# Problem about $\sin(1/x)$ in topology. (open and closed functions)

Let $$f:(0,\infty)\to [-1,1]$$ defined by $$f(x)=\sin(1/x)$$. Show that $$f$$ is continuous but neither open nor closed, where $$(0,\infty)$$ and $$[-1,1]$$ are a subspace of $$\mathbb{R}$$ with usual topology.

First, $$f$$ is continuous, since if $$A \subseteq [-1,1]$$ is a open, then $$f^{-1}(A)$$ is open, where $$A=[-1,1] \lor A=[-1,b) \lor A=(a,1] \lor A=(a,b)$$. If $$-1 \leq a.

But i don't know show that $$f$$ neither open nor closed.

This problem is in General Topology (Schaum):

• @AlexR: Exhibiting an open subset $A$ of $(0,\infty)$ such that $f(A) = [-1,1]$ does not show that $f$ isn't an open mapping since $[-1,1]$ is open in the codomain. Apr 20, 2015 at 15:08
• @kahen I've edited my comment just as you commented :D Apr 20, 2015 at 15:08
• I also do not understand your purported argument as to why $f$ is continuous. Isn't it much easier to argue that $f$ is the composition of continuous maps? Apr 20, 2015 at 15:08

$$f:(0,\infty)\to [-1,1],x\mapsto\sin\frac1x$$ is indeed (as suggested by @AlexR) not closed, since the image of the closed subset $$[1,\infty)$$ is $$(0,\sin1]$$, which is not closed in $$[-1,1]$$.
But $$f$$ is open, since the image of an open interval in $$(0,∞)$$ is one of the following form: $$(a,b), \quad (b,1], \quad [-1, a), \quad [-1,1]$$ and all of them are open in $$[-1,1]$$.
• What is $f([1, \infty))$?
• What is $f((\frac1{2\pi + \epsilon}, 1))$?