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Let $$A=\{(x,\sin(1/x)) \mid x > 0 \} \cup \{(0,y) \mid y\in[-1,1]\} \subset\mathbb{R}^2$$

how to prove that this set is not path-connected(or just not connected)?

Thank you

Stefan Hansen
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user74311
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2 Answers2

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Let $C$ be that sine curve part, and $S$ be the vertical segment. Suppose there exists a path $\omega: [0, 1] \to A$ that connects a point $(0, 1) \in S$ with $(\frac{1}{\pi}, 0) \in C$. Consider the set $U = \{t \in [0, 1]: t \in C \} = \omega^{-1}(C)$. As sine curve is open in $A$, $U$ is an open subset of $[0, 1]$. Let $t_0 = \inf U$. As $U$ is open and $0 \not \in U$ (because $\omega(0) = (0, 1) \in S$, $\omega(t_0) \in S$, but for some $t_1 > t_0$, $\omega((t_0, t_1)) \subset C$. Consider a small open neighbourhood $V \subset A$ of $t_0$ that has infinitely many disconnected components. By continuity of $\omega$, there's some interval $(t_0 - \epsilon, t_0 + \epsilon)$ such that $\omega((t_0 - \epsilon, t_0 + \epsilon)) \subset V$. For some point in $t \in (t_0, t_0 + \epsilon)$, $\omega(t)$ is in a different component of $V$ than $t_0$, which is a contradiction, because $\omega((t_0 - \epsilon, t_0 + \epsilon))$ is connected, as a continuous image of a connected set.

xyzzyz
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  • Out of all the points in S, why did you choose the point (0,1)? Also, you write 'Consider a small open neighbourhood V ... that has infinitely many disconnected components.' - aren't you assuming what you're trying to prove? If it's a subset of A (as you state), then how can there be infinitely many disconnected components? It seems to me as if there should be 2 disconnected components (C and S), or there should be one connected component (that is, A). – Charlie Jun 26 '20 at 00:04
  • It doesn't matter which point you choose in S, since S is path connected. As for your other question, you missed the crucial word "small". I actually made a typo in my answer, so I just edited it. The point is that $\omega(t_0)$ must lie on the horizontal segment, since $t_0 \not \in U$ (as $U$ is open in $[0, 1]$ and $0 \not \in U$). Then, when you take a small enough ball around $\omega(t_0)$ in $A$, small enough to not intersect horizontal lines $(\pm 1, y)$ for $y \in \mathbb{R}$, since horizontal segment is close to infinitely many coils, you get infinitely many disconnected segments – xyzzyz Jun 27 '20 at 16:03
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Suppose this set is path connected. Consider the point $(0,0)$ and the point $(1,\sin(1))$. The only way to link these two points is along the path $(x,\sin(\frac{1}{x}))$. However $\lim_{x \rightarrow 0}\sin(\frac{1}{x})=$DNE. Hence there is no path connecting $(0,0)$ to $(1,\sin(1))$, a contradiction. Therefore the set is not path connected.

Wintermute
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  • To mtiano,

    is it the general way to prove it?

    since the set contains (0,y) -1<=y<=1, i think i have to check every (0,y) to (x,sin(1/x)).

    is it enough to check just 2 points and including all possible link?

    THank you.

     – user74311 Apr 25 '13 at 11:44
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    This is not a proof. Why "the only way to link these two points" is along that path? – xyzzyz Apr 25 '13 at 12:09