The Sierpinski Space.

March 22, 2011

What’s the weakest separation we can have in a topological space?

Well, “no separation” is pretty weak.  But this creates the trivial topology and that’s a bit boring.  So let’s say this:

 

Definition: A topological space X is T_{0} or Kolmogorov if for every two points x,y\in X we have that there exists a neighborhood U such that either x\in U and y \notin U or y\in U and x\notin U.

 

In other words, a space is T_{0} if for every pair of points there is at least one open set which contains one and doesn’t contain the other.  This is a pretty weak separation condition.  Certainly, every Hausdorff space is T_{0}, but there are ones which are even weaker which satisfy this condition.  Let’s try to construct a really easy one.

Since a 1-element set is somewhat trivial, let’s begin with a 2-element set.  Let’s say our set is \{0,1\}.  Now, by the definition of a topology, we need \emptyset, \{0,1\} to be open.  Since we want to have this space be T_{0}, we should probably make an open set that contains 1 but does not contain 0.  So our open sets are:

\tau = \{\emptyset, \{1\}, \{0,1\}\}

which gives us a topology on this set (check this!).  Moreover, this was, by construction, the smallest set such that this topology was neither discrete (as \{0\} is not open) nor trivial.  In addition this space is non-Hausdorff, so this gives us a nice example of a space which has a strictly weaker separation property than any Hausdorff space.  In fact, this set has a name.

 

Definition: Let \{0,1\} be given the topology \tau above.  Then we call this space the Sierpinski Space and it is denoted {\mathbb S}

 

Let’s just state a few nice properties of this.

  • This space fails to be Hausdorff and Regular, but it is Normal since there are no pairs of disjoint closed sets.
  • This space is connected (check this!) and, in fact, it is even path connected.  How do we define a path on this?  (Hint: check to see if sending the first point of our path to 0 and the rest to 1 is continuous.)
  • Trivially, this space is compact.

 

An interesting point arises when we consider convergence, though.  We’ll finish the post with this consideration. 

Take a sequence \{a_{n}\} in {\mathcal S}.  Then this sequence converges to 0.  The proof here is trivial: the only open set containing 0 is the entire space, and so each of the points is in this open set for every n\in {\mathbb N}.  The question is then, when does this sequence converge to 1?  A little bit of thinking will show you that if we have only finitely many 0’s in the sequence then the sequence will converge to 1.  For example, 0,1,0,1,0,1,0,\dots does not converge to 1 (why not?) but it still converges to 0. 

What does the sequence 1,1,1,1,1,\dots converge to?  Are limits unique in this case?  What is the weakest separation we can have to force the limits of a sequence to be unique?

 

I’ll end with a cute exercise: try to iterate our process of making the Sierpinski Space but use three points; say, \{0,1,2\}, then let \tau = \{\emptyset, \{2\}, \{1,2\}, \{0,1,2\}\} be the topology.  What can we say about this space?  What kind of limits do our sequences have? 

Repeat this with n points.  Repeat this with countably infinite points.  Now you have a neat topology on {\mathbb N}.  What kind of things do sequences in this topology converge to?

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: