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

February 16, 2015

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

Filed under: student learning, topology — Tags: connected, path connected, topologist's sine curve — 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.

Comments (5)

5 Comments »

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

Reply
- 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
  
  Reply
  - 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
Drat…I’ll have to fix this up. Thanks. 🙂

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

Reply
- 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
  
  Reply

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College Math Teaching

February 16, 2015