# Topological spaces $(X,\tau)$ where $|\text{Cont}(X,X)| = |X|$

Let $(X,\tau)$ be a topological space. Let $\text{Cont}(X,X)$ denote the set of continuous functions $f:X\to X$.

What can be said about spaces $(X,\tau)$ where $|\text{Cont}(X,X)| = |X|$? For instance, is it impossible that they are zero-dimensional? (Note that trivially spaces with $|\text{Cont}(X,X)| = |X|$ must be infinite, because all constant functions as well as the identity are members of $\text{Cont}(X,X)$.)

• X must be infinite or a point. – Ramiro de la Vega Feb 12 '15 at 19:34
• Dominic, you could include as a second part of the question what you already have implicitly, about the spaces which have no other self-maps but identity and constants (you could ask about different classes of spaces in this, theorems and counter-examples, in this context). – Włodzimierz Holsztyński Feb 13 '15 at 0:06

## 4 Answers

If a Hausdorff space $\ X\$ admits a dense subset $A\subseteq X\$ such that

$$|X|^{|A|}\ =\ |X|$$

then indeed $\ \left|Cont(X\ X)\right|\ =\ |X|$.

This holds in particular for the separable metric spaces of cardinality continuum, as was already noted by Tomek Kania in the first Answer.

• What's nice about this is that it lets us produce examples of such spaces of arbitrarily large cardinality. – Noah Schweber Feb 12 '15 at 23:54
• For Hausdorff spaces, is the converse true? Suppose $|\text{End}_{\text{Top}}(X)| = |X|$, is it true that there is a dense subspace such that...? – Ivan Di Liberti Nov 23 '17 at 10:12

If $X$ denotes the Cantor set, then $\mathfrak{c} = |X|=|{\rm Cont}(X,X)|$, so yes, such spaces can be zero-dimensional. You may replace $X$ with any separable metric space of cardinality continuum. What is your motivation?

Suppose that $\lambda$ is an infinite cardinal with $\lambda^{\aleph_{0}}=\lambda$. Then Saharon Shelah has shown in his paper Existence of endo-rigid Boolean algebras that there is some Boolean algebra $B$ with $|B|=\lambda$ but where the only endomorphisms $f:B\rightarrow B$ are the trivial endomorphisms. When we transfer by Stone duality these facts over to topological spaces, for each infinite cardinal $\lambda$ with $\lambda^{\aleph_{0}}=\lambda$, there is a compact totally disconnected space $X$ with $w(X)=\lambda$ (here $w$ denotes the weight of the space) but where the only continuous maps $f:X\rightarrow X$ are the mappings where there is some partition $A_{1},...,A_{n}$ of $X$ into clopen sets where $f|_{A_{i}}$ is a constant function or where $f|_{A_{i}}$ is the identity function for each $i\in\{1,...,n\}$ (i.e. the only mappings are the trivial mappings). Take note that if $X$ is a compact space, then $w(X)\leq|X|$ and if $X$ is compact and totally disconnected, then $|\mathfrak{B}(X)|\leq|X|$ where $\mathfrak{B}(X)$ is the Boolean algebra of clopen subsets of $X$. Therefore if $X$ is a compact zero-dimensional space where the only mappings $f:X\rightarrow X$ are the trivial mappings, then $|C(X,X)|=|X|$ since there are at most $|X|$ ways to partition $X$ into finitely many clopen sets.

E-D-I-T 1   @RasmusBentmann has pointed out to my error (in the $\,T_0\,$ case (I considered $\ f(X)\$ instead of $X).\$ Thus virtually nothing of my attempt is left. Perhaps Rasmus Bentmann can post his proof if it's ready. I'll have to think a bit more about the situation. Perhaps I'll remove my post unless I can provide a correct result, either one pretty soon.

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E-D-I-T 2   My "proof" was false. Eric Vofsey has provided a correct proof though in a comment below--I hope that Eric will his argument into an Answer (then I'll remove my answer immediately upon seen it, and with a relief).

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THEOREM   $\ \left|Cont(X\ X)\right|\ \ge\ |X|+2\,\$ for every finite topological space $\ X\$ such that $\ |X|\ge 3$.

PROOF   We need to show that there is a continuous function $\ f:X\rightarrow X\$ such that it is not the identity, nor constant.

First, let $\ X\$ be not a $T_0$-space, i.e. there are two different points $\ a\ b\in X\$ such $\ \left|G\cap\{a\ b\}\right|\ne 1\$ for arbitrary open $\ G\subseteq X.\$ Define $\ f:X\rightarrow X\$ by conditions:

• $\ f\,|\,X\!\setminus\!\{a\}\$ is the identity on $\ X\!\setminus\!\{a\}$
• $\ f(a)\, :=\, b$

Then obviously $\ f^{-1}(G)\,=\,G\$ for every open $\ G\subseteq X.\$ Thus $\ f\$ is continuous, it's not an identity on $\ X,\$ and it's not a constant map (since $\ |X|\ge 3$. Thus the theorem holds in this case.

Now let $\ X\$ be a $T_0$-space. Then see the Eric's argument in the comment below (until and if Eric expands his comment into an Answer).

• In the second case, unless $f(a)=a$, we have $f^{-1}(A)=\emptyset$, contradicting what you wrote about $f^{-1}(G)$. In fact, the function $f$ will not always be continuous: if $X=3\to 2\to 1$, $A=\{3\}$ and $f(3)=1$ then $f$ will not be monotonous, hence not continuous ($f^{-1}(\{3,2\}=\{2\})$ is not open). I don't know how to repair the proof but haven't found a counterexample either. – Rasmus Feb 13 '15 at 9:29
• If there was a counterexample, it had to have the fixed-point property, so by a result of Ivan Rival from 1976 it could not contain a "crown." – Rasmus Feb 13 '15 at 9:57
• That is to say, a counterexample $X$ had to be connected and every "convex" subset of $X$ had to be simply connected. – Rasmus Feb 13 '15 at 10:38
• Is think one can do this: choose a chain $C$ of maximal length in $X$. Define $f\colon X\to X$ such that $f(x)$ is the open point in $C\cap\overline{\{x\}}$. If this intersection is empty, set $f(x)$ to be the closed point in $C$. Then $f$ is continuous and is the identity on $C$. Unless, $X$ is totally ordered or discrete, $f$ is not the identity or constant. – Rasmus Feb 13 '15 at 11:08
• If $X$ is finite and not discrete, it contains a 2-point subspace $A$ that is not discrete. Unless $X=A$ is the Sierpinski 2-point space, there will be a nonconstant map $X\to A$ that is not the identity. – Eric Wofsey Feb 13 '15 at 18:48