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I'm studying about manifolds. I'm an engineer and want to grasp the concept of manifold without too deep dive into rigorous mathematics. My question is whethere the "topologist's sine curve" defined as $S = \{(0, y) : y \in (-1,1)\} \cup \{(x, \sin (1/x)) : x \in (0, 2/\pi)\}$ can be a manifold. I tried to find some explnations in books and web. However, what I found are all about (path/local) connectedness. I'm not familiar with topology and not want to study it extensively. So I tried to understand the question with very minimal knowledge.

As far as I understand, to say that a set is a manifold, I need to set a topology (open subsets) on the set. As a subset of $\mathbb{R}^2$, every open set in $S$ should have a form $S \cap A$ for an open set $A \in \mathbb{R}^2$. Any interval in the vertical axis part of $S$, e.g. $I = \{(0, y) : 0 < y < 1/2\}$, cannot be an open set in $S$. It is because every open set in $\mathbb{R}^2$ containing $I$ also contains some other part of $S$ : there is no way to express $I$ in the form of $S \cap A$ where $A$ is open in $\mathbb{R}^2$. Then there is no homeomorphism from an open subset of $S$ containing $I$ to an open set in $\mathbb{R}$. It is because, if there is one, the preimage of an open interval in $\mathbb{R}$, to which $I$ is mapped, must be open but $I$ is not open. So there cannot be a coordinate chart for $I$ and $S$ is not a manifold. This is what I understand. My first question is whether or not this understanding is correct.

The second question is whether $S$ can be a manifold if I give a different topology. I cannot express precisly how to define the topology but intuitively the sine function part is homeomorphic to an open interval in the real line and the vertical part is also homeomorphic to an open interval in the real line, if I consider $S$ as a space of its own, not a subspace of $\mathbb{R}^2$. Is this true? If true, how can I express the topology precisley?

This question is not a duplicate of Topologist's Sine Curve not a regular submanifold of $\mathbb{R^2}$?. I read it. However, the concept of connectedness confuses me. As I said in the first paragraph, I want to understand my question with minimal knowledge on topology such as just the definition of open sets, continuous functions and manifolds. Moreover the linked question does not concern my second question.

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    A space is connected iff it is NOT the union of a pair of disjoint non-empty open sets. A space $X$ is path-connected iff for any $p,q\in X$ there is a continuous $f:[0,1]\to X$ with $f(0)=p$ and $f(1)=q$. We call such an f a path from $p$ to $q$. Path-connectedness implies connectedness,but the two properties are not equivalent..... $S,$ as a sub-space of $\mathbb R^2$, is not path-connected but it is connected, because the "sine" part of $S$ is a connected open subspace of $S$ whose closure is $S.$ – 2017-01-10
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    As the answers mention, if you take away the topology from $S$, all you have left is a "bare" set of cardinality $2^{\aleph_0}$. You can turn that into many manifolds, such as $\mathbb{R}^{17}$, but this is not what you have in mind. I think you want to move away the vertical segment from the rest. So if we consider $$\widetilde{S} = \{(-1, y) : y \in (-1,1)\} \cup \{(x, \sin (1/x)) : x \in (0, 2/\pi)\}$$ then $\widetilde{S}$ is a (non-connected) 1-dimensional manifold. It should be homeomorphic to $\{ -1, +1 \} \times \mathbb{R}$. – 2017-01-10
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    And with the obvious bijection between $S$ and $\widetilde{S}$ you can move the topology of $\widetilde{S}$ over to $S$ if you want. – 2017-01-10

2 Answers 2

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Choosing a specific subset $I$ and showing that there's no coordinate chart for $I$ is not enough. However, you are indeed very close with your idea.

If $S$ is a manifold, then by definition the point $p=(0,0)$ must have a neighborhood which is homeomorphic to $\Bbb{R}^n$. However, $\Bbb{R}^n$ is path connected, and all neighborhoods of $p$ are not. Path connectedness is preserved by homeomorphism. Therefore $S$ cannot be a manifold.

You can consider the same set with a different topology, but as John has pointed out, you are no longer talking about the Topologist's Sine Curve anymore. Any intuition you have about what that topological space "looks like" goes out the window if you introduce a different topology.

For instance, you can give $S$ the discrete topology, which means every subset of $S$ is open. Then for every $p\in S$, $\{p\}$ is a neighborhood of $S$ which is homeomorphic to $\Bbb{R}^0$, so it is a $0$-manifold.

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    Why is "a subset cannot have coordinates" not enough? Would you explain it more? In my understanding, it is a way to express the subset is not homeomorphic to $\mathbb{R}^n$. – 2017-01-10
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    @cleone Because by definition, $S$ is a manifold if for every $p\in S$ *there exists* a neighborhood $U$ of $p$ which is homeomorphic to $\Bbb{R}^n$. It doesn't have to be true that *every* neighborhood of $p$ is homeomorphic to $\Bbb{R}^n$. – 2017-01-10
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    @Mathers thank you. Now I understand it. – 2017-01-10
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    @cleone No problem – 2017-01-10
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The topologist's sine curve $S$ is a subspace of $\mathbb{R}^2$ meaning that it is a subset of $\mathbb{R}^2$ and inherits its topology from the topology of $\mathbb{R}^2$. In order to be a manifold it must be locally Euclidean in the inherited topology. That means that it must also be locally connected but, as noted, it is not locally connected on $0\times[-1,1]$.

You ask whether perhaps some other topology on $S$ would make it a manifold.

Perhaps, but with some topology other than the one it inherits from $\mathbb{R}^2$ it is no longer the topologist's sine curve.

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    I'm still confused with the notions of connectedness. I think they are not so intuitive. I'm not comfortable with "connected but not path connected or locally connected." Anyway, you mean what I said in the second paragraph of my questioin is just another way of saying $S$ is not locally connected. Is it right? – 2017-01-10
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    Everything in our everyday experience which is connected is also path connected and locally connected. It is only in abstract examples such as $S$ where a connected "object" can fail to be path or locally connected. There is no arc from the point $\left(\pi,\sin\left(\frac{1}{\pi}\right)\right)$ to $(0,1)$ in $S$ so it is not path connected, nor is there a connected open set in $S$ containing the point $(0,1)$ so it is not locally connected at $(0,1)$. – 2017-01-10