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Let $A = \{ (x,y) \in \mathbb{R}^{*} \times \mathbb{R}; y=\sin(\frac{1}{x})\}$. I need to do 3 things:

  1. Find the accumulation points of A (this I've done, it's $A \cup \{0\} \times [-1,1]$);
  2. Show that the closure (of A) it's connected. (The closure it's actually, the accumulation points of A)
  3. Show that the closure (of A) it's not path-connected.

I'm having some real troubles to show the properties 2 and 3. Could please someone help? Thank you!

big_GolfUniformIndia
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1 Answers1

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This question is about a variant of the topologist's sine curve. Similar questions have been asked many times in this forum. But strictly speaking, your question is not a duplicate.

Let $A_+ = \{ (x,y) \in (0,\infty) \times \mathbb{R}; y=\sin(\frac{1}{x})\}$ and $A_- = \{ (x,y) \in (-\infty,0) \times \mathbb{R}; y=\sin(\frac{1}{x})\}$. These are homeomorphic copies of $ (0,\infty)$ and $(-\infty,0)$, thus connected (they are even path connected). Since you solved 1., you will agree that $\overline{A_\pm} = A_\pm \cup \{0\} \times [-1,1]$. Now see the accepted answer to Topologist's sine curve is connected to understand that $\overline{A_\pm}$ are connected and hence also $\overline{A} = \overline{A_+} \cup \overline{A_-}$ is connected.

Why is $\overline{A}$ not path connected? Either you use the methods in the answers to e.g. Topologist's sine curve is not path-connected or Is this proof that the topologist's sine curve is not path connected valid, or you directly invoke the theorem that $T = \{ (x,y) \in (0,1] \times \mathbb{R}; y=\sin(\frac{1}{x})\} \cup \{0\} \times [-1,1]$ is not path connected. In fact, if $\overline{A}$ were path connected, then also $T$ would be path connected because the map $$r : \overline{A} \to T, r(x,y) = \begin{cases} (0,y) & x \le 0 \\ (x,y) & 0 \le x \le 1 \\ (1, \sin(1)) & x \ge 1\end{cases}$$ is a continuous surjection.

Paul Frost
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