Could you please check my proof to the following exercise:

Consider the topologists's sine curve $X$ defined by: $$X = \{\ \{(0,0)\}\ \cup \ \{(x,\sin(\frac{1}{x})) \in \mathbb{R}^2 \text{ for } x > 0 \}\}$$ We will prove that $X$ is connected. We start by assuming that $X = U_1 \cup V_1$ where $U_1$ and $V_1$ are open in $X$ and $U_1 \cap V_1 = \emptyset$.

a) Explain why there exist open sets $U,V \subset \mathbb{R}^2$ such that $$U_1 = U \cup X \text{ and } V_1 = V \cup X$$

b) W.l.o.g. we assume that $(0,0) \in U_1$. Prove that there is a $x_0 > 0$ such that $(x_0,\sin(\frac{1}{x_0})) \in U_1$.

c) Explain why $X \setminus \{(0,0)\}$ is connected.

d) Define $U_2 = U_1 \setminus \{(0,0)\}$. Prove that $U_2$ is open in $X \setminus \{(0,0)\}$ and $X \setminus \{(0,0)\} = U_2 \cup V_1.$ Conclude from that that $X \setminus \{(0,0)\} = U_2$ and therefore $X= U_1$.

My take on the exercise:

a) This is precisly the concept of the subspace topology.

b) If $U_1$ would be a singleton it could not be open, therefore there have to be additional elements in $U_1$. By definition of $X$ these have to have the form $(x_0,sin(1/x_0))$.

c) It is the image of the connected set $(0,\infty)$ under the continuous function $$f: x \in (0,\infty) \mapsto (x,\sin(\frac{1}{x}))$$

and therefore connected.

d) Punctured open sets are still open, therefore $U_2$ is open in $X \setminus \{(0,0)\}$. By assumption in b) $(0,0)$ is in $U_1$ and $U_1$ and $V_1$ are assumend to be disjoint, therefore $X \setminus \{(0,0)\} = U_2 \cup V_1$. Since by b) $U_2$ is non-empty and by c) $X \setminus \{(0,0)\}$ is connected we see that $X \setminus \{(0,0)\} = U_2$ (since $U_2$ is open) and so clearly $X = U_1$.

Thanks in advance.

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    $\begingroup$ You mean intersection in a), right? $\endgroup$ – Laz Oct 10 '18 at 23:44
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    $\begingroup$ The step b) needs to find an $x_0$ such that $(x_0, sin(1/x_0))$ belongs to a neighborhood of (0, 0). A singleton may be open in some cases (though not in this case). $\endgroup$ – Miles Zhou Oct 10 '18 at 23:50
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    $\begingroup$ Generally: (1). If $X$ is any space and if $Y$ is a dense subset of $X$ and if $W$ is an open subset of $X$ then $X\cap W=\emptyset \iff W=\emptyset.$ (2). Let $Y$ be a dense connected subspace of a space $X$, and let $X=U\cup V$ where $U,V$ are open disjoint subsets of $X.$ Then $ U_Y=U\cap Y, V_Y=V\cap Y$ are disjoint open subsets of the space $Y$ and their union is $Y,$ so one of $U_Y,V_Y$ is empty. Therefore by (1), one of $U,V$ is empty.... So a space with a dense connected subspace is connected. In your Q, let $Y=X\setminus \{(0,0)\}$... This is not meant as a criticism of your work. $\endgroup$ – DanielWainfleet Oct 11 '18 at 0:43
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    $\begingroup$ General fact: $A$ connected and $A \subseteq B \subseteq \overline{A}$ then $B$ is connected. $\endgroup$ – Henno Brandsma Oct 11 '18 at 5:06

a) Exactly right.

b) You need to show that $(0, 0)$ is not isolated. Note that, if $(0, 0)$ were replaced with the point, say, $(0, 10)$, this would no longer be true, as the open ball $B((0,10); 1)$ would intersect the set at only $(0, 10)$, proving that the singleton set $\lbrace (0, 10) \rbrace$ is open (and indeed clopen, which would make the set disconnected).

Find a sequence of points $(x_n, \sin(1/x_n))$ that converge to $0$ (hint: look at the roots of the function).

c) Good.

d) The logic is all there, but I feel that it's all just a little bit too brief. I feel like, for example, the fact that $U_2 \cap V_1 = \emptyset$ could be explicitly mentioned, given that it is necessary to conclude that $X \setminus \{ (0, 0) \} = U_2$.


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