** A Lie group is a manifold endowed with group

properties (of the sort we (!) have been discussing).

So what is a "manifold". It is a topological space with

certain additional properties. So, you don't know what a topological

space is, huh? Quite, so let's get dirty. Consider a set S.

**Definition**

A class T of subsets of S is said to a be topology on S iff the

following are true:

finite intersection of elements of T (aka subsets of S) are in T;

infinite union or elements of T (aka subsets of S) are in T;

S is in T;

Ø is in T.

The pair (S,T) is referred to as a topological space. (Note that it is

usual to abuse notation hugely and assert, for example, that "X is a

topological space" (when it is, of course), so take note: it means X =

(S,T))

I'll give some examples in a minute, but let's press on a bit.

The elements of T (aka subsets of S) are said to be the open sets in

(S,T). Clearly, elements of (S,T) in the complement of T are the closed

sets in (S,T).

So let's have a couple of examples. Consider the subsets S and Ø of S.

The topology T = {Ø, S} is referred to as the indiscrete (or trivial,

or concrete) topology on S.

Now consider the power set on S, that is all elements and their

combinations. (The power set P(S) for any set with n elements, is 2ⁿ

e.g. for S = {a, b, c}, P(S) = {{a}, {b}, {c}, {a,b}, {a,c}, {b,c},

{a,b,c}, Ø}). This is in fact a topology on S and is referred to as the

discrete topology on S: it is the finest possile topology on S.

Now remember this: elements in the topology T on S are subsets of S.

But elements of S are points, say s and r. So T is a set of sets, so to

speak.

Let (S,T) be a topological space. For some point s in S, an open set in

(S,T) containing s is referred to as a neighbourhood of s.

Confused? Good. Here's something rather reassuring. The real line R (I

trust you all know what that is), where we do our merry LaTexing, is a

topological space, is a manifold, is a ring, is a vector space is a

group........

There are a couple more things I need to say about topological spaces.

Let X and Y be topological spaces (note the abuse of notation I

referred to earlier). A map f: X → Y is said to be continuous iff, for

each x in the domain of f and each y in the codomain, there is a

neighbourhood U of Y (aka open set containing y) such that the

pre-image f-¹(U) (containing x) is open in X. It is (relatively) easy

to show that this corresponds to the usual ε - δ definition of

continuity we learn in school.

There are loads of properties of topological spaces I

could mention, but I think I only need these two before getting to manifolds.

**Definition**

A topological space is said to be connected iff it is *not* the

union of non-empty disjoint sets. Effecitively this means we can roam

over the space without ever falling into a hole.

**Definition**

A topological space X is said to be Haudorff iff, for each pair of

points x, y in X there are open sets U containing x and V containing y

such that U intersect V = Ø. Basically this is is saying that in the

topological space R, for example, any line connecting x and y can be

infinitely subdivided.

As always, I'll try to answer questions (though I am still learning this stuff myself)

Phew, I fancy a beer. Care to join me?

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

A class T of subsets of S

I've never seen this use of the word class. Is there a mathimatical definition behind it, or do you just mean "type"?

There are two words, disjoint and open, which you used. I know their meaning and everything, but I only know them from their use on the reals. Is there a more topological definition?

Is it that any nonempty set S can be put into a topology X by just picking the right set of subsets T?

A topological space is said to be connected iff it is not the

union of non-empty disjoint sets.

Is this definition meaningless for finite topologies?

Can you give an example of a topological space that is not Haudorff?

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**ben****Member**- Registered: 2006-07-12
- Posts: 106

Ricky wrote:

I've never seen this use of the word class. Is there a mathimatical definition behind it, or do you just mean "type"?

I don't think I made it up, but I use it to mean "a collection".

There are two words, disjoint and open, which you used. I know their meaning and everything, but I only know them from their use on the reals. Is there a more topological definition?

As far as disjoint, it means the same here as anywhere: two sets are

disjoint if they have no elements in common.

Regarding "open" this is slightly more subtle: a subset U of S is open

iff it is included in a topology T on S. Note that subsets can be both

open and closed or neither under this definition. To reassure you,

however, in what is called the usual (or standard) topology on the real line (this is the union of open intervals in R), open sets are open

intervals of the form (a, b), just as we always thought. Beware though, in other topologies on R, some open intervals may not be in the topology i.e. they are not open sets in R viewed as a topological

space.

