Let's talk topology

Ben · 06-11-2009

This is nice subject to study. It also has important applications in many areas of applied mathematics, such as physics, chemistry, economics, meteorology and computer science. Probably others too.

Let me say this - you can think of this as a "tutorial" if you must, though I would rather you didn't. It's intended more as a "discussion platform". Also, it will be rather superficial - if more depth is wanted, I may be able to supply it. Finally - I am human (well almost), so I make mistakes. PLEASE correct me if you spot any. So here goes....

We start with a set $S$ of points. We can, if we must, think of these points as being concrete objects like numbers, farmyard animals or the things in my garage, but it's best if we don't - better to keep them as abstract little buggers.

Now the powerset on $S$ , generally written as $\mathcal{P}(S)$ , is the set comprising all the subsets of $S$ .

Read that again: it's important to notice that, whereas $S$ is a set of points, $\mathcal{P}(S)$ is a set of sets of points. Ugh!

Also important is the fact that, by definition, every set has at least 2 subsets - $S$ itself and the empty set $\O$ .

So, for example, if $S = \{a,b,c\}$ , then $\mathcal{P}(S) = \{\{a\}, \{b\}. \{c\}, \{a,b\}, \{a,c\}, \{b,c\}, S, \O\}$ .

A topology $T$ on $S$ is defined as $T\subseteq \mathcal{P}(S)$ , satisfying the following axioms.

the intersection of any finite number of elements of $T$ is in $T$ ;

union of an arbitrary number of elements of $T$ is in $T$ ;

$S \in T$ ;

$\O \in T$

The "indivisible pair" $S(T)$ is called a topological space.

Notice that one can always associate more than one topology to a given set, so let's dispose of two rather uninteresting topologies. The first is the case that $T = \mathcal{P}(S)$ which is called the "discrete topology".

Even less interesting is the so-called "trivial" or "indiscrete" topology $T = \{S,\O\}$ .

Now the elements (sets, recall) in $T$ are called the open sets in $S(T)$ . The closed sets are found as follows.

Suppose $A \subseteq S$ . Then the complement of $A$ in $S$ is simply all of $S$ except $A$ . This is often written as $S\setminus A$ , but I will use $A^c$ .

So, whenever $A$ is open in $S(T)$ id est $A \in T$ , then $A^c$ is closed in $S(T)$ .

Let's return to our simple example. Let $S=\{a,b,c\}$ and suppose that $T = \{\{a\}, \{b\}, \{a,c\}, \,S,\,\O\} \subsetneq \mathcal{P}(S)$ are the open sets in $S(T)$ .

Then $\{a\}^c = \{b,c\},\, \{b\}^c= \{a,c\},\, \{a,c\}^c = \{b\},\, S^c = \O,\,\O^c = S$ are the closed sets in $S(T)$ .

Notice then that while $\{b,c\}$ is closed and not open, and $\{a\}$ is open and not closed, the sets $\{b\},\,\{a,c\},\,S,\, \O$ are both open and closed and the set $\{c\}$ is neither open nor closed. This generalizes.

Wow!

sanctus · 06-11-2009

Quote:

Also important is the fact that, by definition, every set has at least 2 subsets - $S$ itself and the empty set $\O$ .

This is obvious if S is an empty set, but I was wondering if S contains say 2 points then you could say that S contains at leas 4 subsets, the ones above plus the ones including each point only.
Can't you also elaboarte it further then and say that every set S has 2+n! subsets where n is the number of points in S? I.e. 2 from $S$ and $\O$ and then the possible order independent combinations of the elements/points in S...

maddog · 06-11-2009

Quote:

Originally Posted by Ben

A topology $T$ on $S$ is defined as $T\subseteq \mathcal{P}(S)$ , satisfying the following axioms.

...

The "indivisible pair" $S(T)$ is called a topological space.

I would like to understand better the significance of a Topological Space.

Also, what it means when a said space if "compact" or as in "compact cover". This is of
course presuming we have some kind of metric and a Hausdorf space definition where
conitinuity is in force. Thus we have neighborhoods about each member of the space
$S$ such that for any neighborhood of a member of $S$
contains another member of $S$ . Or so I think... Hmmm...

maddog

Qfwfq · 06-12-2009

Quote:

Originally Posted by sanctus

This is obvious if S is an empty set, but I was wondering if S contains say 2 points then you could say that S contains at leas 4 subsets, the ones above plus the ones including each point only.

Actually, the odd-man-out is $\O$ because in this case $S$ coincides with it. If cardinality is 1 then $\O$ and $S$ are the only ones, but distinct. For greater cardinality of course there are all the others.

