I have encountered a proof of the statement that the "The Topologist's sine curve is connected but not path connected" and I am not able to understand some part.

Example 5.2.23 (Topologist’s Sine Curve-I). Let $$ A := \{ (x, \sin(\pi/x)) : 0 < x \leq 1 \} \quad\text{and}\quad B := \{ (0,y) : -1 \leq y \leq 1 \}. $$ Let $X = A \cup B \subset \mathbb{R}^2$ be given the induced metric topology. We claim that $X$ is connected but not path connected.

Let $\gamma \colon [0,1] \to X$ be a path joining $(0,0)$ to $(1,0)$. We write $\gamma(t) = (\gamma_1(t), \gamma_2(t))$. Since $B$ is closed in $X$, the inverse image $\gamma^{-1}(B)$ is closed, $0 \in \gamma^{-1}(B)$. Let $t_0$ be the least upper bound of this closed and bounded set. Obviously, $t_0 \in \gamma^{-1}(B)$. Note that $0 < t_0 < 1$. We claim that $\gamma_2$ is not continuous at $t_0$.

For $\delta > 0$ with $t_0 + \delta \leq 1$ we must have $\gamma_1(t_0 + \delta) > 0$. Hence there exists $n \in \mathbb{N}$ such that $\gamma_1(t_0) < 2/(4n+1) < \gamma_1(t_0 + \delta)$. By the intermediate value theorem applied to the continuous function $\gamma_1$, we can find $t$ such that $t_0 < t < t_0 + \delta$ and such that $\gamma_1(t) = 2/(4n+1)$. Hence $\gamma_2(t) = 1$ and $|\gamma_2(t) - \gamma_2(t_0)| \geq 1$. We therefore conlude that $\gamma_2$ is not continuous at $t_0$.

(Original images here and here.)

My doubt here is whether if :
$t_0$ is Sup{$\gamma^{-1}(B)$},
B={$(0,y)|-1\leq y\leq 1$}
then why $\gamma_2(t_0)<0$ as there is statment that $|\gamma_2(t)-\gamma_2(t_0)|\geq 1$ ?[I know that value of $\gamma_2(t)=1$
Any help will be appreciated.

  • $\begingroup$ Uniform continuity is the key here... $\endgroup$
    – dmtri
    Aug 23, 2018 at 14:49

1 Answer 1


If $\gamma_2(t_0) \le 0$ then the above argument works.

If $\gamma_2(t_0) \ge 0$, then replace $\frac{2}{4n+1}$ everywhere in the argument with $\frac{2}{4n-1}$ so that $\gamma_2(t) = -1$.

In either case you are able to find a $t \in (t_0, t_0 + \delta)$ such that $|\gamma_2(t) - \gamma_2(t_0)| \ge 1$.

  • 2
    $\begingroup$ The main point is, the second part is ignored in the excerpt from the book, and it ought not to be. At least it should've been mentioned in some way. $\endgroup$
    – Arthur
    Aug 22, 2018 at 5:01

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