Topologists Sine Curve problem

Question

Topologist's Sine Curve:

$A:=\{(x,\sin \frac{\pi}{x}):0<x\leq 1\}\cup B:=\{(0,y):-1\leq y\leq 1\}$

I can show that it is connected. Problem is I can't show that it is path connected.

Let $\gamma : [0,1] \rightarrow X$ be a path joining $(0,0)$ to $(1,0)$. We write $\gamma(t) = (\gamma_1(t),\gamma_2(t))$.

The author asks to show that $\gamma _2$ is not continuous at $t_0$ where $t_0$ is the least upper bound of the closed and bounded set $\gamma^{-1} B$

Such a $t_0$ exits since $B$ is closed so $\gamma^{-1} B$ is closed and $0\in \gamma^{-1} B$.Thus we will get a least upper bound of $\gamma^{-1} B$

But I can't proceed next. Any help?

The point $(1,0)$ is not in the set $A \cup B$. Did you mean $(1,\sin(1))$? — Michael, Feb 26 '15 at 02:59
I think the point is that everything in $(t_0, 1]$ maps to a point on the sin curve $A$, and the point $t_0$ must map to a point in $B$ that is not $(0,0)$. — Michael, Feb 26 '15 at 03:09

kobe · Accepted Answer · 2015-02-27T18:22:59.033

Note $0 < t_0 < 1$. Let $\gamma_2(t_0) \le 0$. Given $\epsilon > 0$ with $t_0 + \epsilon \le 1$, $\gamma_1(t_0 + \epsilon) > 0$. For some $k\in \Bbb N$, $\gamma_1(t_0) = 0 < \frac{2}{4k+1} < \gamma_1(t_0 + \epsilon)$. By the intermediate value theorem, for some $t\in (t_0, t_0 + \epsilon)$, $\gamma_1(t) = \frac{2}{4k + 1}$. Then $\gamma_2(t) = \sin \frac{(4k+1)\pi}{2} = 1$. Therefore $|\gamma_2(t) - \gamma_2(t_0)| \ge \gamma_2(t) - \gamma_2(t_0) \ge 1$, and consequently, $\gamma_2$ is discontinuous at $t_0$. A similar argument holds when $\gamma_2(t_0) > 0$.

Topologists Sine Curve problem

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