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A Topological Trifle

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Guitarist

Posted: Fri Jul 11, 2008 8:52 am Post subject: A Topological Trifle

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Posts: 718

So, let's talk about point set topology. First, a point set is, as you might have gathered, a set with no algebraic structure (well, let's say - I don't think that is the most general definition).

OK. Let

be a set, and let $\mathcal{P}(S)$ denote the powerset on

. Then one says that $T \subseteq \mathcal{P}(S)$ is a topology on

iff the following are true:

finite intersections of elements in

(sets, recall) are in

;

arbitrary union of elements in

are in

;

$S \in T$

$\O \in T$ .

The "indivisible" pair S(T)

is called a topological space.

Before continuing, let me hammer this home. Whereas elements in

are points, elements in $T \subseteq \mathcal{P}(S)$ are sets of points.

We shall that this implies that every element in S(T)

is likewise a set.

Shortly we will allow this bit of terminological abuse: it is customary to refer to a topological space simply as, say

(when is is, of course): one says "let

be a topological space", the existence of

and the underlying set being assumed. but we shan't doing that just yet.

OK? So the elements in

(sets, recall), are called the open sets in S(T)

. This may seem a bit weird first time around, but I will explain!

Recall we talked about the complement of a set. The closed sets in S(T)

are those elements in S(T)

which are the complement in

of some set in

. Jane would no doubt prefer to say that the closed sets in S(T)

are elements in the set $S\setminus T$ , and I think on this occasion I would have to agree

Example: Let $S =\{a,b,c\}$ and suppose that $T = \{\{{a\}, \{b\}, \{b,c\}, S, \O\}$ . These are the open sets in S(T)

The closed sets (complements in

) are $\{a\}^c = \{b,c\},\, \{b\}^c = \{a,c\},\, \{b,c\}^c = \{a\},\, S^c = \O,\, \O^c = S$ .

From which I hope you can deduce that a set may open, closed, both or neither. Obviously, $S,\, \O$ are both open and closed in any topology.

Which reminds me; I really ought to give some examples of topologies and topological spaces.

Right. Recall that $\mathbb{R}$ is the set of real numbers Recall also that this set admits of an order. One says that the "standard" (or usual) topology on $\mathbb{R}$ is given by $T = \{a,c\}\cup \{b,d\} \cup \{c,e\},...$ , where these are the open sets in $\mathbb{R}(T)$ .

One calls this the topological real line $\mathbb{R}^1$ (or real line for short), and interprets this as the union of the open intervals $(a,c)\cup (b,d) \cup (c,e)....$ .

Now recall that the union of any arbitrary number of elements in

is in

. So, $(a,b) \cup (c,d) \in T$ . Then $\{(a,b) \cup (c,d)\}^c = [b,c]$ and one calls this a closed set in $\mathbb{R}(T) \equiv \mathbb{R}^1$

Does this make sense, or do I need a bucket of cold water throwing over me?

JaneBennet

Posted: Fri Jul 11, 2008 10:09 am Post subject: Re: A Topological Trifle

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Joined: 06 Apr 2008
Posts: 801

Guitarist wrote:

Jane would no doubt prefer to say that the closed sets in S(T)

are elements in the set $S\setminus T$ , and I think on this occasion I would have to agree

No, I wouldnâ€™t. I would prefer to say that the closed sets in S(T)

are elements in the set $\{S\setminus U:U\in T\}$ . Very Happy

Okay, just nitpicking, but thatâ€™s what I would do.

Quote:

Then $\{(a,b) \cup (c,d)\}^c = [b,c]$

More nitpickig, sorry: $\{(a,b) \cup (c,d)\}^c = (-\infty,a]\cup[b,c]\cup[d,\infty)$ Wink

Anyway, this is a great thread! Iâ€™m looking forward to the next â€œlessonâ€ already. Very Happy

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Guitarist

Posted: Fri Jul 11, 2008 12:12 pm Post subject:

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Joined: 08 Jun 2005
Posts: 718

So, as I don't have a hot date tonight (any volunteers?), let me continue to harangue you.

Let me first dispose of a couple of very uninteresting topologies on any point set.
Recall I said that a topology

on the set

is such that $T \subseteq \mathcal{P}(S)$ .

Then the topology

. where $T = \mathcal{P}(S)$ is called the discrete or concrete topology on this set. It is somewhat pathological.

By the same token, I will have that the topology on

may be simply $T = \{S, \O\}$ . This is called the indiscrete or trivial topology.. This is even less interesting.

Right.

I emphasized that the elements in a topology are sets of points, which is true. It is nonetheless true that the set upon which we are imposing a topology is a set of points.

