Let $(X,\tau)$ be a topological space such that $\tau$ contains no singleton. We say that a map $c:X\to \kappa$, where $\kappa$ is a cardinal, is a coloring for $(X,\tau)$, if for every $U\in \tau\setminus \{\emptyset\}$ the restriction $c|_U$ is non-constant. (Note that this coloring notion comes from hypergraph coloring.)

The chromatic number $\chi(X,\tau)$ of a space $(X,\tau)$ is the smallest cardinal $\kappa$ such that there is a coloring $c:X\to \kappa$.

We have $\chi(\mathbb{R})=2$ when $\mathbb{R}$ is endowed with the Euclidean topology: color $\mathbb{Q}$ with $0$ and $\mathbb{R}\setminus\mathbb{Q}$ with $1$. This works for all spaces having a dense set $D$ such that $X\setminus D$ is also dense. Note that not all connected $T_2$-spaces contain a dense subset with this property.

Given an integer $n>2$ is there a connected Hausdorff space $(X,\tau)$ such that $\chi(X,\tau) = n$?

  • 3
    $\begingroup$ My guess would be that you could build examples using the methods of this paper: arxiv.org/pdf/math/0609090.pdf. I'm not familiar enough with those methods to say for sure -- but maybe it's a start. $\endgroup$ – Will Brian Jan 30 at 14:02
  • 1
    $\begingroup$ @WillBrian. Note that $\kappa$-resolvability is equivalent to $\chi(X)=\kappa$ only for $\kappa=2$ since the restrictions are not required to take all values, only to be non-constant. $\endgroup$ – Ramiro de la Vega Jan 30 at 22:14
  • $\begingroup$ @RamirodelaVega: I did realize that -- I just thought that getting "control" over the structure of the dense subsets of a space (as in the paper) might allow you to control the value of $\chi$. It looks like my intuition was wrong, though -- +1 for the great answer you just posted. $\endgroup$ – Will Brian Jan 30 at 22:23

The answer is no.

A space is called resolvable if it contains two disjoint dense subspaces. Clearly $X$ is resolvable if and only if $\chi(X)=2$. Lets prove by induction on $n \geq 2$ that if $\chi(X) \leq n$ then $X$ is resolvable (and hence $\chi(X)=2$).

The base case $n=2$ is clear so suppose there is a coloring $f:X \to n+1$. Let $V$ be the union of all open $U$ such that the range of $f\upharpoonright U$ is contained in $n$. Then $V$ is open and $f \upharpoonright V$ is a coloring of $V$ in $n$ colors so that $\chi(V) \leq n$ and by induction hypothesis $V$ is resolvable. Now let $W=X \setminus \mathrm{Cl}(V)$. Since $W$ is open, by definition we have that $f^{-1}(\{n\}) \cap W$ is dense in $W$. But since $f$ is a coloring we also have that $f^{-1}(n) \cap W$ is dense in $W$. This shows that $W$ is also resolvable. Since $V \cup W$ is dense in $X$ we get that $X$ is resolvable.

Note that we didn't use that $X$ is Hausdorff, connected or anything else (I guess just that $X$ is crowded so that $\chi(X)$ makes sense).

Also note that we can not generalize the above result to infinite cardinals since there are examples of (Tychonoff) countable irresolvable spaces (see for example Alas O., Sanchis M., Tkačenko M.G., Tkachuk V.V., Wilson R.G., Irresolvable and submaximal spaces: Homogeneity versus σdiscreteness and new ZFC examples, Topology Appl. 107 (2000), 259–273, DOI: 10.1016/S0166-8641(99)00111-X.), and such a space must have chromatic number $\aleph_0$.

  • 1
    $\begingroup$ Formally speaking, you didn't "use" that $X$ is crowded, since the assumption $\chi(X)<\infty$ implies that $X$ is crowded. $\endgroup$ – YCor Jan 31 at 5:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.