Post History
#3: Post edited
by
clemens
·
2026-05-24T20:26:14Z (4 months ago)
- All but one of the spaces under consideration are $T_1$
- ---
- This is not a complete answer, but note that one can straightforwardly prove that the space is either the Sierpiński space or a $T_1$ space! (We know it is $T_0$ by the argumentation in the OP---if it weren't $T_0$, then swapping two topologically equivalent points would give us a non-identity non-constant continuous endofunction.)
- To prove this we shall find a non-constant continuous endofunction for any $T_0$ and non-$T_1$ space, then we shall prove that this endofunction can only be the identity in one case—that of the Sierpiński space.
- We know that any non-$T_1$ space contains distinct points $x$, $y$ such that $x \in \text{cl}(\{y\})$ (so $\{x,y\}$ is homeomorphic to the Sierpiński space, given that our space is $T_0$). But any nontrivial $T_0$ space has a nontrivial continuous map to the Sierpiński space. Specifically, let $O$ be any nontrivial open and $C$ be its complement; then the map that takes $O$ to $y$ and $C$ to $x$ is continuous and non-constant!
- (And we see that, in the case of the Sierpiński space, this map is equal to the identity; in all other cases, the map cannot be equal to the identity.)
- ---
As for the more general question I *suspect* that it depends on the Axiom of Choice, and more particularly that, in certain models of ZF, there are subsets of the reals which have no endomorphisms except constant functions and the identity; but I am not sure of this.
- All but one of the spaces under consideration are $T_1$
- ---
- This is not a complete answer, but note that one can straightforwardly prove that the space is either the Sierpiński space or a $T_1$ space! (We know it is $T_0$ by the argumentation in the OP---if it weren't $T_0$, then swapping two topologically equivalent points would give us a non-identity non-constant continuous endofunction.)
- To prove this we shall find a non-constant continuous endofunction for any $T_0$ and non-$T_1$ space, then we shall prove that this endofunction can only be the identity in one case—that of the Sierpiński space.
- We know that any non-$T_1$ space contains distinct points $x$, $y$ such that $x \in \text{cl}(\{y\})$ (so $\{x,y\}$ is homeomorphic to the Sierpiński space, given that our space is $T_0$). But any nontrivial $T_0$ space has a nontrivial continuous map to the Sierpiński space. Specifically, let $O$ be any nontrivial open and $C$ be its complement; then the map that takes $O$ to $y$ and $C$ to $x$ is continuous and non-constant!
- (And we see that, in the case of the Sierpiński space, this map is equal to the identity; in all other cases, the map cannot be equal to the identity.)
- ---
- As for the more general question I *suspect* that it depends on the Axiom of Choice, and more particularly that, in certain models of ZF, there are subsets of the reals which have no endomorphisms except constant functions and the identity; but I am not sure of this.
- It may also be useful to note that, because (as the OP noticed) the space must be connected, it must be either uncountable or non-regular, as all countable regular spaces are metrizable and hence disconnected.
#2: Post edited
by
clemens
·
2026-05-24T18:20:59Z (4 months ago)
- This is not a complete answer, but note that one can straightforwardly prove that the space is either the Sierpiński space or a $T_1$ space! (We know it is $T_0$ by the argumentation in the OP---if it weren't $T_0$, then swapping two topologically equivalent points would give us a non-identity non-constant continuous endofunction.)
- To prove this we shall find a non-constant continuous endofunction for any $T_0$ and non-$T_1$ space, then we shall prove that this endofunction can only be the identity in one case—that of the Sierpiński space.
- We know that any non-$T_1$ space contains distinct points $x$, $y$ such that $x \in \text{cl}(\{y\})$ (so $\{x,y\}$ is homeomorphic to the Sierpiński space, given that our space is $T_0$). But any nontrivial $T_0$ space has a nontrivial continuous map to the Sierpiński space. Specifically, let $O$ be any nontrivial open and $C$ be its complement; then the map that takes $O$ to $y$ and $C$ to $x$ is continuous and non-constant!
- (And we see that, in the case of the Sierpiński space, this map is equal to the identity; in all other cases, the map cannot be equal to the identity.)
- ---
- As for the more general question I *suspect* that it depends on the Axiom of Choice, and more particularly that, in certain models of ZF, there are subsets of the reals which have no endomorphisms except constant functions and the identity; but I am not sure of this.
- All but one of the spaces under consideration are $T_1$
- ---
- This is not a complete answer, but note that one can straightforwardly prove that the space is either the Sierpiński space or a $T_1$ space! (We know it is $T_0$ by the argumentation in the OP---if it weren't $T_0$, then swapping two topologically equivalent points would give us a non-identity non-constant continuous endofunction.)
- To prove this we shall find a non-constant continuous endofunction for any $T_0$ and non-$T_1$ space, then we shall prove that this endofunction can only be the identity in one case—that of the Sierpiński space.
- We know that any non-$T_1$ space contains distinct points $x$, $y$ such that $x \in \text{cl}(\{y\})$ (so $\{x,y\}$ is homeomorphic to the Sierpiński space, given that our space is $T_0$). But any nontrivial $T_0$ space has a nontrivial continuous map to the Sierpiński space. Specifically, let $O$ be any nontrivial open and $C$ be its complement; then the map that takes $O$ to $y$ and $C$ to $x$ is continuous and non-constant!
- (And we see that, in the case of the Sierpiński space, this map is equal to the identity; in all other cases, the map cannot be equal to the identity.)
- ---
- As for the more general question I *suspect* that it depends on the Axiom of Choice, and more particularly that, in certain models of ZF, there are subsets of the reals which have no endomorphisms except constant functions and the identity; but I am not sure of this.
#1: Initial revision
by
clemens
·
2026-05-24T18:03:03Z (4 months ago)
This is not a complete answer, but note that one can straightforwardly prove that the space is either the Sierpiński space or a $T_1$ space! (We know it is $T_0$ by the argumentation in the OP---if it weren't $T_0$, then swapping two topologically equivalent points would give us a non-identity non-constant continuous endofunction.)
To prove this we shall find a non-constant continuous endofunction for any $T_0$ and non-$T_1$ space, then we shall prove that this endofunction can only be the identity in one case—that of the Sierpiński space.
We know that any non-$T_1$ space contains distinct points $x$, $y$ such that $x \in \text{cl}(\{y\})$ (so $\{x,y\}$ is homeomorphic to the Sierpiński space, given that our space is $T_0$). But any nontrivial $T_0$ space has a nontrivial continuous map to the Sierpiński space. Specifically, let $O$ be any nontrivial open and $C$ be its complement; then the map that takes $O$ to $y$ and $C$ to $x$ is continuous and non-constant!
(And we see that, in the case of the Sierpiński space, this map is equal to the identity; in all other cases, the map cannot be equal to the identity.)
---
As for the more general question I *suspect* that it depends on the Axiom of Choice, and more particularly that, in certain models of ZF, there are subsets of the reals which have no endomorphisms except constant functions and the identity; but I am not sure of this.