The topologists sine curve is the set:

$$ T =\left\{\left(x, \sin\left(\frac{1}{x}\right)\right): x \in (0,1) \right\} \cup\{(0,y):y\in \mathbb{R}\}$$

A common (and fun!) exercise is to prove that the set is connected. The usual answer goes like this:

(1) the graph of the function $f(x) = \sin\left(\frac{1}{x}\right)$ is the graph of a continuous function over a connected domain, so there is an homeomorphism between its domain and the graph, thus the graph is connected

(2) $(0,0)$ is a limit point of the graph, so if we add it to the set connectedness is preserved

(3) $T$ is then the union of the set constructed in the last step and the $y$ axis, being the union of two non disjoint connected sets, $T$ is also connected.

I am interested in solving the problem directly at step 1 by finding a continuous map from some simple connected set to $T$.

My (wrong) candidate would be: Let $M = \{(0,y) \in \mathbb{R}^2 : y \in \mathbb{R} \}\cup\{(x,0) \in \mathbb{R}^2 : x \in (0,1) \}$. Let $g:M\to T$ defined as:

$$g(x,y) = \begin{cases} (x,\sin\left(\frac{1}{x}\right)) & \mbox{if $0\lt x \lt 1$,}\\\ (x,y) & \mbox{otherwise}\end{cases} $$

The function would "transform" a connected subset of the $x-y$ axis into $T$, but now I see that my construction is not continuous*. Could we change $g$ in any way to make it continuous? Or maybe there is another connected (and simple!) set for which there is a continuous map into $T$?

*we could get a sequence $(X_k,0)$ in $M$ going to $(0,0)$ but for which $g(X_k,0)$ is not going to $g(0,0)$

** I have no definition for "simple", I hope it makes sense

  • $\begingroup$ Another set that has also been called the topologists' sine curve is $C=A\cup (\{0\}\times [-1,1])$ where $A=\{(x,\sin 1/x): 0<x<1\},$ which can be shown to be connected by your method, or by observing that $A$ is a dense connected sub-space of $C,$ which implies that $C$ is connected. $\endgroup$ May 10, 2018 at 7:27
  • $\begingroup$ Another example of a connected but not path-connected and not locally-connected sub-space of $\Bbb R^2$ is $[\; (\bigcup_{n\in \Bbb Z^+}(A(n)\cup B(n)\;]\; \bigcup \;[\;\Bbb R\times \{0\}\;] $ where $A(n)$ is the closed line-segment joining $(0,1)$ to $(n, 1/(n+1)) $ and where $B(n)=(-\infty,n]\times \{1/(n+1)\}.$ $\endgroup$ May 10, 2018 at 7:43

1 Answer 1


No. It's not possible to get a continuous surjection from your $M$ to $T$, since $M$ is path-connected and $T$ is not. More generally, any space with a continuous surjection to $T$ will have to not be path-connected. Since $T$ is one of the most basic examples of a connected space that is not path-connected, I highly doubt there is any connected space with a continuous surjection to $T$ which you would consider "simpler".

  • $\begingroup$ Thanks! Path-connectedness is really the concept that I was missing here. But now I think we could go for another direction: if we still want to use topological invariance over maps to prove the connectedness of T, now we know that continuous functions are not a good tool for that because they also preserve path - connectedness. Do you know if there is any other "kind" of map that we could use? $\endgroup$ May 10, 2018 at 4:41
  • $\begingroup$ There is not really any other kind of map that is natural and useful for topological spaces. Basically, I don't believe that there is any way to make your proposed alternate proof strategy work. $\endgroup$ May 10, 2018 at 4:44
  • $\begingroup$ Great( though a little disappointing). Better than keep hitting the wall. $\endgroup$ May 10, 2018 at 5:03

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