Is it that any nonempty set S can be put into a topology X by

just picking the right set of subsets T?

Hm. I know what you mean, but strictly speaking *elements* of T

are *subsets* of S. But yes, often we can choose a topology in the

way you suggest, but not always.

A topological space is said to be connected iff it is not

the union of non-empty disjoint sets.Is this definition meaningless for finite topologies?

Do you ask this because I said that infinite union of open sets is

open? I guess I meant infinite *number of unions*, not the union

of an *infinite number of sets* So the answer is no, it's not

meaningless.

Can you give an example of a topological space that is not

Haudorff?

Ha! Good question. Non-Hausdorff spaces tend to be trivial or

pathological, although I can think of an important exception. The

topology T on S denoted by T = {Ø, S} is not Hausdorff: all points in S are included in the same open set, namely S.

The important exception, which I didn't really want to get into (not

yet anyway) is the quotient topology: for some equivalence relation ~

on X, the topology that arises from a partition into equivalence

classes of some Hausdorff space X need not be Hausdorff.

*Last edited by ben (2006-08-17 03:04:09)*

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**ben****Member**- Registered: 2006-07-12
- Posts: 106

OK,

I want to say a bit more about maps between topological spaces, because they are important.

But first let me say a bit more about the Hausdorff property, as it

tripped me up big time in some exercise I was given early on.

Let X be a topological space, and let x, y, z be points in X. Now X is

Hausdorff if I can find some subset A containing x and some subset B

containing y such that A ∩ B = Ø. Now let z be in the neighbourhood

(aka open set) C.

It need not follow that A ∩ C = Ø, nor that B ∩ C = Ø. In other words,

neighbourhoods of x which do not intersect with those of y may well

intersect with those of z. As it happens we can do something rather

simple about this, but let's not go there just now. Let's just say that

the neighbourhood of x that is disjoint from that of y may not be

disjoint from that of z. But if X is Hausdorff there *will* be a

neighbourhood of x, maybe not A, that is disjoint from that of z.

Right. Continuous maps. Recall that f: X → Y is continuous iff for each

x in X, if there is some open set δ in Y containing f(x) there is an

open set ε in X containing x.

Suppose that f is a bijective function. Now if it is the case that

there is also a continuous map f-¹: Y → X (same definition of continuity), then f is said to be a hom**e**omorphism. I bolded the e there to emphasize this is not a homomorphism. In fact, a little thought suggests it is the topological space equivalent to the concept of isomorphism we meet in group theory.

This is the central theme of topology that I want to get across. It

implies that, for example, viewed as topological spaces, the circle,

the triangle and the rectangle are homeomorphic: one can be

continuously deformed into another and back again, without cutting or

gluing. (It gives rise to the joke I'm sure you've all seen about the

topologist who doesn't know the difference between her coffee mug and her doughnut)

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

A topological space is said to be connected iff it is not the

union of non-empty disjoint sets. Effecitively this means we can roam

over the space without ever falling into a hole.

I'm still trying to get this definition of connected through my head. You say "iff it is not...". What is the "it" in that sentence? S? T? A topological space is a set combined with a set of its subsets, but you talk about it here as if its one set.

Regarding "open" this is slightly more subtle: a subset U of S is open

iff it is included in a topology T on S. Note that subsets can be both

open and closed or neither under this definition.

This definition does seem to comply with any idea of open that I have. Nor do I see how a set could be open, closed, or neither under this definition. Is the definition of closed the negation of this?

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**ben****Member**- Registered: 2006-07-12
- Posts: 106

Ricky wrote:

I'm still trying to get this definition of connected through my head. You say "iff it is not...". What is the "it" in that sentence? S? T? A topological space is a set combined with a set of its subsets, but you talk about it here as if its one set.

I'm sorry if I gave the impression we are here dealing with two

separable entities. We are not. "Combined" is a misleaing way of

putting it.

Look. As you yourself point out in your excellent intro to sets, a set is

simply a bunch of things with no intrinsic structure. The notation (S,T) is very rarely used in practice, and merely says we have here a

set S with some additional structure denoted by T.

