The topologist's sine curve is by definition the closure $\overline{S}$ in $\mathbb{R}^2$ of the set $S$ given by $$S = \{ \ x \times \sin \frac{1}{x} \ \colon \ 0 < x \leq 1 \ \}.$$

Let $V = \{ 0 \} \times [-1, 1]$. Then $\overline{S} = S \cup V$.

How to show that $\overline{S}$ is not locally connected?

Although this question has been asked here, I haven't been able to grasp the argument mentioned there.

I have yet to study this document though.

By definition, a topological space is said to be locally connected at a point $x \in X$ if for every open set $U$ in $X$ such that $x \in U$, we can find a connected open set $V$ in $X$ such that $x \in V \subset U$. If $X$ is locally connected at each of its points, then $X$ is locally connected.

At which point(s) does $\overline{S}$ fail to be locally connected and how? A rigorous argument is needed.

My effort:

Let $-1 < b < 1$, and let $0 < \delta < \frac{1}{2} \min ( b +1 , 1 - b )$. Let $p = 0 \times b \in \mathbb{R}^2$, and let $$B(p; \delta) = \{ \ x \times y \in \mathbb{R}^2 \ \colon \ \sqrt{ (x-0)^2 + (y-b)^2 } < \delta \}.$$

Let's choose $b$ such that $S \cap B(p; \delta) \neq \emptyset$. This is possible since. for any $k \in \mathbb{N}$, the function $\varphi(x) = \sin \frac{1}{x}$ for $x > 0$. takes on all the values in the closed interval $[-1, 1]$ if we vary $x$ throughout the closed interval $[ \frac{1}{2 (k+1) \pi }, \frac{1}{2k \pi }]$. So if we take $k$ large enough, we can take $x$ sufficiently close to $0$ and $\sin \frac{1}{x}$ will still traverse the entire closed interval $[-1, 1]$. So we can choose $b$ such that $S \cap B(p; \delta) \neq \emptyset$.

Is this reasoning correct?

Then $B(p;\delta)$ is open in $\mathbb{R}^2$. So $\overline{S} \cap B(p; \delta)$ is open in $\overline{S}$.

Now $\overline{S} \cap B(p; \delta)$ can be wirtten as the union of two disjoint sets, viz. $S \cap B(p; \delta)$ and $V \cap B(p; \delta)$.

Now $V \cap B(p; \delta)$ is open in $\overline{S}$. Am I right?

Now we need to show that $S \cap B(p; \delta)$ is also open. Suppose that the point $p_0 = x_0 \times y_0 \in S \cap B(p; \delta)$. Then $x_0 > 0$, $y_0 = \sin \frac{1}{x_0}$, and $$\sqrt{ x_0^2 + (y_0-b)^2} < \delta.$$

Let's take a real number $\epsilon$ such that $0 < \epsilon < \frac{1}{4} \min ( x_0, \delta - \sqrt{ x_0^2 + (y_0-b)^2} )$. Then the open ball $B(p_0 ; \epsilon) $ does not intersect $V$.

So $\overline{S} \cap B(p_0; \epsilon)$ is contained in $S \cap B(p; \delta)$ so that $S \cap B(p; \delta)$ is open in $\overline{S}$. Am I right?

Is the above reasoning correct? If so, have I been able to formulate it properly?

If this reasonging is flawed, then where do I need to look?

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  • $\begingroup$ @Brian M. Scott, can you please have a look at my post and then give your valuable feedback? $\endgroup$ – Saaqib Mahmood May 9 '15 at 6:17

The set $\overline{S}$ has essentially two different kind of points: $1.$ those on the $y$-axis and $2.$ those who lie on the curve $x\mapsto\sin (x^{-1})$ for $x>0$. Lets compare these points in terms of their basis neighbourhoods.

  1. Note that if we have a point $v=(0,y)\in \overline{S}$ then the intersection of any sufficiently small open ball $B(v,r)$ with $\overline{S}$ is disconnected. You can use the following reasoning for instance. Let $v=(0,y)\in \overline{S}$ and $r<1$ be fixed. For every $k\in\mathbb{N}$ the function $\varphi(x)=\sin(x^{-1})$ takes all values between $-1$ and $1$ on the interval $I_{k}= (\frac{1}{2\pi (k+1)},\frac{1}{2\pi k}]$. Since $\frac{1}{2\pi k}\to 0$ as $k\to\infty$, there exists $k_{r}\in\mathbb{N}$ so that for all $k\geq k_{r}$ we have $\varphi(I_{k})\cap B(v,r)\neq \emptyset$. By the same token, since $r<1$, then for all $k\geq k_{r}$ we find $x_{k}\in I_{k}$ so that $\varphi(x_{k})\in \varphi(I_{k})\setminus B(v,r)$. Take all such $x_{k}$ and denote the set by $A_{k}$. Now, for example, $I:=(0,\frac{1}{2\pi k_{r}}]\setminus \bigcup_{k\in\mathbb{N}}A_{k}$, is a disconnected set and homeomorphic to graph of $\varphi$ when restricted to the set $I$. So the graph of $\varphi$ is disconnected when restricted to $I$. Take then a union of this set with the $y$-axis piece $(\{0\}\times [-1,1])\cap B(v,r)$ and the resulting set remains disconnected, and in fact this is precisely $B(v,r)\cap\overline{S}$. Now $\overline{S}$ can not be locally connected at $v$: All sufficiently small open balls are disconnected.

  2. We keep the same notations as above for this case. For the points on the curve: if $v=(x,\varphi(x))$ for $x>0$, then $x\in I$ for some open connected interval $I$ where $\varphi$ restricts to a homeomorphism to its graph over $I$. The projection to the first component of any small enough open ball $B(v,r)\cap\overline{S}$ is a connected open subinterval of $I$. Since this projection map is precisely the above mentioned homeomorphism, then these balls are a connected open neighbourhood basis on $v$, proving local connectedness on $v$.

Please notify me if I overlooked some detail or if there is a mistake/typo somewhere. The reasoning should be fairly intuitive.

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  • $\begingroup$ how is the set $I$ disconnected? I'm referring to the set $I$ where you wirte, "Now, for example, $I \colon = $...". $\endgroup$ – Saaqib Mahmood May 9 '15 at 4:53
  • $\begingroup$ @SaaqibMahmuud: For each $k$, we have $\emptyset\neq A_k\neq I_k$, so $I$ is just a countable union of disjoint disconnected sets. $\endgroup$ – T. Eskin May 10 '15 at 19:41

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