So, I will make the following assertion: for any open set $U \in S(T)$ (i.e. any set $U \subseteq T$ ) that contains the point

, I will call

a neighbourhood of

. (Note that some writers require the qualification "open neighbourhood"). Notice that $U \ni x$ need not be unique, so evidently, if $U \ni x,\, V \ni x,\, U \cap V \ne \O$

Now the level of abstraction climbs a little.

You may think it silly, but in any sort of space other than a metric space (and topological spaces may have a metric, but they don't need to), we need to be very sure what we mean when we say that 2 points are the same or different.

This opportunity is provided us by the so-called "separation axioms" of topology. These go by the catchy names of $T_0,\,T_1,\,T_2,..$ up to 4.

In fact, the only separation axiom of any real interest is the T_2

axiom. Any topological space satisfying this separation axiom is called a Hausdorff space, which I will explain another time (since my "hot date" - the bloke next door - has just arrived)

Guitarist

Posted: Sat Jul 12, 2008 9:42 am Post subject:

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Joined: 08 Jun 2005
Posts: 718

Well, OK, I see now I missed out something important, but since I started wittering about Hausdorff spaces, I had better finish.

A topological space

is called a "Hausdorff" space iff, for any $x \ne y$ , and neighbourhoods (open sets) containing these points, I may find neighbourhoods $U_x \ni x,\, U_y \ni y$ such that $U_x \cap U_y = \O$ .

Conversely, I will say that, if no such construction exists, then x=y

.

Note this well: if $U_x \cap U_y = \O$ and $U_y \cap U_z = \O$ this does not imply that $U_x \cap U_z = \O$ , i.e this property is not transitive.

However, Hausdorff asserts that, nonetheless, there will be neighbourhoods $V_x \ni x,\,V_z \ni z$ such that $V_x \cap V_z = \O$ .

OK so far?

Let me go back to the bit that should have followed the introduction of closed vs, open sets. Here, our intuition will be a reliable guide, provided we keeps our heads.

Again let

be a topological space with $A \subseteq X$ (notice I haven't specified whether it is open or closed)

Then the interior of

is the largest open set contained in

, and is written A^o

. Equivalently, we may say that A^o

is the union of all open sets in

. Evidently A^o

is open in

, and if

is open,

, not otherwise..

Conversely (in a manner of speaking), the closure of

, written

, is the smallest closed set containing

. Equivalently, A^-

is the intersection of all closed sets of which

is a subset. Equally evidently, A^-

is a closed set, and if

is closed, A = A^-

, not otherwise..

I now define the boundary $\partial A$ of

as $\partial A = A^- \setminus A^o$ .

This is merely a fancy way of saying that the boundary of a closed set is included in the set, the boundary of an open set is not.

The demonstration that this must be true is easy.

Anyone want a pop? Umm, I have just seen a question looming that I don't quite know how to answer: what if

is both open and closed. Hmm... I'd better hit those books!

JaneBennet

Posted: Sat Jul 12, 2008 9:55 am Post subject:

Forum Ph.D.

Joined: 06 Apr 2008
Posts: 801

Guitarist wrote:

This is merely a fancy way of saying that the boundary of a closed set is included in the set, the boundary of an open set is not.

Guitarist wrote:

I have just seen a question looming that I don't quite know how to answer: what if

is both open and closed. Hmm... I'd better hit those books!

The boundary of a â€œclopenâ€ set (a set thatâ€™s both open and closed) is empty. (For a clopen set A, A^o=A^-=A

and $A\setminus A=\O$ .) So to qualify your first statement: The boundary of a closed set is included in the set; the boundary of an open set that is not closed is not. Wink

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Proves its worth by fighting back.
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Chemboy

Posted: Sat Jul 12, 2008 10:28 am Post subject:

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Joined: 01 Jul 2006
Posts: 1061
Location: NY

Thank you for starting this thread, Guitarist. Unfortunately I'm going to be away for the next week, and without internet access. And then for about 5 days after that, though at that point I may have internet access. Hopefully I'll find time to give everything you've covered so far a good read-through tonight and get any questions I have posted before I leave tomorrow morning. Thanks again for doing this.
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Guitarist

Posted: Sat Jul 12, 2008 10:47 am Post subject:

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Joined: 08 Jun 2005
Posts: 718

Chemboy: You got it! I shall be minimal on this subject for the next 5 days, so you don't get overwhelmed on your return. Have a good trip!

Miss Bennet: I would take it to be the greatest possible delicacy on your part if you never again, ever, used the word "clopen" in my presence; it is regarded in polite circles as being most frightfully vulgar.