By analogy, recall we agreed a group is a set with an operation. The formal way of denoting this is (G, ·). A vector space is a set V together with its

underlying field F. We would write (V, F). And so on. These guys are

not the "combination" of two entities. So the "it" is the topological

space, a single enity; S specifies what set we are dealing with, T

tells us what the topology on that set is.

But it is a piece of formalism you don't need to worry too much about, we normally say S is a topological space. If we care what particular topology we are dealing with, we would specify it, but usually we don't care.

This definition does seem to comply with any idea of open that I have. Nor do I see how a set could be open, closed, or neither under this definition. Is the definition of closed the negation of

this?

I take it you meant it does *not* accord with your idea of open?

Well, as I tried to point out, in the topological space R (the real

line) with the standard topology i.e. the way we normally think about

R, open intervals (a, b) are open sets, closed intervals

[a, b] are closed sets. The standard topology on R is the union of all

open intervals e.g {(a, c) U (b, d).....}. These intervals are the

elements in T, aka open subsets of R. This much you will recognise. So. What about [a, b)? Open, closed, neither, both?

But not all spaces admit of the notion of an interval (technically,

this requires a metric; b is in the interval (a, c) iff a < b <

c) so we need an alternative definition.

Take for example what I called the discrete topololgy on S = {a, b}, T = {{a}, {b}, {a, b}, Ø}. {a} is in T, and therefore open, {b} is in T, also open. But {b} is the complement of {a} in {a, b} (i.e. {b} = {a, b} - {a}) and is also closed, likewise {a} is the complement of {b}, so it is also closed. So in this topology all subsets are both open and closed.

Note carefully: all topologies on S have as elements S and Ø, so these are open subsets of S. But S is the complement of Ø, and vice versa, so S and Ø are always both open and closed.

I hope you can see this definition of openness *includes but is notrestricted to* the one we all grew up with. I could say something

about set boundaries if you want, but that will require something of a

detour. We'll see.

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**Ricky****Moderator**- Registered: 2005-12-04
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Back to connected:

A topological space is said to be connected iff it is not the

union of non-empty disjoint sets. Effecitively this means we can roam

over the space without ever falling into a hole.

I do not believe this definition is correct. (1, 2) and [2, 4) are disjoint and non-empty. But (1, 2) U [2, 4) = (1, 4). Are you telling me that (1, 4) is disconnected?

The standard topology on R is the union of all open intervals e.g {(a, c) U (b, d).....}

Then you are using the "real number" definition of open to define the topology of R. So in other words, the topological definition of open is not an equivalent of the real number definition of open. Correct?

With this, it makes sense if we define the topology of R to be the collection of all open sets.

So. What about [a, b)? Open, closed, neither, both?

Neither, since [a, b) is not in the topology and neither is the complement of [a, b).

Edit: Also, to avoid confusion, please don't go on with new things, as I have some things on post #4, but I want to wait till the above is cleared up.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**ben****Member**- Registered: 2006-07-12
- Posts: 106

Ricky wrote:

Back to connected:

(1, 2) and [2, 4) are disjoint and non-empty.

yes.

But (1, 2) U [2, 4) = (1, 4).

Oh? This is a contradiction. Union of disjoint intervals can never generate another interval.

Are you telling me that (1, 4) is disconnected?

I supppose I should blame myself, but the notion of connectedness refers to the whole space, not to particular elements of it.

Then you are using the "real number" definition of open to define the topology of R. So in other words, the topological definition of open is not an equivalent of the real number definition of open. Correct?

Other way round. I am generalizing in order to capture the real line definition as a special case of open so that it applies equally to topological spaces with no sense of order. And yes, the two defintions are equivalent.

With this, it makes sense if we define the topology of R to be the collection of all open sets.

Well union rather than collection, but yes. Good.

So. What about [a, b)? Open, closed, neither, both?

Neither, since [a, b) is not in the topology and neither is the complement of [a, b).

Excellent

Edit: Also, to avoid confusion, please don't go on with new things, as I have some things on post #4, but I want to wait till the above is cleared up.

