Let $X$ be the real line with the usual topology. Then clearly $|C(X)| = c = |X|$ and on the other hand $|X| = 2^{\aleph_0}$. Now my question is as in the title: Is there a Tychonoff space $X$ of cardinality not of the form $2^\alpha$ such that $|C(X)| = |X|$ (where $C(X)$ is the set of all real valued continuous functions on $X$). Any reference or help would be appreciated.


2 Answers 2


The answer is yes.

Let $\kappa$ be any singular strong limit cardinal of uncountable cofinality, such as the cardinal $\beth_{\omega_1}$ for a specific example, and let $X=\kappa+1$, the ordinals up to and including $\kappa$ itself. Under the order topology, this is a compact Hausdorff space. Note that $|X|=\kappa\neq 2^\alpha$ for any $\alpha$, since $\kappa$ is a strong limit cardinal. But meanwhile, every continuous function $f:X\to\mathbb{R}$ is eventually constant, in order that it is continuous at the final point $\kappa$, because it has uncountable cofinality. So every function in $C(X)$ is determined by a restriction of it $f\upharpoonright\gamma:\gamma\to\mathbb{R}$ for some $\gamma\lt\kappa$. Since $\kappa$ is a strong limit, there are only $\kappa$ many such restricted functions, and so $|C(X)|=\kappa=|X|$, as desired.

Update. If the GCH fails in a convenient way, we can make a much smaller example. Suppose $2^\omega=\omega_1$ and $2^{\omega_1}\gt\omega_2$, and let $X=\omega_2+1$. This is a compact Hausdorff space of size $\omega_2$. Notice that any continuous function $f:X\to\mathbb{R}$ is eventually constant and indeed takes at most countably many distinct values (since otherwise we can find a point $\gamma$ of cofinality $\omega_1$ with $f\upharpoonright\gamma$ not eventually constant, which will violate continuity at $\gamma$.) It follows that $C(X)$ has size $\omega_2^\omega$, which has size $\omega_2$ under our assumptions. So this is a case where $|C(X)|=|X|=\omega_2$, but $\omega_2\neq 2^\alpha$ for any $\alpha$.

  • $\begingroup$ +1 Nice. If $Y = \kappa$ ($\kappa$ as above) then what is the cardinality of $C(Y)$? Is it $\kappa$ !? $\endgroup$
    – user37834
    Commented Sep 14, 2013 at 13:52
  • $\begingroup$ @Silvi, the same argument works even if you don't have the final point, since every continuous function $\kappa\to\mathbb{R}$ is eventually constant, because $\kappa$ has uncountable cofinality. So we would also have $|C(Y)|=\kappa$. But in this case, $Y$ would not be compact, although still Tychonoff. $\endgroup$ Commented Sep 14, 2013 at 14:06

$\textbf{A simple counterexample}$ The following counterexample is somewhat similar to Joel David Hamkins' counterexample, but it is still worth mentioning. Let $D$ be a discrete space and let $D\cup\{\infty\}$ denote the one-point-compactification of $D$. If $f\in C(D\cup\{\infty\})$, then for each $\epsilon>0$ there is a finite subset $A_{\epsilon}\subseteq D$ where if $x\not\in A_{\epsilon}$, then $|f(x)-f(\infty)|<\epsilon$. In particular, if $A=\bigcup_{n}A_{1/n}$, then $f(x)=f(\infty)$ outside $A$. Therefore, the function $f$ only depends on the countable set $A$($|D|^{\aleph_{0}}$ choices), the values on the set $A$($2^{\aleph_{0}}$ choices), and the value at $\infty$($2^{\aleph_{0}}$ choices). We conclude that there are $|D|^{\aleph_{0}}$ choices for the function $f$. If $|D|^{\aleph_{0}}=|D|$, then $|D\cup\{\infty\}|=C(D\cup\{\infty\})$.

$\textbf{Constructing related counterexamples}$ The following facts may be useful in constructing related counterexamples.

First of all, if $C^{*}(X)$ denotes the set of all bounded continuous functions from $X$ to $\mathbb{R}$, then $|C(X)|=|C^{*}(X)|$. Furthermore, $C^{*}(X)\simeq C^{*}(Y)$ if and only if $\beta X=\beta Y$ where $\beta X$ denotes the Stone-Cech compactification of $X$. In particular, if $|X|\leq|C^{*}(X)|\leq|\beta X|$, then there is a set $Y$ with $X\subseteq Y\subseteq\beta X$ and $|Y|=|C^{*}(X)|$. In this case, we have $|C^{*}(Y)|=|C^{*}(X)|=|Y|$.

If $X$ is a space, then let $w(X)$ denote the weight of the space $X$. Let $k(X)$ denote the least cardinal such that every open cover of $X$ has a subcover of cardinality less than $k(X)$.

