Continuum = compact connected set.

Suppose that $U$ and $V$ are nonempty disjoint open subsets of $[0,1]^2$. Is there necessarily a continuum $K\subseteq [0,1]^2$ that divides $U$ and $V$? More precisely, is there a continuum $K$ such that $[0,1]^2\setminus K$ is the union of two disjoint open sets $T$ and $W$ with $U\subseteq T$ and $V\subseteq W$.

It seems like this is a classical result but I don't know.

Note: You may not be able to choose $K$ to be an arc because for certain open $U$ and $V$ it could have to be something like the topolgist's sine curve.

  • $\begingroup$ What do you mean with "continuum"? A set of cardinality $\mathfrak{c}$? $\endgroup$ – Del Oct 27 '16 at 20:57
  • $\begingroup$ But why is $[0,1]^2\setminus (U\cup V)$ connected? $\endgroup$ – Forever Mozart Oct 27 '16 at 21:01
  • $\begingroup$ You should define continuum here. $\endgroup$ – zhw. Oct 27 '16 at 21:01
  • $\begingroup$ @ForeverMozart It is not connected in general. $\endgroup$ – Del Oct 27 '16 at 21:02
  • $\begingroup$ @Del right, I see because $U$ and $V$ could be like tubes. $\endgroup$ – Forever Mozart Oct 27 '16 at 21:03

No. For example, consider these two sets: $$U = \{ x \in [-1,1]^2: |x| \in (1/8, 1/7) \cup (1/4, 1/3) \}$$ $$V = \{ x \in [-1,1]^2: |x| \in (1/10, 1/9) \cup (1/6, 1/5) \}$$ where $|\cdot |$ denotes the usual Euclidean norm.

  • 1
    $\begingroup$ ok I see. I was hoping it was true... $\endgroup$ – Forever Mozart Oct 27 '16 at 21:14

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