If $(X,\tau)$ is a topological space, then we say $A\subseteq X$ is a fiber if there is $f:X\to X$ continuous and $y\in X$ such that $A = f^{-1}(\{y\})$. For any $T_1$-space it is clear that fibers are closed.

In $\mathbb{R}$ the converse holds: all closed sets are fibers.

Question. Is there a connected $T_2$-space $(X,\tau)$ with $|X|>1$ such that there is a closed subset $A\subseteq X$ that is not a fiber?

  • $\begingroup$ Perhaps it would be more interesting to look for a closed and also connected set which is not a fiber? However, if a strongly rigid space contains a closed connected subset, other than itself and a singleton, it would be an answer.. $\endgroup$ – erz Mar 10 '18 at 1:46

Yes. See Kannan and Rajagopalan - Constructions and applications of rigid spaces, I (MSN), particularly their construction 2.2.4, which gives a strongly rigid connected Hausdorff space $Y$. In particular, two-point subsets cannot be fibers.

  • $\begingroup$ Beautiful - great idea to link this with rigidity! $\endgroup$ – Dominic van der Zypen Mar 9 '18 at 14:13

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