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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 $-1 \leq a<b \leq 1$.

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

This problem is in General Topology (Schaum): enter image description here

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  • $\begingroup$ @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. $\endgroup$
    – kahen
    Apr 20, 2015 at 15:08
  • $\begingroup$ @kahen I've edited my comment just as you commented :D $\endgroup$
    – AlexR
    Apr 20, 2015 at 15:08
  • $\begingroup$ 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? $\endgroup$
    – kahen
    Apr 20, 2015 at 15:08

2 Answers 2

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$f:(0,\infty)\to [-1,1],x\mapsto\sin\frac1x$ 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 $[-1,1]$.

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Hint

  • What is $f([1, \infty))$?
  • What is $f((\frac1{2\pi + \epsilon}, 1))$?
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