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In Michael Coornaert's book 'Topological Dimension and Dynamical Systems' he mentions the existence of a set $X$ in the plane that is totally separated, but not zero-dimensional. He says it is due to Sierpinski, but the paper he cites is in French. The example doesn't seem to be in Steen & Seebach; is anyone familiar?

Here, 'totally separated' means that the quasicomponents of $X$ are points. Equivalently, for any $x, y \in X$ there is a separation of $X$ into disjoint open sets $U, V$ such that $x \in U$ and $y \in V$.

The Sierpinski paper is here: https://eudml.org/doc/212954

A zero-dimensional space in this context can equivalently mean a space with a clopen basis, a space all of whose points have a clopen basis, or a space with arbitrarily fine covers by disjoint open sets.

EDIT: David Hartley below has described Sierpinski's space satisfying the desired requirements. I've attached a (bad) drawing.

enter image description here

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    $\begingroup$ What notion of dimension are you using? $\endgroup$ – Lorenzo Najt Jul 29 '18 at 21:35
  • $\begingroup$ Since we are in the separable metric case, any of the standard definitions will do. Covering, small ind, big ind or scattered. $\endgroup$ – John Samples Jul 29 '18 at 21:43
  • $\begingroup$ I was wondering if the Cantor set counted as an example. $\endgroup$ – Lorenzo Najt Jul 29 '18 at 21:47
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    $\begingroup$ But can you also provide, for sake of completeness, an explicit description for zero dimensionality? $\endgroup$ – Berci Jul 29 '18 at 21:48
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    $\begingroup$ Can you post a link to the paper by Sierpinski? $\endgroup$ – Rob Arthan Jul 29 '18 at 21:53
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The example in the Sierpinski paper quoted is quite different from his carpet, except it also uses lots of rectangles. Here's a brief sketch of the construction. It uses repeated applications of the following procedure, illustrated on the rectangle with corners at $(\pm1,\pm1)$. Let $R(2n-1)$ be the rectangle in the top-left quarter bounded by $y=0,y=1,x=-(2^{-(2n-2)})$ and $x=-(2^{-(2n-1)})$. Let $R(2n)$ be the matching rectangle in the bottom-right quarter (rotate 180 about origin). The same procedure, scaled appropriately, is repeated in each $R(n_1)$ to give rectangles $R(n_1,n_2)$ and again within those, and so on, giving $R(n_1, n_2,..,n_k)$ for any natural k.

Let $S_k$ be the union of all these rectangles with k indices and P be the intersection of all $S_k$. Let Q be the set containing the centres of all the defined rectangles and $E = P \cup Q$.

$E$ is totally separated but not zero-dimensional. In particular, any clopen set containing the centre of the outer rectangle has the mid-points of the top and bottom sides in its closure in $\mathbb R^2$ (they are not in $E$) and so has diameter at least 2. Hence the centre point cannot have a nbhd. basis of clopen sets.

Adding one of those mid-points to $E$ forms $E_1$ which cannot be totally separated but remains totally disconnected.

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  • $\begingroup$ Should the exponents for the $x$-coordinates be negative? $\endgroup$ – John Samples Jul 31 '18 at 3:19
  • $\begingroup$ Ok, this space works! Thanks! The details for the non-zero-dimensionality are a bit unpleasant; do you prefer to post them or should I go ahead? $\endgroup$ – John Samples Jul 31 '18 at 4:21
  • $\begingroup$ Yes, they should be negative. I'll change them. (I tried to simplify the original as I'm not much good with Tex, but I slipped up there.) $\endgroup$ – David Hartley Jul 31 '18 at 8:07
  • $\begingroup$ I didn't make it clear that this is not my example, just a "translation" of the one in the Sierpinski paper. I'll have a go at the proof if you like, but it will have to wait to tonight. $\endgroup$ – David Hartley Jul 31 '18 at 9:01
  • $\begingroup$ if you can I would appreciate it, I have been MSEing on nothing but a tablet this summer (touch screen keyboard). I deserve a new badge haha $\endgroup$ – John Samples Aug 1 '18 at 3:48

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