College Math Teaching

February 16, 2015

Topologist’s Sine Curve: connected but not path connected.

Filed under: student learning, topology — Tags: , , — collegemathteaching @ 1:01 am

I wrote the following notes for elementary topology class here. Note: they know about metric spaces but not about general topological spaces; we just covered “connected sets”.


I’d like to make one concession to practicality (relatively speaking). When it comes to showing that a space is path connected, we need only show that, given any points x,y \in X there exists f: [a,b] \rightarrow X where f is continuous and f(a) = x, f(b) = y . Here is why: s: [0,1] \rightarrow [a,b] by s(t) = a + (b-a)t maps [0,1] to [a,b] homeomorphically provided b \neq a and so f \circ s provides the required continuous function from [0,1] into X .

Now let us discuss the topologist’s sine curve. As usual, we use the standard metric in R^2 and the subspace topology.

Let S = \{(t, sin(\frac{1}{t}) | t \in (0, \frac{1}{\pi} \} . See the above figure for an illustration. S is path connected as, given any two points (x_1, sin(\frac{1}{x_1}), (x_2, sin(\frac{1}{x_2}) in S , then f(x) = (x, sin(\frac{1}{x}) is the required continuous function [x_1, x_2] \rightarrow S . Therefore S is connected as well.

Note that (0,0) is a limit point for S though (0,0) \notin S .

Exercise: what other limit points does S that are disjoint from S ?

Now let T = S \cup \{ (0,0) \} , that is, we add in the point at the origin.

Fact: T is connected. This follows from a result that we proved earlier but here is how a “from scratch” proof goes: if there were open sets U, V in R^2 that separated T in the subspace topology, every point of S would have to lie in one of these, say U because S is connected. So the only point of T that could lie in V would be (0,0) which is impossible, as every open set containing (0,0) hits a point (actually, uncountably many) of S .

Now we show that T is NOT path connected. To do this, we show that there can be no continuous function f: [0, \frac{1}{\pi}] \rightarrow T where f(0) = (0,0), f(\frac{1}{\pi}) = (\frac{1}{\pi}, 0 ) .

One should be patient with this proof. It will go in the following stages: first we show that any such function f must include EVERY point of S in its image and then we show that such a function cannot be extended to be continuous at (0,0) .

First step: for every (z, sin(\frac{1}{z})), there exists x \in (0,\frac{1}{\pi} ] where f(x) = (z, sin(\frac{1}{z}) ) Suppose one point was missed; let z_0 denote the least upper bound of all x coordinates of points that are not in the image of f . By design z_0 \neq \frac{1}{\pi} (why: continuity and the fact that f(\frac{1}{\pi}) = (\frac{1}{\pi}, 0) ) So (z_0, sin(\frac{1}{z_0}) cuts the image of TS into two disjoint open sets U_1, V_1 (in the subspace topology): that part with x-coordinate less than and that part with x-coordinate greater than x = z_0 . So f^{-1}(U_1) and f^{-1}(V_1) form separating open sets for [0,\frac{1}{\pi}] which is impossible.

Note: if you don’t see the second open set in the picture, note that for all (w, sin(\frac{1}{w})), w > z_0 one can find and open disk that misses the part of the graph that occurs “before” the x coordinate z_0 . The union of these open disks (an uncountable union) plus an open disk around (0,0) forms V_1 ; remember that an arbitrary union of open sets is open.

Second step: Now we know that every point of S is hit by f . Now we can find the sequence a_n \in f^{-1}(\frac{1}{n \pi}, 0)) and note that a_n \rightarrow 0 in [0, \frac{1}{\pi}] . But we can also find b_n \in f^{-1}(\frac{2}{1 + 4n \pi}, 1) where b_n \rightarrow 0 in [0, \frac{1}{\pi}] . So we have two sequences in the domain converging to the same number but going to different values after applying f. That is impossible if f is continuous.

This gives us another classification result: T and [0,1] are not topologically equivalent as T is not path connected.

February 14, 2015

No, I don’t “learn more from my students than they do from me”: BUT…..

I admit that I chuckled when a famous stand up comic said: “”New Rule: Any teacher that says, ‘I learn as much from my students as they learn from me,’ is a sh***y teacher and must be fired.””

Yes, I assure you, when it comes to subject matter, my students had bloody well learn more from me than I do from them. 🙂

BUT: when it comes to class preparation, I find myself learning a surprising amount of material, even when I’ve taught the class before.
For example, teaching third semester calculus (multi-variable) lead me to thinking about some issues and to my rediscovering some theorems presented a long time ago and often not used in calculus/advanced calculus books. THAT lead to a couple of published papers.

And, given that my teaching specialty has morphed into applied mathematics, teaching numerical analysis has lead me to learn some interesting stuff for the first time; it has filled some of the “set of measure infinity” gaps in my mathematical education.

So, ok, this semester I am teaching elementary topology. Surely, I’d learn nothing new though I’d enjoy myself. It turns out: that isn’t the case. Very often I find myself starting to give a proof of something and find myself making (correct) assumptions that, well, I last proved 30 years ago. Then I ask myself: “now, just why is this true again?”

One of the fun projects is showing that the topologist’s sine curve is connected but not path connected (if one adds the vertical segment at x = 0). It turns out that this proof is pretty easy, BUT…I found myself asking “why is this detail true?” a ton of times. I drove myself crazy.

Note: later today I’ll give my favorite proof; it uses the sequential definition of continuity and the subspace topology; both of these concepts are new to my students and so it is helpful to find reasons to use them, even if these aren’t the most mathematically elegant ways to do the proof.

This is why I proved the Intermediate Value Theorem using the “least upper bound” concept instead of using connectivity. The more they use a new concept, the better they understand it.

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