Let $(X,\tau)$ be a topological space.

We say that $(X,\tau)$ is zero-dimensional with respect to the Lebesgue covering dimension (zd1) if every open cover of the space has a refinement which is a cover of the space by open sets such that any point in the space is contained in exactly one open set of this refinement.

Moreover, $(X,\tau)$ is zero-dimensional with respect to the small inductive dimension (zd2) if it has a base consisting of clopen sets.

Is there a space that is (zd1) but not (zd2)?

EDIT: I accepted the small and correct example given by Gabriel below; it works for the reason that $X\in \mathcal{V}$ for every open cover $\cal V$. It would be great to see an example of a space $(X,\tau)$ that is (zd1) but not (zd2), and such that $X$ has a cover $\cal V$ such that $X\notin \mathcal{V}$.

  • 4
    $\begingroup$ There is an excellent treatment of the three classical dimension functions including theorems on when they coincide and counter-examples for when they don't in chapter 7 of Engelking's "General Topology". I haven't had time to check whether your specific question is answered there but you might want to take a look. $\endgroup$ – report Apr 21 '15 at 9:03
  • 2
    $\begingroup$ and he has written a whole book about this subject. $\endgroup$ – report Apr 21 '15 at 9:08
  • $\begingroup$ I think one would really like an example that is at last $T_2$. $\endgroup$ – report Apr 21 '15 at 12:23
  • $\begingroup$ Note that Engelking incorporates the $T_3$ condition in his definitions of dimension to circumvent the fact that in the absence of enough open sets they can be vacuous. $\endgroup$ – report Apr 21 '15 at 12:39
  • 1
    $\begingroup$ zero-dimensionality+$T_{1}$ with respect to Lebesgue covering dimension is known as ultraparacompactness while zero-dimensionality with respect to small inductive dimension is commonly known as simply zero-dimensionality. See my answer at mathoverflow.net/a/134184/22277 for more information on the relation between ultraparacompactness and zero-dimensionality. $\endgroup$ – Joseph Van Name Apr 21 '15 at 17:38

Take the Sierpiński space $X=\{0,1\}$ with $\tau=\{\{\},\{0\},\{0,1\}\}$.

| cite | improve this answer | |
  • $\begingroup$ Is there also an example of such a space $X$ such that there is an open covering $\cal V$ with $X\notin \mathcal{V}$? $\endgroup$ – Dominic van der Zypen Apr 21 '15 at 12:10
  • 2
    $\begingroup$ @DominicvanderZypen: What about the disjoint sum of the Sierpiński space with, say, a singleton? $\endgroup$ – Emil Jeřábek Apr 21 '15 at 12:18
  • $\begingroup$ OK that's right... Thanks @EmilJeřábek $\endgroup$ – Dominic van der Zypen Apr 21 '15 at 12:22

I will show that there are no $T_1$ examples (note that Gabriel´s example is $T_0$ but not $T_1$).

Let $X$ be a $T_1$ space and suppose that it has covering dimension $0$. We will show that the clopen subsets of $X$ form a base. For this, fix an open $U \subseteq X$ and a point $p \in U$ and consider the open cover $\mathfrak{A}=\{U, X\setminus \{p\}\}$. By hypothesis there is a refinement $\mathfrak{B}$ of $\mathfrak{A}$ which consists of pairwise disjoint open sets. Let $C=\bigcup\{W\in \mathfrak{B} : p \in W\}$. Note that $X \setminus C= \bigcup\{W\in \mathfrak{B} : p \notin W\}$ so that $C$ is clopen. Moreover since $\mathfrak{B}$ refines $\mathfrak{A}$ we have $W\subseteq U$ whenever $p \in W \in \mathfrak{B}$, and hence $C \subseteq U$.

| cite | improve this answer | |
  • $\begingroup$ It turns out that one can get by with this result with subfitness which is a pointfree separation axiom which is weaker than $T_{1}$. We say that a frame $L$ is subfit if whenever $x\not\leq y$ there is some $c$ with $x\vee c=1\neq y\vee c$. In fact, if $L$ is a subfit frame such that if $x\vee y=1$, then there is some complemented element $a$ with $a\leq x,a'\leq y$, then $L$ has a basis of complemented elements. $\endgroup$ – Joseph Van Name Apr 21 '15 at 17:58

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.