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Let $X$ be a topological space and let $f:X\rightarrow X$ be a continuous self-morphism of topological spaces. Let $Y$ be a closed $f$-stable subset of $X$, that is, suppose $f(Y)\subseteq Y$. Consider the additional condition that $f^{-1}(Y)=Y$. Is there a terminology for this situation in topological dynamics? I am not sure if there exists a terminology for this, but I am tempted to say $f$ isolates $Y$ if: 1) $Y$ is $f$-stable, and 2) $f^{-1}(Y)=Y$.

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  • $\begingroup$ Adding on to Alexandre's answer: although I am not sure about the origins of the term "completely invariant", the only person that I know of who regularly uses this term is Stankewitz. See, for instance, arxiv.org/abs/math/9810090 and also rstankewitz.iweb.bsu.edu/numcomp.pdf where the term is defined in the background sections. If you can't find a cite-able original reference, maybe you can send him an email and ask where he saw first it. $\endgroup$
    – Vidit Nanda
    Commented Dec 27, 2012 at 21:06

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The commonly accepted term is "completely invariant". A set which is mapped to itself is called simply "invariant" and a stronger property to coincide with its preimage is called complete invariance.

Sometimes "complete invariance" refers to a weaker property that a) the set is invariant, and b) the full preimage is contained in the set.

EDIT. On your further questions: For a reference, see for example the survey "Dynamics of analytic transformations", Leningrad Math. J. (1990). (It is avalable on my web site).

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  • $\begingroup$ Thank you! Can you suggest a good reference in order to cite this terminology? Also, if $f(Y)\subseteq Y$ and $f^{-1}(Y)\subseteq Y$ then $f^{-1}(Y)=Y$, so I am not sure why you used the term "weaker property"? $\endgroup$
    – Mahdi Majidi-Zolbanin
    Commented Dec 27, 2012 at 20:58
  • $\begingroup$ Hi Alex, I think Mahdi is right - for the exponential map, $f^{-1}(\mathbb{C})=\mathbb{C}$ ... indeed, $f(Y)\subset Y$ is equivalent to $Y\subset f^{-1}(Y)$. $\endgroup$
    – Lasse Rempe
    Commented Dec 28, 2012 at 16:19
  • $\begingroup$ Of course, sorry:-) $\endgroup$
    – Alexandre Eremenko
    Commented Dec 28, 2012 at 17:19

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