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#1 Re: Euler Avenue » Lie groups anyone? » 2006-08-23 09:35:31

Ok.  So you are saying the open intervals are a basis for the standard topology on R.  Which is exactly the same as saying the topology is the collection of open sets in R since all open sets in R can be expressed as the union of open intervals.

Sorry.  Inconsistency in my notation.  I meant T to be the whole space.  What you said is better.

#2 Re: Euler Avenue » Lie groups anyone? » 2006-08-22 20:19:28

Thought I'd throw in my 2p.  Where topological spaces come from and how there are really two ways of saying a set is open. Sorry if you already know all this.

The usual definition of an open set in R (before thinking about topological spaces) is:
A is open if for all x in A we can find ε>0 such that {y in R : |x-y|<ε} is a subset of A. 
So given any point in S (no matter how close to the edge) there's always room to move around without leaving S.  Obviously if really close to the edge it won't be much room but it'll always exist.

The next step is to ask what an open set in something other than R is.  We used |x-y| in the definition of open but that won't work in general as | | only applies to R (and R² etc).  That's where metric spaces come in.  All we really need is a notion of "distance between two points" to be able to say a set is open. 
A metric space is a set S together with a function d:SxS→R satisfying:
d(x,y)≥0 for all x,y in S
d(x,y)=0 iff x=y
d(x,y)=d(y,x) for all x,y in S
d(x,z)≤d(x,y)+d(y,z) for all x,y,z in S (the triangle inequality)
Basically, it satisfies all the stuff you'd expect a "distance" function to satisfy.
The metric space would be written (S,d). 
See that d(x,y) = |x-y| defines a metric on R.

So now we can define "open set" in general.  If A is a subset of (S,d) then it's open if for all x in A we can find ε>0 such that {y in S : d(x,y)<ε} is a subset of A

We can also talk about continuity:
Let (A,a) and (B,b) be metric spaces.  f:A→B is continuous if for all x in A and for all ε>0 there exists δ>0 such that y is in A and d(x,y)<δ => d(f(x),f(y))<ε

One thing to note is that there can be different metrics on A and B and f may be continuous with respect to one pair of metrics but discontinuous with respect to another.  Similarly, a set may be open in one metric and not open in another.  So you have to be careful.  For example.
Define a metric on R by d(x,y) = 0 if x=y and 1 if not.  Check this is a metric (called the discrete metric). What are the open sets? Let A be a subset of R. For any x in A take ε=½.  Then {y in R : d(x,y)<ε} = {x} which is clearly a subset of A.  So by definition A is open.  Since A is ANY subset of R we have proved that under this metric any set is open in R.  This is clearly not true under the usual modulus metric.

Open sets still satisfy the properties they do in R. ie:
If A and B are open sets then their intersection is open. (Finite intersection is still open)
The union of any arbitrary collection of open sets is open.  (This allows for the union of an infinite number of open sets)
The empty set and the whole space are open.

Next step is to throw away the metric.  The idea is that the metric isn't important.  It's the behaviour of open sets under intersection and union (above) that's important.  This is how a topological space is defined.  You simply decide which subsets you want to call open and as long as they satisfy the above you can talk about continuous maps and compactness and conectedness or whatever, just as you could in a metric space.

Continuity can be stated in terms of the open sets alone:
Let A and B be topological spaces.  f:A→B is continuous iff whenever U is open in B, f-¹(U) is open in A. 
Applying this definition to metric spaces you can show that it agrees with the usual ε-δ definition.  Topological spaces are just an abstraction of that.  Again, this depends on the topologies.  May be continuous with one pair and not with another.

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 the problem came when deciding whether [1,2) was open or closed or neither or both. 
"It's neither because it's not one of the sets in the topology".  "So the topology is defined by taking the open sets and the open sets defined by the topology... kinda circular".  Something like that.  Well this is where two senses of the word open come into play.  The standard topology is defined by taking the open sets and the open sets are defined in the usual way (with the modulus).  You could define a different topology with the sets looking like [a,b).  Then the set [m,n) would be open in the topology but neither open or closed in the usual sense.
When talking about general topological spaces the phrase "U is open" just means "U is one of the sets in the collection that makes the topology" or "U is one of the sets in the topology".

Oh.  If A is a subset of T then A is closed in T if its complement T\A is open.

And remember the discrete metric.  And how all subsets are open under it.  If S is a set and the collection of subsets in our topology is ALL the subsets of S (all subsets are "open") it's called the discrete topology on S.  Mentioned in the first post. 

We can define a topology from a metric by taking our collection of subsets to be those which are open under the metric.  The metric is said to "induce" the topology on the set.  On the other hand, if we are given a topological space and we can find a metric that would induce it, that topology is said to be "metrizable".

Um.  I think I'll go now.  Hope this helped someone. smile

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