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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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The correct definition is $S = \{(x,\sin(1/x))\mid 0 < x \le 1\}$, edit. –  Martín-Blas Pérez Pinilla Feb 7 at 8:27
    
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. –  hengxin Feb 7 at 14:29

1 Answer 1

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

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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? –  hengxin Feb 7 at 14:37
    
You are right. I was thinking in the curve $x\sin(1/x)$... If $y\ne 0$ is required, the reason is more subtle... –  Martín-Blas Pérez Pinilla Feb 7 at 16:26

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