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?
Totally disconnected implies base of closed sets?
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0For the case that $X$ is compact, the result mentioned by Dylan is shown in this interesting blog post: [Thoughts about connectedness (Totally disconnected spaces)](http://drexel28.wordpress.com/2010/03/28/thoughts-about-connectedness-totally-disconnected-spaces/). – 2012-06-01
1 Answers
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).
Several examples can be found in pi-base.
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
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 $(a_1,\dots,a_{k-1},\frac {i+1}k,0,0,\dots)\notin U$ 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).
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0Sure. +1 for a nice answer. – 2012-05-31