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In Munkres's Topology, it is claimed that "The topologist's sine curve" is not locally connected without further explanation (See Example 3 of Section 25 "Components and Local Connectedness", 2nd edition).

The topologist's sine curve: Let $S$ denote the following subset of the plane. $$S = \{ (x, \sin(1/x)) \mid 0 < x \le 1\}.$$ The set $\bar{S}$ is called the topologist's sine curve, which equals the union of $S$ and the vertical interval $0 \times [-1,1]$.

An explanation that it is not locally connected can be found here.

The topologist's sine curve is not locally connected: take a point $(0, y) \in \bar{S}, y \neq 0$. Then any small open ball at this point will contain infinitely many line segments from $S$. This cannot be connected, as each one of these is a component, within the neighborhood.

I am able to catch the basic idea of the explanation, except that:

My problem: Why is the origin $(0,0) \in \bar{S}$, as a counterexample, ruled out?

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    $\begingroup$ The correct definition is $S = \{(x,\sin(1/x))\mid 0 < x \le 1\}$, edit. $\endgroup$
    – Martín-Blas Pérez Pinilla
    Feb 7, 2014 at 8:27
  • $\begingroup$ I was following the Munkres's Topology, which usually uses $a \times b$ instead of $(a,b)$. Is this non-standard (I am a beginner of general topology who rely entirely on this book)? Anyhow, I will adopt the $(a,b)$ notation here to be consistent. $\endgroup$
    – hengxin
    Feb 7, 2014 at 14:29
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    $\begingroup$ It is interesting that this question still isn't answered. $\endgroup$
    – ashman
    Dec 22, 2014 at 22:18

2 Answers 2

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The most likely reason is that it is less clear what happens in neighborhoods of $(0,0)$ compared to what happens in neighborhoods of $(0,y)$ for $y\neq 0$. The author is only trying to argue that the space as a whole is not locally connected so does not care whether or not the space is locally connected at $(0,0)$. The author likely felt (either correctly or incorrectly) that more of an argument would be required to demonstrate the lack of local connectivity at this point.

But, you can rest assured knowing that the space is in fact not locally connected at $(0,0)$ and the same argument can be applied here.

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Draw the intersection of a (0,0)-centered ball with the set. How many "chunks" you see?

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  • $\begingroup$ I still cannot find out the difference between the $(0,0)$-centered ball and other balls centered at $(0,y)$ with $y \neq 0$. Could you show me more hints? $\endgroup$
    – hengxin
    Feb 7, 2014 at 14:37
  • $\begingroup$ You are right. I was thinking in the curve $x\sin(1/x)$... If $y\ne 0$ is required, the reason is more subtle... $\endgroup$
    – Martín-Blas Pérez Pinilla
    Feb 7, 2014 at 16:26
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    $\begingroup$ We want to find a neighbourhood which does not contain any connected open neighbourhoods. How would considering only open balls help? We proved that open balls around $(0,0)$ are not connected. What about other neighbourhoods? (I know open balls form a base but can't recall how it helps and was annoyed by not being able to recall) $\endgroup$
    – Hrit Roy
    Nov 21, 2020 at 16:03

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