Fair enough, it's tough going at first, I know. Hang in there, I promise it's good stuff.

P.S. Just had an e-mail from mate suggesting the following. A quick scan suggests it's first class. Look http://oldwww.ballarat.edu.au/~smorris/topology.htm

*Last edited by ben (2006-08-18 09:35:04)*

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**Ricky****Moderator**- Registered: 2005-12-04
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But (1, 2) U [2, 4) = (1, 4).

Oh? This is a contradiction. Union of disjoint intervals can never generate another interval.

Can you name a real number that is in (1, 4) that is not in (1, 2) U [2, 4)? I believe you are confusing disjoint with separated.

Other way round. I am generalizing in order to capture the real line definition as a special case of open so that it applies equally to topological spaces with no sense of order. And yes, the two defintions are equivalent.

But look at how you defined the topology on R:

The standard topology on R is the union of all open intervals e.g {(a, c) U (b, d).....}

So the topology of R is the union of open intervals, and an interval is open if it is within the topology of R. Seems rather circular.

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**ben****Member**- Registered: 2006-07-12
- Posts: 106

Ricky wrote:

Can you name a real number that is in (1, 4) that is not in (1, 2) U [2, 4)? I believe you are confusing disjoint with separated.

The problem here is that there is no topology on R that has both these intervals as open sets (or closed for that matter). Disjoint means that (1, 2) ∩ [2, 4) = Ø, which is clearly the case here.

However, if you don't like my "negative" definition of connected (I promise you it is correct), try this (which I was trying to spare you, as it is less intuitive). A topological space is connected iff the only sets both open and closed are Ø and S (the underlying set).

So the topology of R is the union of open intervals, and an interval is open if it is within the topology of R. Seems rather circular.

No. A **set** is open if is is an element in the standard topology on R. It so happens that open sets in this topology are open intervals.

The point here is that we start with our intuition about openness in the real line, and try to find a formulation which is true here *and also where the notion of an interval makes no sense*

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**Ricky****Moderator**- Registered: 2005-12-04
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The problem here is that there is no topology on R that has both these intervals as open sets (or closed for that matter). Disjoint means that (1, 2) ∩ [2, 4) = Ø, which is clearly the case here.

However, if you don't like my "negative" definition of connected (I promise you it is correct), try this (which I was trying to spare you, as it is less intuitive). A topological space is connected iff the only sets both open and closed are Ø and S (the underlying set).

Every place I look (including my real analysis book) says that connected is when one set intersected with the closure of the other set is non-empty. In that way:

closure of (1, 2) interesect [2, 4) = [1, 2] intersect [2, 4) = {2}, which is non-empty, and thus, (1, 2) U [2, 4) is connected.

Another definition from mathworld says that a set is connected if it is not the union of two disjoint **open** sets. Again, this would mean that (1, 2) U [2, 4) is connected as there aren't two open sets which don't overlap and cover all numbers, so to say.

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**ben****Member**- Registered: 2006-07-12
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Ricky wrote:

closure of (1, 2) interesect [2, 4) = [1, 2] intersect [2, 4) = {2}, which is non-empty, and thus, (1, 2) U [2, 4) is connected.

Sure, but singletons are always closed (they may in addition be open, but they are **never** open and not closed) so, first we cannot be sure they are in some topology (we are seeking a general definition here, remember). Second, if they are, as a topological space is connected iff the sets that are both open and closed are S and Ø, then any space with singletons in the topology must be disconnected.

(By the way, if you're using Windows on a PC, use alt+239 (number pad) for ∩ and alt+0472 for Ø. I can send the full list of non-LaTex codes for Windows if anyone wants)

Another definition from mathworld says that a set is connected if it is not the union of two disjoint

opensets. Again, this would mean that (1, 2) U [2, 4) is connected as there aren't two open sets which don't overlap and cover all numbers, so to say.

This is perfectly correct, as we expect from Wolfram. But we are not doing set theory! You are again resisting the defintion of openness in the theory of topological spaces. A set is open if it is an element in the topology. Period.

Of *course* R (standard topology) is connected, that's where this line of thinking starts. But sets of the form (a, b) **are** in this topology, and sets of the form [c, d) are not, as you said yourself 2 days ago.

