I sought for a topological space that is homeomorphic to its own power, e.g. $X \cong 2^X$ for the discrete two-point space $2$. Of course, $2^X$ has the compact-open topology.
Here's my approach: First I probed its underlying set. Simple argument; treat the base $2$ as the power set operation. That is, let $X_0 = \emptyset$, let $X_{k+1} = \mathcal{P}(X_k)$, and take the categorical direct limit, where the morphism $f_{k, k+1} : X_k \to X_{k+1}$ is given as $f_{k, k+1}(x) = \{x\}$. That gives $X = \emptyset \cup \mathcal{P}(\emptyset) \cup \mathcal{P}^2(\emptyset) \cup \mathcal{P}^3(\emptyset) \cup \cdots$, resulting in some countably infinite set.
But that doesn't give any information about the topological structure of $X$. I tried probing some topological properties of $X$, and concluded that $X$ is:
- zero-dimensional Hausdorff
- not discrete
- not compact
- locally compact (proved using Ascoli's theorem)
How can I explicitly identify the topology of $X$? A subbasis would suffice, as long as it doesn't involve circular reasoning.
Best Answer
There is no such space, by the Lawvere fixed point theorem, because there is a map $2 \to 2$ with no fixed points. ($\text{Top}$ is not a cartesian closed category but the proof only requires that the specific exponential used in the argument exists, and in fact it only requires that continuous maps $X \to 2^X$ can be identified with continuous maps $X \times X \to 2$. I am assuming that your use of the notation $2^X$ implies that this holds.)