# Non-discrete $T_2$-space $(X,\tau)$ with $2^{|X|}$ retracts

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 $$|\text{Retr}(X)| = 2^{|X|}$$?

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.

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