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Any $\ T_0$ space that has a base consisting of closed (hence clopen) sets is totally disconnected. Does a totally disconnected space necessarily have a base consisting of closed sets?

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I think the converse is true if you add in "locally compact Hausdorff". –  Dylan Moreland May 7 '12 at 5:05
    
Willard's General Topology has one counterexample in exercise 29B. See here. There are others; c.f. Counterexamples in Topology Steen and Seebach. –  David Mitra May 7 '12 at 13:13
    
For the case that $X$ is compact, the result mentioned by Dylan is shown in this interesting blog post: Thoughts about connectedness (Totally disconnected spaces). –  Martin Sleziak Jun 1 '12 at 4:57
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1 Answer 1

Several examples of this kind are mentioned at standard places where to look for counterexamples in general topology.

Wikipedia article on totally disconnected spaces mentions Erdős space. The same space is also mentioned as Example 6.2.19 in Ryszard Engelking: General Topology, Heldermann Verlag, Berlin, 1989. The proof I give bellow is essentially the same as in Engelking's book.

If we look at Steen-Seebach: Counterexamples in Topology we find out from Figure 9, p. 32 that further examples of $T_2$-spaces, which are totally disconnected but not zero-dimensional should be Examples 72 (Rational extension in the plane), 79 (Irregular lattice topology), 113 (Strong ultrafilter topology), 127 (Roy's lattice subspace), 129 (Knaster–Kuratowski fan a.k.a. Cantor's teepe).


Erdős space

$\newcommand{\ve}{\varepsilon}\newcommand{\norm}[1]{\lVert{#1}\rVert}$Let us consider the space $$X=\ell_2\cap \mathbb Q^{\mathbb N}=\{(x_i); \sum {x_i}^2<\infty, (\forall i) x_i\in\mathbb Q\}$$ of all sequences of rational numbers which belong to $\ell_2$. We endow this space with the metric derived from the usual $\ell_2$ norm, i.e. $$d(x,y)=\norm{x-y}_2 = \sqrt{\sum (x_i-y_i)^2}.$$

Erdős space is totally disconnected: If $a\ne b$ then we have at least one coordinate such that $a_i\ne b_i$. W.l.o.g. we assume $a_i<b_i$. We choose an arbitrary irrational number $z$ such that $a_i<z<b_i$. The sets $U_{i,z}=\{x\in X; x_i<z\}$ and $V_{i,z}$ are open, disjoint and their union is the whole space $X$. (For any $x\in U_{i,z}$ and $\ve<z-x_i$ the $\ve$-ball around $x$ is a subset of $U_{i,z}$. The same argument shows that $V_{i,z}$ is open.) So we have found two clopen subset of $X$ such that one of them contains $a$ and the other one contains $b$. Therefore the connected components of $X$ are singletons, i.e. it is totally disconnected.

Erdős space is not zero-dimensional: Let $V=B(0,1)=\{x\in X; \norm{x}_2\le 1\}$. We will try to show that no open neighborhood $U$ of $0$ such that $U\subseteq V$ is clopen.

By induction we define a sequence $(a_k)$ of rationals such that for the sequence given by $$x_k=(a_1,\dots,a_k,0,0,\dots)$$ we have $x_k\in U$ and $d(x_k,X\setminus U)\le\frac1k$.

For $k=1$ we can choose $a_1=0$.

If $a_1,\dots,a_{k-1}$ are already chosen, we consider the numbers $\frac i{k}$, $i=0,\dots,k$ as possible candidates for $a_k$. If we choose $a_k=\frac ik$ in a such way that $(a_1,\dots,a_{k-1},\frac ik,0,0,\dots)\in U$ and $(a_1,\dots,a_{k-1},\frac ik,0,0,\dots)\notin U$ then it is clear that $x_k\in U$ and $d(x_k,X\setminus U)\le\frac1k$.

Now let $x=(a_k)$. Since $\norm{x_k}_2^2 = \sum_{i=1}^k a_i^2 < 1$, we have that $\sum_{i=1}^\infty a_i^2 \le 1$ and $x\in X$.

We also see that the sequence $x_k$ converges to $x$, and thus $x\in \overline U$.

The condition $d(x_k,X\setminus U)\le\frac1k$ implies that there is a sequence $(v_k)$ of elements of $X\setminus U$ such that $d(x_k,v_k)\le\frac1k$. Since $x_k\to x$, we see that also $v_k\to x$ and thus $x\in\overline{X\setminus U}$.

We have shown that $\overline U \cap \overline{X\setminus U}\ne\emptyset$, therefore $U$ cannot be clopen. $\hspace{2cm}\square$

For the origin of this example, I quote from Engelking's book:

Example 6.2.19 was described in Erdős' paper [1940] (the first example of a separable metric space with similar properties was defined by Sierpinski in [1921]; Sierpinski's space is, moreover, completely metrizable).

  • Erdős, P.: The dimension of the rational points in Hilbert space, Ann. Math. 41 (1940), 734-736. jstor renyi.hu

  • Sierpinski, W.: Sur les ensembles connexes et non connexes, Fund. Math. 2 (1921), 81-95. matwbn

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@Thomas Thanks for noticing the typos, I've corrected these two. –  Martin Sleziak May 31 '12 at 10:39
    
Sure. +1 for a nice answer. –  Thomas E. May 31 '12 at 10:40
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