TITLE: Compactness of the set.
QUESTION [0 upvotes]: I have given that the set 

$$\left \{ (x,y)\in \mathbb{R^2}: x>0,y=\sin\frac{1}{x} \right \}\bigcap \left \{(x,y) \in \mathbb{R^2}: x>0,y=\frac{1}{x} \right \}$$

and i have to tell does  it is compact or not.I know that any set which is closed and bounded in $\mathbb{R^2}$ is compact. But here it is very dificult for me understand this how can i find there intersection ,seems like these two functions only can be same at $\infty$
Please help

REPLY [3 votes]: Notice that for $x > 0$, we have $\left| \sin x \right| < x$. Therefore, $\sin x < x$ for $x > 0$.
Now, for $x > 0$, $\dfrac{1}{x} > 0$ and infact, the function $\dfrac{1}{x} : \left( 0, \infty \right) \rightarrow \left( 0, \infty \right)$ is a bijection.
Hence, there is no such $x \in \mathbb{R}$ so that $\sin \dfrac{1}{x} = \dfrac{1}{x}$. This proves that your set is empty and hence compact.
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