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.


  1. How do you argue that the sequence a_n goes to zero. Surely I could define my hypothetical path f by letting it be constant on the first half of the interval and only then trying to run over the sine curve?…

    Comment by Andrew. — November 28, 2016 @ 6:07 pm

    • f(0) = 0 by hypothesis. …f is the path where f(0) = (0,0) and f(1/pi) = (1/pi, 0). So f(a_n) =(1/(npi),0) goes to (0,0)

      Comment by blueollie — November 28, 2016 @ 8:27 pm

      • Hi blueollie. I agree that f(0) = (0,0), and that f(a_n) = (1/(npi),0). But I don’t think this implies that a_n should go to zero.

        Comment by Andrew. — November 29, 2016 @ 6:18 pm

  2. Drat…I’ll have to fix this up. Thanks. šŸ™‚

    Comment by blueollie — November 29, 2016 @ 6:33 pm

    • To show that the image of f must include every point of S, you could just compose f with projection to the x-axis. Then you have a continuous function [0,1/pi] to itself that is the identity on the endpoints, so it must be onto by the intermediate value theorem. It then follows that f must be onto.

      Comment by Andrew. — August 21, 2017 @ 1:10 pm

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