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In this exercise $I = [-1, 1]$ and $\mathcal{E}_1$ is the euclidean topology. I was looking at the function: $f:(\mathbb{R},\mathcal{E}_1) \rightarrow (I,\mathcal{E}_1)$ defined by: $$ f(x) = \begin{cases} \sin(\frac{1}{x}) & x \neq 0 \\ 0 & x = 0 \end{cases} $$ This function is not continuous but I was wondering if it is open or not. By definition a function is open is for every open set in the starting topology I get an open set, I think that is the case here.

What does happen if I change $I = [-1, 1]$ with $\mathbb{R}$?

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    $\begingroup$ f seems to have local extrema. so it doesn't seem open to me. $\endgroup$ Commented Feb 18, 2018 at 21:58

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To determine whether a function $f:X \rightarrow Y$ is open, it suffices$^\dagger$ to determine whether it's open on a basis for the topology on $X$. In this scenario, single open intervals $(a,b)$ serve as a basis for $\mathbb{R}$. So let's think about what the image of such an interval looks like.

This consideration breaks down into two cases: whether a complete oscillation between the two extrema ($-1$ and $1$) is made in $f\Big((a,b) \Big)$. If one is, then every value in $[-1,1]$ is hit. Since $[-1,1]$ is the whole space, it is open, so $(a,b) \mapsto \{\text{open set}\}$. If a complete oscillation is not made, then the extrema of $f\Big((a,b) \Big)$ will be either an open interval looking like $\Big(\sin(1/a), \ \sin(1/b) \Big)$—or the other way around—or a half-open interval with $\sin(1/a)$ or $\sin(1/b)$ on the open end and an extremum on the closed end. In either case, these are open sets in $[-1,1]$.

So $f$ is open because basis elements in $\mathbb{R}$ (open intervals) get sent to open sets in $[-1,1]$ under the subspace topology inherited from $\mathbb{R}$. Notice that this conclusion is quite dependent on the choice of target space. If we'd considered this same function as $f: \mathbb{R} \rightarrow \mathbb{R}$, then $f$ would not be open because $[-1,1]$ is not open in $\mathbb{R}$.


$^\dagger$This is due to the fact that $\displaystyle f \left( \bigcup_k U_k \right) = \bigcup_k f(U_k)$.

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$f$ is not open, since $f((0,1))=[-1,1]=I.$ We can pretty much aply the same reasoning to all of $\mathbb{R}$, so that f is not continuous neither open.

EDIT: That'd be the case if $f$ was defined from $\mathbb{R}$ to $\mathbb{R}$. How this is not the case, turns out that $f$ is open :v

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If $f$ is open then for every $x$, and for every open set $U$ such that $x \in U$, there exists an open set $V$ such that $f(x) \in V$ and $V \subseteq f(U)$.
Take $x=2/\pi$. Take the open interval $(2/\pi - \varepsilon, 2/\pi + \varepsilon)$.
Since $f(x) = 1$ we would need to find some $\delta$ such that $$(1 - \delta, 1+\delta) \subseteq f((2/\pi - \varepsilon, 2/\pi + \varepsilon)) \subseteq [-1,1],$$ a contradiction.


Edit: This is actually for $f: \mathbb R \to \mathbb R$.
If we consider $f: \mathbb R \to I$, then $I$ is open...

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