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If $(X,\tau)$ is a topological space, we call $A\subseteq X$ a retract if there is a continous map $r:X\to A$ such that $r(a) = a$ for all $a\in A$ (we assume $A$ to be endowed with the subspace topology inherited from $X$). By $\text{Retr}(X)$ we denote the collection of retracts of $X$.

Is there a non-discrete, infinite $T_2$-space $(X,\tau)$ such that $|\text{Retr}(X)| = 2^{|X|}$?

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Yes.

The space of rational numbers $X=\mathbb{Q}$ is an instance.

We can view $X$ as a countable union of countably many disjoint copies of $\mathbb{Q}$.

Any nonempty subset $A$ of those copies (that is, taking all or none of each copy) is a retract of $X$, since we can map the unused copies to a fixed copy, and this is continuous. And there are $2^{\aleph_0}$ many such $A$, so we've got enough retracts.

A similar idea works with copies of other spaces, and one can make uncountable examples this way.

More examples:

  • Every infinite ordinal $\lambda$, under the order topology. Any closed subset $A\subset\lambda$ is a retract, since you can map each $\alpha<\lambda$ to the next element of $A$, that is, the smallest $\beta\in A$ with $\alpha\leq\beta$. This is continuous. And there are $2^{|\lambda|}$ many closed subsets of $\lambda$.

  • The long rational line, $\left([0,1)\cap\mathbb{Q}\right)\cdot\omega_1$. This has size $\omega_1$, and yet every closed subset of $\omega_1$ gives rise to a retract (by taking the point $0$ in that interval). So there are $2^{\omega_1}$ many retracts.

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    $\begingroup$ Thanks for your additional examples - and welcome back after your break! $\endgroup$ – Dominic van der Zypen Dec 12 '18 at 13:44
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    $\begingroup$ Thanks very much for the welcome. This has been my first post since August; I've been away much longer than I had expected. $\endgroup$ – Joel David Hamkins Dec 12 '18 at 13:46

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