## 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.

## Number of Elements in a Topology.

### October 14, 2011

The following question came up today after a brief and somewhat unsatisfying talk I gave on the intuition behind topologies:

Question: Given $n\in {\mathbb N}$, can we construct a topology $\tau$ such that $\tau$ has exactly $n$ elements?

The solution wasn’t obvious to me until I looked at the pattern of numbers on the back of a thing of Dried Papaya Chunks.  For $n = 1, 2$ we can simply take $\emptyset$ and $\{1\}$ to be our sets, but for $n = k$ we need to be a bit more clever.  But not much more.  Consider:

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

This is a topology on $\{1,2,\dots, k-1\}$ with exactly $k$ elements.

In these examples, the elements which are nested inside each other and is somewhat reminiscent of the "particular point" topology — the topology in which a set is open if and only if it contains some predetermined point (or is empty).  In the finite case, how many elements are in the particular point topology?  Let’s do a few examples.

If our universe is $\{1\}$, there are two open sets in the particular point topology (empty set and the whole set).

If our universe is $\{1,2\}$, then we have three open sets in the particular point topology.

If our universe is $\{1,2, 3\}$, then we have five open sets in the particular point topology.

If our universe is $\{1,2,3,4\}$, then we have nine open sets in the particular point topology.

If our universe is $\{1,2,3,4,5\}$, then we have seventeen open sets in the particular point topology.

If we exclude the empty set, it seems like we will have $2^{k-1}$ elements in our particular point topology.  Why is this?

(Hint: Take the last example above with seventeen open sets and suppose the element 5 is your particular point.  Write down all the open sets containing 5.  Now take your pen and scribble out the 5’s in each of those open sets — do the resulting sets look familiar?  Hm.)

Neat.  If you take other finite topologies, are there any neat patterns that arise when you increase the number of digits?  Is it possible to associate some sort of polynomial or invariant with finite topologies?  With infinite topologies?  Would this even be useful?