You are not logged in.
Having fun, guys? Some of you, over the age of say, oh, 13 for example, may remember I suggested creating this sub-forum for serious mathematical discussion.
Well done! You turned it into a chat room. Great. This whole forum fails in its objective.
I am aware that I am talking to myself, but this is common. So I will
continue down this road in the way that makes sense to me.....So.
Recall my definition of a manifold. Let M be a manifold, with p a point in M and U and V be neighbourhoodsof p. Let the point p be in W = U ∩ V, and let f: U → Rⁿ and g: V → Rⁿ be homeomorphisms. Then (U, f) and (V, g) are charts on M.
Let {x¹, x²,.....,xⁿ} = {a} be a coordinate set on the chart (U, f) and
{y¹, y²,....,yⁿ} = {b} be a coordinate set on (V, g). How do we relate
these guys, as we must, since p lives in both. (By the way, the indices
are not exponents, just tags)
Now f and g are homeomorphisms, that is, they have continuous inverses.
So f-¹: Rⁿ → U and g: V → Rⁿ. Then the composite (g · f-¹): Rⁿ → Rⁿ,
that is, we now have a way of "comparing" {a} and {b}, the two
different coordinate systems which apply equally to our point p.
Now, if we consider what (V,g) will "look like" from (U, f) we see that
each component of {b} will a function h of each component of {a}, like
this:
y¹ = h¹(x¹, x²,....,xⁿ)
y² = h²(x¹, x²,....,xⁿ)
..............................
yⁿ = hⁿ(x¹, x²,....,xⁿ)
Now obviously, the value of each h is completely dependent on the
numerical value of the corresponding y, so we may as well have
y¹ = y¹(x¹, x²,....,xⁿ)
y² = y²(x¹, x²,....,xⁿ)
..............................
yⁿ = yⁿ(x¹, x²,....,xⁿ)
And going the other way, that is, what (U, f) looks like from (V, g) we
simply substitute x for y, and therefore see that the coordinates on U
and V are functions on the point p.
This is an important, but not very profound, result. It gives you the whole flavour of manifolds, which are hugely important.
Anyway, I'm a bit bit tired of talking to myself, so I'm going to quit, unless I can think of something that will interest us all equally.
Until then, xxx
In case you thought my babbling about road maps was mere analogy, let me try and show that it is much stronger than that.
I remind you that a neighbourhood of a topological space M is an open
set U containing some point p in M.
We define a homeomorphism f: U → Rⁿ. This simply says that, for each point p we can find a neighbourhood sufficiemtly small that it is indistinguishable from some Euclidean space. Think of your road map here.
Note that, topological spaces, being sets of a sort, don't have
dimensions. But locally, M "inherits" the dimension n from from Rⁿ.
Nice thing about Euclidean space is we can throw down coordinates.
Let's do that here. Then we can call the pair (U, f) a chart of
dimension n, with U now veiwed as the coordinate neighbourhood of p.
Note that f(p) is now a collection of real numbers (it's called an
n-tuple, if you care) which are simply the coordinate components of p.
And if M can be completely covered by a collection of such charts, called
appropriately enough, an atlas, we will say that M is a topological
manifold (or manifold for short).
Now I said "covered". It should be easy enough to see that this implies
some notion of overlap between adjacent charts, and we will stipulate
then that each point p, q,..... in M lies in at least one
coordinate neighbourhood. (Recall here that most road maps overlap at
their edges). We see that this gives us a connected space, and let's
stipulate the Hausdorff property, just to be complete.
Note that each chart has its own coordinate set, locally deterimed by
the topology on M. Under these circumstances, consider the point p
which lies in two overlapping charts U and V. We see that p can be
described by two different coordinate systems, those on U and V, so it
must be possible to relate these to each other.
Let's try that now.....
On second thoughts, let's not. I guess that's enough to be going on with. Later, if anyone wants.
Hey, welcome unorthadog! Hope you find what you're looking for here. Stick around, post crazy ideas to try on your students, and see.
-ben-
Ok, let's get to the heart of this. I'm not going to do any math today, just to orientate you a little. Well...a manifold is a topological space which can be covered by coordinate neighbourhoods. That's enough of that.
So. Ever used a road map? Do you think that Earth is flat? Well, your road map certainly is, so what's going on?
First consider the problem of trying to represent an entire 2-sphere on a plane surface. We are familiar with such maps, and we see, for example, in most there is is an infinite distance between Alaska and Siberia (is this a good thing?). Similarly, we lose information at the poles. We have the additional (and potentially more serious) problem that distance and angle cannot easily be transferred from the sphere to the plane.
And yet, we are happy to navigate using our road map. What's going on? Well, all the map makers have done is broken our sphere (Earth) into small patches where the difference between a curved surface and a flat one is ignorably samll. Some of you smarties out there will immediately think calculus - and you would be correct!
All we need do now is find a sensible way of relating a map of one small patch to that of a neighbouring small patch in such a way that we don't lose sight of the fact that our Earth is really a 2-sphere.
If there exists a toplogical space where this an be done, then (loosely) we'll say it is a manifold.
Try here for some fun along these lines
SsEe I was trying to take things sowly, for the benefit of some very talented members who are less accustoumed to abstact algebra. But please feel free to correct my errors
Anyway, Ricky (and all others who might be interested, of course) you invited me to proceed.
I said a while back that the notion of homeomorphism is a central theme
in topology. Now seems an appropriate time to revisit this.
Let X and Y be sets, and let f: X → Y. Recall that f is a surjection
(an onto map) if, for every y in Y there is at least one x
in X such that f(x) = y. Also that f is an injection (a one-to-one map)
if, for any y = f(x) there is at most one x in X for which
f-¹(y) = x. If f is both surjective and injective, it is referred to as
a bijection, and we may assume there is some g: Y → X such that g(f(x))
= x for all x in X. g is thus an inverse of f, and one writes (g · f): X → X.
Now using the definition of continuity of a function between
topological spaces that we agreed, if both f and g are continuous inverses of each other, one says that they are homeomorphisms, and that X and Y are
homeomorphic.
Effectively (remebering our new definition of continuity) this means that we are entitled to think of each neighbourhhod of X as equivalent to some neighbourhood of Y.
And as the neighbourhoods of x in X and y in Y are the open sets in their
respective topologies, it follows that there is a 1 to 1 correspodence
between these topologies: X and Y are thus topologically equivalent.
Topologically equivalent spaces share what are known as topological
properties. We have only mentioned the Hausdorff and conectedness
properties, but there are others, as SsEe hinted at.
And if we say that any closed, connected and Hausdorff n-dimensional
surface is topologically equivalent to any other closed, connected and
Hausdorff n-dimensional surface, we are simply showing off, and really
mean that any circle is homeomorphic to any triangle and any rectangle!
We can think of this as meaning that each can be changed into the other
without cutting or gluing. When (if!) we get to talk about manifolds,
this will be an incredibly important notion.
Do you know if there is a way to type alt codes on a laptop with no number pad?
There will be a way, I am sure. Experiment, that's how I found the codes I posted.
Let X, Y be topological spaces. If f: X → Y, then f is continuous at x if for every open set ε containing y = f(x), there is an open set δ containg x such for every ε in Y, f-¹(ε) is open in X (f-¹(ε) is the preimage in X of ε in Y),
In my rush to get off last night, I forgot to add the following qualifier......"and δ subset f-¹(ε), or f(δ) subset ε".
SsEe Excellent post. I am aware that it is common to start with a metric space and then chuck out the metric. It is not the approach I favour, I prefer to work by analogy rather than deconstruction, but it's merely a personal preference.
Not all spaces are metrizable, of course. In fact the ones I wanted to taalk about are not. A couple of points, however.
There was some confusion over the "standard" topology on R. We know what an open set in R looks like. If we take our topology to be the collection (that is collection, not union) of usual open sets we get the standard topology on R
I think you misread my post. I had said that open sets in the standard topology on R is all unions of open intervals in R, not the union of open sets. (Hmm, looking back, I see I didn't quite say that, but nearly!)
If A is a subset of T then A is closed in T if its complement T\A is open.
This is misleading, as I used T for the topology on the set S. Under this nomenclature, all sets in T are open. Are you using T for the whole space? Let (S,T) be a topological space. A subset A of (S,T) is closed in (S,T) if its complement (S,T) - A is open i.e. in T.
Oh? Then you are lot smarter than me. Again, I'm working till 11pm UK, but I really want to say a bit more about homeomorphism before moving on. But first assure me that you understand this generalization of the usual δ/ε defintion of continuity of a function.
Let X, Y be topological spaces. If f: X → Y, then f is continuous at x if for every open set ε containing y = f(x), there is an open set δ containg x such for every ε in Y, f-¹(ε) is open in X (f-¹(ε) is the preimage in X of ε in Y), Gotta run now. Oh remember, "open set containing x" = "neighbourhood of x"
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.
For those running Windows, the following work on this forum
Activate NumLock, hold down the Alt key while entering the following codes from the number pad.
Note that the glyph appears after Alt is released.
α 224
ß 225
Γ 226
π 227
Σ 228
σ 229
µ 230
τ 231
Φ 232
Θ 233
Ω 234
δ 235
∞ 236
φ 237
ε 238
∩ 239
≡ 240
± 241
≥ 242
≤ 243
≈ 247
° 248 e.g. x°
· 250 e.g. p·q
√ 251
ⁿ 252 e.g xⁿ
¹ 0185 e.g. x¹
² 253 e.g. x²
³ 0179 e.g. x³
ª 166 e.g. xª
0134
0131
¼ 0188
½ 0189
→ 282
← 283
é 386
× 0471
Ø 0472
÷ 0503
0135
I say you were asserting transitivity, but it's a minor point.
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.
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.
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
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.
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.
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 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.
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?
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
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
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 not
restricted 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.
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 homeomorphism. 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)
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.
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?
Look at this excellent intro for starters: http://www.mathsisfun.com/forum/viewtopic.php?id=4041
Well yes and no. Remember (why should you) the example I gave: log(xy) = log(x) + log(y)? Let f(x) = log(x), this becomes f(x·y) = f(x) + f(y), so f is an isomorphism.
You see it's not so much a matter of re-naming the group elements, it can be shuffling them around (basically we learn this from Dr Cayley) and (possibly) changing the operation, as above, provided that the equality f(x·y) = f(x)*f(y) holds - here the centre dot and the star can be any operation we choose.
But yes, group theorists don't get too exercised about the difference between say, {0,1} and {-1,+1}. They are clearly equal only "up to isomorphism" , but that's usually all these guys care about.
On a slightly deeper note: strict equality is devilish hard to pin down. In what sense, say, is the abstraction 1 + 1 identically the same as the abstraction 2. I come from alpha centuri, I can see they are not the same objects, but you might convince me they are isomorphic!!!