Closed maps in terms of "thickening fibers" Recently, I have asked for intuition for the characterization of compactness of a space $X$ in terms of the closedness of the projection $X\times Y\rightarrow Y$. The wonderful answer I have received includes the following statment:

  
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*A continuous function $f:X\rightarrow Y$ is closed iff for each $y\in Y$ and each open $U\subset X$ with $f^{−1}[{y}]\subset U$ there is an open neighbourhood $V$ of $y$ such that $f^{−1}[V]\subset U$.
  

It seems to me that continuity is not relevant here. Saying $f$ is closed is equivalent to saying $f(U^c)^c$ is open for every open $U\subset X$. But $f(U^c)^c$ is the set of elements of $Y$ whose fibers are all contained in $U$, so to say it is open is equivalent the second part of the statement.
Is this a silly mistake on my part or is continuity really irrelevant here?

Continuity of $f$ is indeed irrelevant here. Your proof is very short and elegant. And, also not quite unimportant, it is correct.
