# Proving that the topologists's sine curve is connected

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 $$X = U_1$$.

• You mean intersection in a), right? – Laz Oct 10 '18 at 23:44
• 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). – Miles Zhou Oct 10 '18 at 23:50
• 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. – DanielWainfleet Oct 11 '18 at 0:43
• General fact: $A$ connected and $A \subseteq B \subseteq \overline{A}$ then $B$ is connected. – 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 $$X \setminus \{ (0, 0) \} = U_2$$.