Two notions of zero-dimensionality for topological spaces

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 $X\notin \mathcal{V}$.

• 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. – report Apr 21 '15 at 9:03
• and he has written a whole book about this subject. – report Apr 21 '15 at 9:08
• I think one would really like an example that is at last $T_2$. – report Apr 21 '15 at 12:23
• 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. – report Apr 21 '15 at 12:39
• 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. – Joseph Van Name Apr 21 '15 at 17:38

Take the Sierpiński space $X=\{0,1\}$ with $\tau=\{\{\},\{0\},\{0,1\}\}$.
• Is there also an example of such a space $X$ such that there is an open covering $\cal V$ with $X\notin \mathcal{V}$? – Dominic van der Zypen Apr 21 '15 at 12:10
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 $C \subseteq U$.
• 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. – Joseph Van Name Apr 21 '15 at 17:58