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I have recently made use of the following generalization of a continuous function, which seems simple enough it ought to have been used before, but I cannot find any references.

We will say a function $f$ has a semi-continuity property if $f^{-1}(U)$ contains a non-empty open set whenever $U$ is a non-empty open set.

Is this a studied property? If $f:X\to Y$ and $Y$ is disconnected, then this disagrees with usual continuity but does provide some nice properties. For instance, if $D$ is a dense set, then $f(D)$ is dense.

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  • $\begingroup$ The empty set is open. Do you mean $f^{-1}U$ contains a non-empty open set whenever $U$ is a non-empty open set? $\endgroup$ Aug 5, 2014 at 21:35
  • $\begingroup$ I do. The question has been corrected. $\endgroup$ Aug 5, 2014 at 21:42
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    $\begingroup$ I wouldn't call this "semi-continuity", since upper and lower semi-continuity have nothing to do with this. $\endgroup$ Aug 5, 2014 at 22:11
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    $\begingroup$ Here's a note: A sufficient condition for $f : X \to Y$ to have this property is that there is an open set $U \subset X$ such that $f$ is continuous on $U$, and $f(U) = f(X)$. If so, then on $X \setminus U$, $f$ can be as discontinuous as it likes. $\endgroup$ Aug 6, 2014 at 2:07
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    $\begingroup$ @Shamisen The property that I have used is if $(X,T)$ and $(Y,S)$ are dynamical systems with $f\circ T = S\circ f$ and $f$ is onto, then if $f$ has this property, $(X,T)$ minimal implies $(Y,T)$ is minimal. Though this follows quickly from the density property. $\endgroup$ Aug 6, 2014 at 6:28

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This property has been studied and it goes by the name somewhat continuous.

FYI, I was led to this discovery by its mention in the middle of p. 92 of Zbigniew Piotrowski's 1987 survey paper A survey of results concerning generalized continuity on topological spaces, and the google search I give above confirmed that this property has been studied a bit.

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I don't know if it's been studied, but here's an example to show how pathological functions with your property can be. Take $X = Y = (0,1)$. Enumerate the nonempty open intervals in $Y$ with rational endpoints as $I_n$, $n \in \mathbb N$. The only constraints we put on $f$ are that $f([1/(n+1),1/n)) \subseteq I_n$. Then $f$ satisfies your property.

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