College Math Teaching

January 25, 2015

An interesting topological space

Filed under: advanced mathematics, topology — Tags: , — collegemathteaching @ 2:46 pm

I am teaching an undergraduate topology course this semester. While we are still going through basic set theory, I’ve been racing ahead and looking at examples. Yes, Counterexamples in Topology (Steen and Seebach) is an excellent reference. In fact, I have two copies! 🙂

One space that caught my eye is Alexandroff Square. Take the usual closed $[0,1] \times [0,1]$ square in the plane. Now if $(x,y)$ is a non-diagonal point, let a local basis be open segments of the form $\{x \} \times (y - \epsilon, y + \epsilon )$, that is, small open vertical line segments that miss the diagonal. Open sets that include diagonal points $(x,x)$ are open horizontal strips $[0,1] \times (x + \epsilon, x - \epsilon)$ MINUS a finite number of vertical line segments $\{x_i \} \times (x + \epsilon, x - \epsilon)$.

This topological space is compact, Hausdorff, regular (that is, $T_3$) and normal (that is, $T_4$)

(quick reminder: Hausdorff (or $T_2$) means that any two points lie in disjoint open sets, Regular means that a point and a disjoint closed set can be separated by mutually disjoint open sets, and normal means that two disjoint closed sets can be separated by mutually disjoint open sets.)

One unusual aspect (to me) about this topology is how different the open sets are; there should be a way of characterizing this property. In a sense, the collection of open sets isn’t homogeneous.

I’ve decided to play with a simpler example that is based on the Alexandroff square:

consider the closed interval $[-1,1]$ For all $x \neq 0$ use the discrete topology. For $0$ use the entire interval minus any finite set as a local basis. Then one obtains many of the same features of the Alexandroff square, for similar reasons.

So, what do I mean by “different types” of open sets for different points?
This might work: let $x, y \in X$. Now I’d say that the open neighborhoods for $x, y$ are similar if, for every open neighborhood $U_x$ containing $x$ there is a continuous bijection $f|U_x \rightarrow X$ where $f(x) = y$ and $f(U_x)$ is an open neighborhood of $y$. That is, $f|U_x$ is a homeomorphism.

Let’s turn to the Alexandroff square for a minute. If $p$ is not a diagonal point, choose $U_p$ to be vertical open line segment. Now let $q$ be a diagonal element; every open $U_q$ has an open subset which is homeomorphic to the standard 2-disk in the plane (usual topology). Hence no local homemorphism from $U_p$ to $U_q$ can exist.