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The function $g\colon (0,\infty)\to \mathbb{R}$, defined by $g(x)=\frac{1}{x}$ sends an open interval to an open interval. The sine function $h\colon \mathbb{R}\to [-1,1]$, $h(x)=\operatorname{sin}x$ is an open map.

So I think that the composition $f\colon (0,\infty)\to [-1,1]$, $f(x)=\operatorname{sin}(\frac{1}{x})$ is an open map, but I found a statement that the function $f$ is continuous, but neither open nor closed.

I also directly tried to prove that $f$ is an open map, and I think that the image of an open interval in $(0,\infty)$ is one of the following form:

$$(a,b), \quad (b,1], \quad [-1, a), \quad [-1,1]$$

Why is the function $f$ not open?

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    $\begingroup$ @AnneBauval I believe it is just confusion about what OP actually means - to quote Wikipedia: Warning: Many authors define "open map" to mean "relatively open map" (for example, The Encyclopedia of Mathematics) while others define "open map" to mean "strongly open map". In general, these definitions are not equivalent so it is thus advisable to always check what definition of "open map" an author is using. $\endgroup$
    – Lucas
    Feb 12 at 12:31
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    $\begingroup$ I think @AdamRubinson is wrong. $f$ does not map $(0,a)$ to $[1,1]$. $\endgroup$ Feb 12 at 12:32
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    $\begingroup$ @Lucas Since the codomain $[-1,1]$ is specified, there is no ambiguity or possible confusion. The OP's function is open. Anyway, Wikipedia's definitions of "strongly open" and "relatively open" coincide here, since $f$ is onto. Wikipedia mentions this obvious fact just after its warning. $\endgroup$ Feb 12 at 12:34
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    $\begingroup$ @AnneBauval I agree, I was just trying to explain where AdamRubinson came from - maybe its OPs unfortunate use of $h:\mathbb{R} \to [−1,1], h(x)=\sin x$ (where it should have been $h: \mathbb{R}\to[−1,1], x \mapsto \sin x$ $\endgroup$
    – Lucas
    Feb 12 at 12:38
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    $\begingroup$ I can see no difference between these 2 definitions of $h$ nor how they could cause Adam's mistake, which seems more basic: $f(0,a)\ne\{1\}$. $\endgroup$ Feb 12 at 12:42

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Edit: As Anne Bauval found out, the source is indeed erroneous. See her answer for details and a better explanation, with picture of the source. I'll leave my original answer nonetheless:

After a discussion in the comments, I have now decided to post an answer to this question.

The function you describe (in a bit of unfortunate notation) is $f:(0,\infty) \to [-1,1], x \mapsto \sin(1/x)$. Since you specify the codomain, you presumably mean the subspace topology on $[-1,1]$ and therefore the function is an open map, as @AnneBauval correctly pointed out (this is also the only way that $\sin(x)$ can be open in our situation).

The difference with your source likely comes from the fact that if you specify $[-1,1]$ as your codomain with the subspace topology then $[-1,1]$ is an open set, whereas the source you looked at presumably used $f:(0,\infty) \to \mathbb{R}, x \mapsto \sin(1/x)$, in which case $(0,a)$ gets mapped to $[-1,1]$ for every positive $a$, which is of course not open in $\mathbb{R}$ (with the usual topology).

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  • $\begingroup$ @AnneBauval Sorry, I switched $f$ and $h$ and have edited my answer. I guess weather the notation is good is a matter of taste, I have never seen this notation before. So much confusion on such a simple question. Its best we lay this to rest. $\endgroup$
    – Lucas
    Feb 12 at 13:25
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    $\begingroup$ Actually this post was a duplicate. I found it preferable to edit an answer in the older post. I also added a screenshot of the source, which is surprisingly erroneous. (Contrarily to our guess, it mentions explicitely $[-1,1]$ as the codomain but still claims the map is not open). $\endgroup$ Feb 12 at 13:27
  • $\begingroup$ @AnneBauval Well, this is now very offtopic, but here we go. In my experience (and this may be different in other countries) the notation $f(x)=$ is used when the domain and codomain are left implicit in cases they don't really matter, and the notation $f:X \to X, x \mapsto f(x)$ specifies everything. I was only thinking of the possibility that at a glance, Adam Rubinson saw $f(x) =$ and assumed the codomain to be $\mathbb{R}$. Of course, this all does not really matter, as you said, everything is explicitly specified, there should be no confusion. I was just offering a possibility. $\endgroup$
    – Lucas
    Feb 12 at 13:53
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    $\begingroup$ Thank you sincerely for your time! $\endgroup$ Feb 12 at 13:59
  • $\begingroup$ Thank you very much !! $\endgroup$
    – user156937
    Feb 12 at 14:04

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