Nested Kind of Topology.

October 16, 2011

In the last post, I posted a topology on \{1,2,3,\dots, k-1\} where the open sets were

\{\emptyset , \{1\}, \{1,2\}, \{1,2,3\},\dots, \{1,2,3,\dots, k-1\}\}

You can check that this is, in fact, a topology and it, in fact, has exactly k elements.  But this is also a really nice topology for intro-topology students.  Why?  Because it has some strange properties which are easy to get your hands on and prove!  Let’s show some of these.  For the sake of brevity, let’s call this topology P.

 

P is not Hausdorff.

This is easy to see.  Pick the elements 1 and 2.  The only set which contains 1 but not 2 is \{1\}, but there is no set which contains 2 but not 1. 

 

P is T_{0}.

Two points are topologically indistinguishable if they have exactly the same neighborhoods; a T_{0} space is one where each point is topologically distinguishable.  It is trivial to check that this holds.

 

P is connected.  (!)

This is somewhat surprising to me (especially because of how disjoint it "seems"!) but if we attempt to find a separation we find that it is impossible: every open set contains 1 except the empty set, so, in particular, no two open sets have trivial intersection unless one is empty.

 

A function f:{\mathbb R}\rightarrow P is continuous if and only if the sets of all x which map to n\in \{1,2,3,\dots, k-1\} are open (in the standard topology on {\mathbb R}

This is simply taking preimages.  I mention it because actually DRAWING the function on the "plane" (where the "x-axis" is {\mathbb R} and the "y-axis" is \{1,2,3,\dots, k-1\} we get something which does not at all look continuous (unless it is constant —).  This is perhaps a good exercise for intro topology students to work out so that they can see that not all continuous maps "look" continuous.  Of course, there are many other ways to see this using other spaces.

 

Limits?  What about the sequences in this set?  Let’s remind ourselves what convergence is:  \{x_{n}\}\rightarrow x if for every open set U containing x there is some N such that for all n\geq N we have x_{n} \in U.  Note then that:

\{1,1,1,1,1,\dots\} converges to 1.

\{2,2,2,2,2,\dots\} converges to 2.

and so on.  But then we have

\{1,2,1,2,1,\dots\} which converges to 2, but not to 1 (why?).

\{1,2,3,1,2,3,\dots\} which converges to 3, but not to 2 or 1 (why?).

and so forth.  In fact, if we have an infinite number of the numbers n_{1}, n_{2}, \dots, n_{r}\in \{1,2,\dots, k-1\} then this sequence will only converge to \max(n_{1}, n_{2}, \dots, n_{r}) which is kind of a neat characterization.  In fact, this takes care of most (all?) of the sequences. 

 

Limit Points?  A limit point p of a subset A is a point such that for every open U containing p we have A \cap (U-\{p\}) \neq \emptyset

Consider the subset A = \{1\} and let’s see if there are any limit points.  What about 2?  Well, we have that \{1,2\} - \{2\} = \{1\} so the intersection of this with A is nontrivial.  Similarly, the other open sets containing 2 have nontrivial intersection with A which includes more than the element 2.  What about 3?  Same deal.  And 4?  Yes.  So we can now ask: what’s the closure of \{1\}

Consider A = \{1,2\} and we will start looking at the element 3.  Is 3 a limit point?  For the same reasons above, it is.  So what is the closure of this set? 

Let’s be a little adventurous and ask what about A = \{1, 3\}.  For the same reasons as above, 4,5,6\dots, k-1 are all limit points, but what about 2?  In fact, it is.

Last, (and this will complete our "characterization" of the closures of our subsets) what about A = \{2,3\}.  Clearly we will have 4, 5, 6, \dots, k-1 as limit points, but the odd one out is 1.  And 1 is not a limit point (why not?).  That’s a bit sad.  So, taking the least element in the subset allows us to describe the closure: it’s the set of every element greater than or equal to it.

 

Closed sets.  It’s a bit interesting to think up what sets we got from taking the closures of each subset, and then to just start with our open sets and take compliments.  Do we get the same thing?  Neato.

 

Compactness?  A bit trivial in this case.  Why?  (Maybe extending to infinite things would make this cooler…)

 

Here’s something weird, too: what is a path in this space?  What is a loop?  How could we construct something like the fundamental group for this?  Homology groups?  To those with more experience, these questions might be easy; but it is not obvious to me (at least at this point!) how these constructions should go. 

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One Response to “Nested Kind of Topology.”

  1. Tyler said

    James, I haven’t thought about this much, but take the elements 1 to k-1 and create a poset with the reverse ordering. In fact you have a lattice, and even a continuous lattice since it is finite (continuous in the sense of continuous dcpo = domain). The Scott topology for this lattice is the same topology you have listed (think upper sets). Interestingly, due to Scott’s Continuous Lattices paper, continuous lattices are injective spaces. A space D is injective if for any space Y with subspace X, a continuous function from X to D can be extended to a continuous function from Y to D.

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