Quote:

Originally Posted by sanctus

Can't you also elaboarte it further then and say that every set S has 2+n! subsets where n is the number of points in S? I.e. 2 from $S$ and $\O$ and then the possible order independent combinations of the elements/points in S...

I always thought it's more like $2^n$ , and with this count including $\O$ and $S$ . Each element is either in or out; if they're all in it's the whole of $S$ , if they're all out it's $\O$ .

Quote:

Originally Posted by maddog

I would like to understand better the significance of a Topological Space.

OK, for the moment we could be precognitive and say that it's an abstraction of some things which are definable in terms of a metric. Actually it's the notion of "open set" that can be so defined and then shown to have those properties. The metric is therefore said to induce a topology, called the topology of the metric.

Ultimately, it all leads to there being some aspects of some things that are topological rather than metric ones, in the sense that they don't change when the critter is "deformed with continuity", but of course this has to yet be defined here.

Quote:

Originally Posted by maddog

Also, what it means when a said space if "compact" or as in "compact cover". This is of
course presuming we have some kind of metric and a Hausdorf space definition where
conitinuity is in force. Thus we have neighborhoods about each member of the space
$S$ such that for any neighborhood of a member of $S$
contains another member of $S$ .

Gosh give the bloke a chance to get that far!

Ben · 06-12-2009

My thanks to Qfwfq for answering these queries. But let me emphasize this: the cardinality (i.e. number of elements in) of the powerset $\mathcal{P}(S)$ will be $2^n$ when the cardinality of $S$ is $n$ , but the topology on $S$ will have cardinality $\le 2^n$ .

So let's see what we have so far: a set from which we formed all possible subsets, and labelled them as open, closed, both or neither. What's the point of that? Well, none really, so, to follow maddog, we need to give some additional structure.

But first I have to explain a bit about what it means for a set to be open or closed.

No, even firster, I have to give a familiar example. Consider the real numbers $\mathbb{R}$ as a set. Now sets are just an unordered jumble of things, just like the things in my garage; here they are called real numbers.

But it is an axiom, unprovable as far as I am aware, that the reals are a total order, that is, for any pair of elements $a,\,b \in \mathbb{R}$ then $a \le b \Rightarrow b \ge a$ (roughly speaking). Then by our axioms for an allowable topology on $\mathbb{R}$ we must have that the union of arbitrary open subsets of $\mathbb{R}$ is itself open.

Standard terminology calls the open sets in $\mathbb{R}$ as $(a,b),\, a < b$ , and the union of all such open sets will thus be the open set $(-\infty, \infty)$ . This union is called the (topological) real line $R^1$ , which must be open since $\pm \infty$ isn't "really" a real number. This is called the standard topology on the reals.

OK? Oh, and if you have trouble convincing yourself that $(1,2) \cup (10,20),\,\, (1,2) \cap (10,20)$ are open, I suggest you follow Qfwqf's prof's advice and take up drinking!

Let's adopt a perfectly standard abuse of notation: since we almost never care which particular topology we are talking about, nobody, but nobody uses $S(T)$ to denote a topological space. They simply say "Let $X$ be a topological space".

Let's now see what it means for a set to be open or closed. Let $A \subseteq X$ be a subset (open or closed) of a topological space. Define the interior of $A$ as the largest open set contained in $A$ , and write this as $A^o$ . Then quite clearly, if $A = A^o$ then $A$ is open. ("largest" here means the union of all open sets entirely within the subset)

Now define the closure of $A$ as the smallest closed set that contains $A$ . Write this as $A^-$ . Obviously, if $A = A^-$ then $A$ is closed. (And "smallest" here means the intersection of all closed sets that completely contain the subset)

Finally define the boundary of $A$ as $\partial A \equiv A^-\setminus A^o$ (where the back-slash is the set theoretic version of arithmetic "minus".

Now you're going to ask what is the boundary of a set that is both open and closed? What if it is neither? Ah, wait!

I trust the suspense won't kill you.....

maddog · 06-12-2009

Quote:

Originally Posted by Qfwfq

Gosh give the bloke a chance to get that far!

Oops.

Did I jump the gun...

maddog

Ben · 06-14-2009

Anyway, one of the principal objects of study here is called a neighbourhood. The definition couldn't be more simple:

Let $X$ be a topological space. Then a neighbourhood of $x \in X$ is any open set containing $x$ .

One writes $U_x$ . Notice that usually, not always, each point will have more that one neighbourhood.

Now this might seem a little strange, but we are going to need a way to specify what exactly we mean when we say that 2 points in $x,\,y\in X$ are the same or different. To see why, notice that not every topological space has a metric. When it has, this is given by a map, say, $d:X \times X \to \mathbb{R}$ and one says that, whenever $d(x,y)=0 \Rightarrow x =y$

In the case that no such simple metric is available we need an alternative definition. This is given us by the family of so-called "separation axioms", of which there are 5, called by the catchy names $T_0,\,T_1, ..., T_4$ . These are of increasing "stringency", in the sense that, if a space satisfies $T_2$ it of necessity satisfies $T_1,\,T_0$ etc.

Sane people only use $T_2$ , which goes as follows.

If, for any 2 points $x,\, y \in X$ there exist neighbourhoods such that $U_x \cap U_y = \O$ , we will say that these 2 points are topologically distinguishable i.e. not equal in our topological space. And conversely.

Notice this crucial fact. We may have $U_x \cap U_y =\O$ and $U_y \cap U_z= \O$ , but this does NOT imply that $U_x \cap U_z = \O$ . In other words this property is not transitive, which reminds us of the triangle inequality for metric spaces.

But $T_2$ guarantees there will always be neighbourhoods $V_x \cap V_z = \O$ .

A topological space with this property is called a Hausdorff space.

So, to continue; roughly speaking, a space will be connected if there is at least one continuous path between any 2 points. More precisely:

A topological space that cannot be written as the union of 2 non-empty disjoint sets is said to be connected.

Alternatively, in a connected space, the only sets that are both open and closed are the base set and the empty set, say $S$ and $\O$ . It is a fun exercise to bring these two definitions into register.

Qfwfq · 06-15-2009

Quote:

Originally Posted by Ben

Let $X$ be a topological space. Then a neighbourhood of $x \in X$ is any open set containing $x$ .

Wait a minute, I protest! A neighborhood is not necessarily open!

So long as $U_x$ has an open subset containing $x$ , it's a neighborhood of it. IOW $x$ musn't belong to the frontier of $U_x$ , which can however comprise any amount of its frontier and even be fully closed!

Note that this makes it equivalent to define the interior of a set $A$ as being the set of all points of which $A$ is a neighborhood. In topology, so many roads lead to Rome.

Ben · 06-15-2009

Quote:

Originally Posted by Qfwfq

Wait a minute, I protest! A neighborhood is not necessarily open!

Well, I grant you 2 things; first my definition was rather imprecise, and second there seems to be no agreement on this in the literature.

Let's go with this compromise, which I copy from Bishop & Goldberg Tensor analysis on Manifolds

Quote:

Originally Posted by Bishop & Goldberg

A neighborhood of $x \in X$ is any $A \subsetneq X$ such that $x \in A^o$ . In particular, any open set containing $x$ is a neighborhood of $x$

(my bolding).

I will happily concede my original definition was wrong iff it can be shown that, for any pair, say, of closed sets $A \ne B$ that $A^o = B^o$ . I doubt it, but I am not completely sure. Someone help me out here!

Quote:

In topology, so many roads lead to Rome.

How true.

While I am here, I promised this

Quote:

Originally Posted by me

A topological space that cannot be written as the union of 2 non-empty disjoint sets is said to be connected.

Alternatively, in a connected space, the only sets that are both open and closed are the base set and the empty set, say $S$ and $\O$ . It is a fun exercise to bring these two definitions into register.

So, here it is.

Suppose that $X$ is connected by our second definition. Let $A \subsetneq X, \, B \subsetneq X$ be open. Let $X = A \cup B$ , and let $A \cap B = \O$ (recall this is the the definition of disjointness).

Then of necessity, $B = A^c \in X,\, A = B^c \in X$ , therefore $A$ and $B$ are also closed, which by our first definition means that $X$ is not connected, a contradiction, and therefore our two definitions coincide.

PS Yikes!! I am losing my marbles, this not what I promised at all. Recall I defined the boundary of an arbitrary set $A$ by $\partial A = A^- \setminus A^o$ .

If $A$ is both open and closed, then $A = A^- = A^o \Rightarrow \partial A = A^- \setminus A^o \equiv A \setminus A = \O$
And what if a set is neither open NOR closed?

Exercise for readers.

Qfwfq · 06-16-2009

Bishop & Goldberg say exactly what I did, they just add the example of an open neighborhood.

Quote:

Originally Posted by Ben

I will happily concede my original definition was wrong iff it can be shown that, for any pair, say, of closed sets $A \ne B$ that $A^o = B^o$ . I doubt it, but I am not completely sure. Someone help me out here!

That clearly can't be, but wouldn't be necessary for stating that a neighborhood needn't be open.

Any set comprising a neighborhood of a point (as a subset) is also a neighborhood of the same point.

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