$\mathbf{Proposition}$ $C(X)\leq((w(X)^{<k(X)})^{\aleph_{0}}$

$\mathbf{Proof}$ Let $\mathcal{B}$ be a basis for $X$ of cardinality $w(X)$. Let $U$ be a cozero set. Then there are open sets $U_{n}$ with $\overline{U_{n}}\subseteq U$ and $\bigcup_{n}U_{n}=U$. Since $\mathcal{B}$ is a basis for $X$, there is a subset $\mathcal{C}\subseteq\mathcal{B}$ with $\bigcup\mathcal{C}=U$. By compactness, there is some subset $\mathcal{C}_{n}\subseteq\mathcal{C}$ with $|\mathcal{C}_{n}|<k(X)$ and where $\overline{U}_{n}\subseteq\bigcup\mathcal{C}_{n}$. In particular, we have $U=\bigcup_{n}\bigcup\mathcal{C}_{n}$. However, since each $\mathcal{C}_{n}$ chooses less than $k(X)$ many elements of $\mathcal{B}$, there are at most $(w(X)^{<k(X)})^{\aleph_{0}}$ many choices of $\bigcup_{n}\bigcup\mathcal{C}_{n}$. Since $U$ ranges over all cozero sets, there are at most $(w(X)^{<k(X)})^{\aleph_{0}}$ cozero sets. Since every continuous real-valued function is determined by the cozero sets $f^{-1}(-\infty,r)$ where $r$ ranges over all rational numbers, there are at most $(w(X)^{<k(X)})^{\aleph_{0}}$ continuous real-valued functions on $X$. $\mathbf{QED}$

It can be shown using a standard compactness argument that if $X$ is a compact Hausdorff space, then $|w(X)|\leq|X|$. Combining this fact with the above result, we obtain the following corollary

$\textbf{Corollary}$ Suppose that $X$ is a compact Hausdorff space. Then $|C(X)|\leq |w(X)|^{\aleph_{0}}\leq|X|^{\aleph_{0}}$.

There is also a lower bound of the number of continuous functions on a completely regular space that depends on a cardinal invariant which I shall call the saturation. If $X$ is a space, then let $s(X)$ be the least cardinal such that if $\mathcal{U}$ is a collection of $s(X)$ nonempty open sets, then $U\cap V\neq\emptyset$ for distinct $U,V\in\mathcal{U}$.

$\textbf{Proposition}$ If $X$ is a completely regular space, then there are at least $\sum_{\kappa<s(X)}\kappa^{\aleph_{0}}$ many functions in $C^{*}(X)$.

$\textbf{Proof}$ Suppose that $\kappa<s(X)$. Then there is a collection $\mathcal{U}$ of pairwise disjoint open sets. For each $U\in\mathcal{U}$ choose some point $x_{U}\in U$. Then let $f_{U}:X\rightarrow[0,1]$ be a function where $f_{U}(x_{U})=1$ and $f_{U}(x)=0$ for each $x\in U^{c}$. Take note that there are $\kappa^{\aleph_{0}}$ injective maps from $\omega$ to $\mathcal{U}$. For each injective $j:\omega\rightarrow\mathcal{U}$, let $F_{j}:X\rightarrow\mathbb{R}$ be the function defined by $\sum_{n}\frac{1}{n^{2}+1}f_{j(n)}$. It is clear that each function $F_{j}$ is distinct. Furthermore, each $F_{j}$ is continuous being the uniform limit of continuous functions. Therefore the family $(F_{j})_{j}$ is a family of $\kappa^{\aleph_{0}}$ continuous functions. Since $\kappa^{\aleph_{0}}\leq|C^{*}(X)|$ for all $\kappa<s(X)$, we conclude that $\sum_{\kappa<s(X)}\kappa^{\aleph_{0}}\leq|C^{*}(X)|$ as well.

Using the following corollary, we immediately obtain my counterexample and Joel David Hamkins' counterexample.

$\textbf{Corollary}$ Suppose $X$ is a compact Hausdorff space with $|X|=|X|^{\aleph_{0}}=\sum_{\kappa<s(X)}\kappa^{\aleph_{0}}$. Then $|C(X)|=|X|$.

  • $\begingroup$ Thanks to you and @JoelDavidHamkins a lot. So fruitful. Seems like a small course on $C(X)$. I must confess that I have believed wrongly that the cardinal of idempotents is equal to $2^A$ where $A$ is the cardinal of connected components of $X$ But I think the correct thing is that the cardinal is equal to the cardinal of clopen sets. Forgive me if I am asking something trivial but what is the cardinality of the set of idempotents in these four examples? Thanks anyway. $\endgroup$
    – user39982
    Commented Sep 15, 2013 at 11:20
  • 1
    $\begingroup$ Yes. The idempotents in $C(X)$ are in a one-to-one correspondence with the clopen sets. Furthermore, for a compact totally disconnected space $X$, the cardinality of the collection of all clopen sets is equal to the weight $w(X)$ and is therefore bounded above by $|X|$. If $X$ is a compact ordinal space or the one-point compactification of a discrete space, then there are $|X|$ clopen subsets of $X$. $\endgroup$ Commented Sep 15, 2013 at 18:39
  • $\begingroup$ So I have to restate my problem. I am looking for a Tychonoff space $X$ such that for any finite subset $F$, $|C(X\backslash F)| = |X| \not = |T|$ (where $T$ is the set of clopen subsets of $X$). I would appreciate your points and suggestions on this one. $\endgroup$
    – user39982
    Commented Sep 15, 2013 at 19:30
  • $\begingroup$ any idea? By the way is it possible to ask the question (mentioned above) in a new post? (I am new and I don't know that much about the regulations here. My new question is very similar to the new one mentioned above. so I am not sure if they accept this as a whole new question) Thanks in advance. $\endgroup$
    – user39982
    Commented Sep 16, 2013 at 7:28
  • $\begingroup$ @Niki. It will probably be best to ask an entirely new question in a new post since the new question seems significantly different from the old one. $\endgroup$ Commented Sep 16, 2013 at 13:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.