Or. In set theory we would say [c, d) is open on the left and closed on the right i.e. [c, d) is both open and closed, right? As it happens there is a topology on R based on sets of this form (look up lower limit topology, or Sorgenfrey line), but we are talking for the moment about the standard topology on R.

Are you getting grumpy with me?

*Last edited by ben (2006-08-19 03:53:24)*

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**Ricky****Moderator**- Registered: 2005-12-04
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Of course not.

I'm not sure if we're on the same page anymore, so let's start fresh.

Is (1, 2) U [2, 4) connected?

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**ben****Member**- Registered: 2006-07-12
- Posts: 106

Yes.

OK, I rechecked all my sources, and they all agree that, in the first definition of connectedness I gave, namely that a space is conncted iff it is not the union of disjoint non-empty sets, I should have inserted the qualifier "open sets". Sorry if that caused confusion.

*Last edited by ben (2006-08-19 20:58:43)*

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**Ricky****Moderator**- Registered: 2005-12-04
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Ok, now we're getting some where. By making them open, it agrees with every definition of connected that I have read.

Now onto defining the topology of R.

No. A set is open if is is an element in the standard topology on R. It so happens that open sets in this topology are open intervals.

How do we determine if an interval is open though?

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**ben****Member**- Registered: 2006-07-12
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Ricky wrote:

How do we determine if an interval is open though?

An interval is open if is doesn't contain its endpoints, as you very well know. If you are leading me make a similar assertion for open sets, forget it. Let me do this instead.

For *any* set A, the closure clA is the intersection of all closed sets containing A. In other words, it is the smallest closed set containing A.

The interior of A, intA is the union of all open sets contained in A, i.e. it is the largest open set contained in A.

We therefore have A subset clA, and intA subset A.

The boundary of A, bdA = clA - intA.

You may easily check that this coincides with the equivalent definitions for intervals.

Just to give you a flavour of how things work in topology, let me offer you this:

I claim that bdA = clA ∩ clAc, where Ac is the complement of A.

I am given bdA = clA - intA. Let A be open, therefore Ac is closed. (the proof works equivalently if A is closed.

I set clAc ∩ bdA = clAc ∩ (clA - intA) = clAc ∩ clA - clAc ∩ intA.

Now, if Ac is closed, Ac = clAc (Ac is the smallest closed set containing Ac)

Since intA is a subset of A, and

A ∩ Ac = Ø, by definition, I have that clAc ∩ intA = Ø, so

clAc ∩ bdA = clAc ∩ clA.

Now bdA is closed, and Ac is closed, so bdA is a subset of Ac = clAc, so

clAc ∩ bdA = bdA, and I have that

bdA = clAc ∩ clA.

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**Ricky****Moderator**- Registered: 2005-12-04
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An interval is open if is doesn't contain its endpoints, as you very well know. If you are leading me make a similar assertion for open sets, forget it. Let me do this instead.

That definition just seems to simple to me. But I guess it works. So open intervals make up the topology of R, and the union of open intervals are also open, which means it agrees with everything I know about R so far.

I have no objections/questions to your last post, but let me go through post 4 in a bit.

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**Ricky****Moderator**- Registered: 2005-12-04
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Let X be a topological space, and let x, y, z be points in X. Now X is

Hausdorff if I can find some subset A containing x and some subset B

containing y such that A ∩ B = Ø. Now let z be in the neighbourhood

(aka open set) C.

Let z be in the neighbourhood of what? Or is C the neighbourhood fo z?

It need not follow that A ∩ C = Ø, nor that B ∩ C = Ø. In other words,

neighbourhoods of x which do not intersect with those of y may well

intersect with those of z. As it happens we can do something rather

simple about this, but let's not go there just now. Let's just say that

the neighbourhood of x that is disjoint from that of y may not be

disjoint from that of z. But if X is Hausdorff there will be a

neighbourhood of x, maybe not A, that is disjoint from that of z.

Assuming Hausdorff, if A ∩ B = Ø, must there exist some neighborhood C such that A ∩ C = Ø and B ∩ C = Ø?

