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Inspired by this question we ask the following question.

Note that the comment conversation and answers to the above question imply that

There are two complementary subsets of the unit sphere of $\ell^2$ which are contractible and dense.

I wonder if this situation mentioned above can hold for a finite dimensional continuum

Namely:

Is there a finite dimensional continuum $X$ which possesses a subset $A$ such that both $A$ and $X\setminus A$ are dense and contractible?

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    $\begingroup$ Does this qualify as example: $X:=[0,1]\times[0,1]$, $A:=(\mathbb Q\cap[0,1])\times[0,1)\cup[0,1]\times\{0\}$ ? $\endgroup$
    – Pietro Majer
    Commented Aug 18 at 0:16
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    $\begingroup$ A further question: can e.g. the unit ball of $\mathbb R^2$ be decomposed into 3 dense contractible sets ? $\endgroup$
    – Pietro Majer
    Commented Aug 18 at 10:49
  • $\begingroup$ @PietroMajer Thanks for your comment and very interesting answer $\endgroup$
    – Ali Taghavi
    Commented Aug 18 at 14:34
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    $\begingroup$ @PietroMajer An amazing point: A telepathy: I was thinking to the same question and assigning a natural number $n$ as the maximum valuse of the number of partition of a continum: Then I thought that re your example in comment possibly can be reconstructed to 3 partition. If we can write $Q$ as disjoint union of two dense subset...and apply it to $\mathbb{Q}\cap [0,1]$ but I realized this idea does not work since we have a connected interval at the ground $\endgroup$
    – Ali Taghavi
    Commented Aug 18 at 14:41
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    $\begingroup$ @PietroMajer So it is unlike that one produce a 3 partition. does this define a topological invariant somewhat related to dimension theory? $\endgroup$
    – Ali Taghavi
    Commented Aug 18 at 14:54

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A fancy decomposition of the closed unit disk $X$ of $ \mathbb{C}$ into two dense contractible sets. Let $D_1$, $D_2$ be a $2$-partition of the interval $[0,1]$ into dense sets. Consider the disjoint sets:

$A_1:= \{z\in X \setminus[-1,0) :|z|\in D_1\} \cup(0,1] $ $A_2:= \{z\in X \setminus(0,1] :|z|\in D_2\} \cup[-1,0) . $ Both are dense and contractible , and $X=A_1\cup A_2$.

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  • $\begingroup$ Very interesting answer Thank you very much. I guess in definition of $A_2$ you mean $D_2$ . the contractibility then comms from the contractibility of logarithm branch. $\endgroup$
    – Ali Taghavi
    Commented Aug 18 at 14:50

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