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Show that the following set is connected subset of $\mathbb R^2$ :

$$\left\{\left(x,\sin \frac{1}{x}\right)\in \mathbb R^2|0<x<\infty\right\}\cup\left\{(0,0)\right\}$$

Attempt : Here $A=\left\{\left(x,\sin \frac{1}{x}\right)\in \mathbb R^2|0<x<\infty\right\}$ is connected as, $f:(0,\infty) \to \mathbb R^2$ defined by $f(x)=\left(x,\sin \frac{1}{x}\right)$ is continuous and $(0,\infty)$ is connected. Again $B=\{(0,0)\}$ is also connected. $A\cap B=\emptyset$. So how can I prove that $A\cup B$ is connected ?

If $A$ and $B$ are connected and $A\cap B\not=\emptyset$ then $A\cup B$ is also connected. But it is a sufficient condition. So I can't conclude that $A\cup B$ is disconnected.

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    I don't think that it's true in general that if $A$ and $B$ are connected, then $A \cup B$ is connected. – MathMajor Apr 10 '16 at 03:35
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    http://math.stackexchange.com/questions/317125/topologists-sine-curve-is-connected – Andres Mejia Apr 10 '16 at 03:36
  • As noted, $(0,0$ is the problem area. Assume that $A \bigcup B$ is disconnected. Then there exists neighborhoods $U$ of $B$ and $V$ of $A$ so that $U \bigcap V= \emptyset$, however, $0$ is a limit point, so there is a contradiction. Intuitively, the problem is that every neighborhood of $(0,0$ intersects $A$ somewhere else. – Andres Mejia Apr 10 '16 at 03:38
  • You essentially need to show that $(0,0)$ is a limit point of $f((0,\infty))$. Can you find a sequence $x_n\to 0$ with $f(x_n)=0$? – MPW Apr 10 '16 at 03:50
  • If $A$ is connected and $A\subseteq B \subseteq \overline{A}$ then $B$ is connected too. Use that $(0,0)$ belongs to the closure of the graph for $x>0$. – Mirko Apr 10 '16 at 04:05

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