This is the central theme of topology that I want to get across. It

implies that, for example, viewed as topological spaces, the circle,

the triangle and the rectangle are homeomorphic: one can be

continuously deformed into another and back again, without cutting or

gluing.

Given a unit sphere, triangle with sides of 1, and square with sides of 1, you're claiming it's possible to set up a homeomorphism between each, right? Can you actually do so? What would the function be?

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**ben****Member**- Registered: 2006-07-12
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Ricky wrote:

Let z be in the neighbourhood of what? Or is C the neighbourhood fo z?

Yes

Assuming Hausdorff, if A ∩ B = Ø, must there exist some neighborhood C such that A ∩ C = Ø and B ∩ C = Ø?

No, I don't think you may assume transitivity here. {a, b} ∩ {c, d} = Ø, {a, b} ∩ {d, e} = Ø, {c, d} ∩ {d, e} ≠ Ø. You (may) need a different neighbourhood for each pair of points. Hausdorff only allows pairwise disjunction (is that a word?)

Given a unit sphere, triangle with sides of 1, and square with sides of 1, you're claiming it's possible to set up a homeomorphism between each, right? Can you actually do so? What would the function be?

Any continuous function f with a continuous inverse. This deserves more thorough treatment than I shall have time for today.

I have no objections/questions to your last post, but let me go through post 4 in a bit.

Of course, take your time. It's tough until you get the feel for it, after that it flows along nicely

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**Ricky****Moderator**- Registered: 2005-12-04
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Assuming Hausdorff, if A ∩ B = Ø, must there exist some neighborhood C such that A ∩ C = Ø and B ∩ C = Ø?

No, I don't think you may assume transitivity here. {a, b} ∩ {c, d} = Ø, {a, b} ∩ {d, e} = Ø, {c, d} ∩ {d, e} ≠ Ø. You (may) need a different neighbourhood for each pair of points. Hausdorff only allows pairwise disjunction (is that a word?)

I'm not talking about transitivity.

Transitivity would be:

If A ∩ B = Ø and B ∩ C = Ø, then A ∩ C = Ø

What I'm asking is this statement gauranteed by Hausdorff:

If A ∩ B = Ø, then there must exist *some* C such that A ∩ C = Ø and B ∩ C = Ø.

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**ben****Member**- Registered: 2006-07-12
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I say you were asserting transitivity, but it's a minor point.

Ricky wrote:

What I'm asking is this statement gauranteed by Hausdorff:

If A ∩ B = Ø, then there must exist

someC such that A ∩ C = Ø and B ∩ C = Ø.

Why? You cannot, in general, assume that C disjoint from A is disjoint from B, as per my example. It might be, but Hausdorff doesn't guarantee it.

*Last edited by ben (2006-08-21 10:28:52)*

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**Ricky****Moderator**- Registered: 2005-12-04
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I think you are confusing *there exists* with *for all*. I'm just talking about the existence of a single set (or more) with such a property. Not that every set has this property.

Either that, or you're wording is confusing me to think that.

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**Ricky****Moderator**- Registered: 2005-12-04
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The answer: No. If A U B = universal set, then there exists no C such that A ∩ C = Ø and B ∩ C = Ø, where C is non-empty.

If C can be empty, the answer is a trivial yes.

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**ben****Member**- Registered: 2006-07-12
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Hmm. I begining to wonder if I'm the right person to be explaining this stuff, as I seem to have obscured a couple of rather simple and very familiar concepts.

Let me try this. Let R be the real line in its familiar form, and let a ≠ b be any two points. We know that the line segment conncting a and b can be infinitely subdivided. This is the Hausdorff property of R.

Now move to R². Here we have to consider the infinite subdivision of lines connecting a, b and c taken pairwise. If this can be done, R² is Hausdorff. If we move to Rⁿ (n > 2) we can continue in this fashion, remembering that we can only make pairwise connections. The formulation for the Hausdorff property I gave captures this rather precisely, with the advantage that it requires no notion of distance between points. C'est tout!

The other thing we like about R is that the line connecting a and b is unbroken. This is the connected propery. Similar considerations to that given for Hausdorff (i.e. omitting all reference to distance, intervals etc) lead to the (corrected) definition I gave.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

I'm with you so far with everything in this thread. Keep going if you wish.

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