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 , can we construct a topology such that has exactly 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 we can simply take and to be our sets, but for we need to be a bit more clever. But not much more. Consider:
This is a topology on with exactly 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 , there are two open sets in the particular point topology (empty set and the whole set).
If our universe is , then we have three open sets in the particular point topology.
If our universe is , then we have five open sets in the particular point topology.
If our universe is , then we have nine open sets in the particular point topology.
If our universe is , then we have seventeen open sets in the particular point topology.
If we exclude the empty set, it seems like we will have 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?