Hehe!

JaneBennet

Posted: Sat Jul 12, 2008 11:03 am Post subject:

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Joined: 06 Apr 2008
Posts: 801

Guitarist wrote:

I would take it to be the greatest possible delicacy on your part if you never again, ever, used the word "clopen" in my presence; it is regarded in polite circles as being most frightfully vulgar.

Iâ€™m sorry, I havenâ€™t a clue what youâ€™re talking about. Confused

_________________

A problem worthy of attack
Proves its worth by fighting back.
(Piet Hein)

Did You Know?
Fact of the day: Old English

Chemboy

Posted: Sat Jul 12, 2008 7:54 pm Post subject:

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Joined: 01 Jul 2006
Posts: 1061
Location: NY

Guitarist wrote:

Chemboy: You got it! I shall be minimal on this subject for the next 5 days, so you don't get overwhelmed on your return. Have a good trip!

Thanks. I'm going to print what's been posted so far so I can look over it while I'm away. Smile

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"There is a kind of lazy pleasure in useless and out-of-the-way erudition." -Jorge Luis Borges

Chemboy

Posted: Tue Jul 22, 2008 8:03 pm Post subject:

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Posts: 1061
Location: NY

I have internet access now and I'll be home Thursday, so feel free to fire up this thread again. Smile

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"There is a kind of lazy pleasure in useless and out-of-the-way erudition." -Jorge Luis Borges

Guitarist

Posted: Thu Jul 24, 2008 8:21 am Post subject:

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Joined: 08 Jun 2005
Posts: 718

OK, Chemboy, let's try and pick up where we left off.

Recall we talked about the "Hausdorff property". This is one of a set of properties called "topological properties", and I now want to mention just a couple more such properties.

A topological space

is said to be connected iff it cannot be written as the union of two non-empty disjoint sets. The intuitive content here should be clear: if $X = A \cup B$ , and $A \cap B = \O$ , then I cannot "move" from a point in A to a point in B without falling into a chasm.

This is, in fact, a rather antiquated (but perfectly serviceable) definition. The better definition is as follows: a topological space

is said to be connected iff the only subsets of

that are both open and closed are

and $\O$ .

These two definitions are easily brought into register: 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 both open and closed, and

is not connected by either definition.

The other topological property I want to mention is compactness. Again, there are two parallel definitions, and I offer the oldest, and most intuitive, first.

A set

is said to be compact iff, for every sequence in

that has a limit, that limit is found in

.

Again, the content should be clear: a compact set is "self-contained" with respect to limits.

The grown-up version of the same property will require a bit of a detour, which I am going to leave for now.

Let me know, anyone, if the above is less than clear

JaneBennet

Posted: Thu Jul 24, 2008 3:16 pm Post subject:

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Joined: 06 Apr 2008
Posts: 801

Definition: If $\left<x_1,x_2,x_3\ldots\right>$ is a sequence of points in a topological space X, a point $x\in X$ is called a limit of the sequence iff given any open set U containing x, we can find a natural number N such that for all natural numbers n>N

, $x_n\in U$ .

Note that such a limit x may not be unique. It would be unique in a Hausdorff space, but not for non-Hausdorff spaces in general.
_________________

A problem worthy of attack
Proves its worth by fighting back.
(Piet Hein)

Did You Know?
Fact of the day: Old English

Chemboy

Posted: Thu Jul 24, 2008 9:38 pm Post subject:

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Joined: 01 Jul 2006
Posts: 1061
Location: NY

Blah...I have questions but I need time to do a really thorough work-through of the material and actually formulate what questions I need to ask. I'm sticking with it though, keep forging ahead. Just wanted you to know I'm into it despite my lack of on-topic posts.
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"There is a kind of lazy pleasure in useless and out-of-the-way erudition." -Jorge Luis Borges

Guitarist

Posted: Fri Jul 25, 2008 7:31 am Post subject:

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Joined: 08 Jun 2005
Posts: 718

OK, Chemboy, got that. I am reluctant to proceed until I hear your questions, otherwise you will be in over your head and then lose interest.

In order to encourage your questions, let me say this. Some of this stuff is quite hard on first encounter, and you are likely to need some help.

Please don't be shy about asking - I know I am not the best explainer in the world. I am also a terrible show-off, so feed my ego!!!

Faldo_Elrith

Posted: Fri Jul 25, 2008 5:54 pm Post subject:

Forum Freshman

Joined: 02 Jul 2008
Posts: 76

Guitarist wrote:

A set

is said to be compact iff, for every sequence in

that has a limit, that limit is found in

That's interesting. I've never seen compact spaces being defined this way before.

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