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Attempt at solution:

  1. I tried to show that there must be a family of disjoint open sets such that $F =\cup_{n \in \mathbb{N}}\alpha_n$. i.e $\forall n \in \mathbb{N}, \alpha_n$ is an open interval.

  2. Assuming that 1 is true, given that $F=\cup_{n \in \mathbb{N}}b_n$, where each $b_n$ is closed, it follows that we have a contradiction since $\forall n \in \mathbb{N} \exists m \in \mathbb{N},b_n \subset \alpha_m$ and so $\cup_{n \in \mathbb{N}}\alpha_n\neq \cup_{n \in \mathbb{N}}b_n$

I'm having trouble showing that 1. is true, but right now I think it's true.

tomasz
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    I guess you may want closed intervals to be non-trivial (not a point). And you 1 is basically saying that "[0,1] minus a countable set is open", which is not true. – AlgRev Nov 16 '15 at 20:29
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    There is a more general question and answer here: http://math.stackexchange.com/questions/6314/is-0-1-a-countable-disjoint-union-of-closed-sets – B. S. Thomson Nov 16 '15 at 20:45
  • And, doesn't the problem, as posed, ask for the family to contain at least two members? – B. S. Thomson Nov 16 '15 at 21:10
  • @B.S.Thomson I know the answer to that question. But, this is a different question. If you can show that the answer to the other question implies the answer to this question...I'd like to see such a proof. It isn't obvious to me. –  Nov 16 '15 at 21:10
  • @B.S.Thomson Good point. In fact, I require that the family contains an infinite number of members. –  Nov 16 '15 at 21:11
  • A countable set is, of course, a union of a countable disjoint family of closed sets. So the point of the problem, as opposed to the more general problem of showing that $[0,1]$ is not a nontrivial union of countably many disjoint closed sets is either: (a) perhaps we are to use the general result and just make sure that any family of disjoint intervals is countable (easy) or (b) there is some different easier proof that somehow exploits the fact that we are considering closed intervals and not just closed sets. See the other answer posted. – B. S. Thomson Nov 17 '15 at 01:15
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    The notation for set difference is either $\setminus$ or (rarely) $-$. Writing $/$ instead is really confusing: it makes it look like a quotient. – tomasz Nov 17 '15 at 01:22

2 Answers2

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This question and its answers answer this question too. Suppose we have a disjoint family (of more than one) closed interval $F_i, i \in I$ (which we assume to be all non-trivial) with $[0,1] \setminus (\cup_i F_i)$ countable, say $\{a_n: n \in \mathbb{N}\}$. Then the $F_i$ (of which there are at most countably many, as each interior contains a different rational) plus the sets $\{a_n\}$ are all connected and form a disjoint cover of the continuum $[0,1]$ by closed sets, and this cannot happen by the referenced result by Sierpiński.

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Henno Brandsma
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I thought, initially, that this problem, which is a very special case of Sierpinski's theorem, would not have a simple solution and any direct proof would probably just copy details from the general case. But here is a sketch which should work if I have the details right.

We suppose that there is disjointed sequence of closed subintervals $\{[a_i,b_i]\}$ so that the set $$C=[0,1]\setminus \bigcup_{i=1}^\infty [a_i,b_i] $$ is countable. We need to have at least two intervals in the sequence to obtain a contradiction. Turn your attention to the interiors and define $$ G = \bigcup_{i=1}^\infty (a_i,b_i). $$ This is an open set and since we are assuming the intervals are disjoint, this is exactly the sequence of components of $G$. The set $$P = [0,1]\setminus G$$ is a closed countable set by our hypothesis. (It contains $C$ and the endpoints of the intervals only.)

But $P$ has no isolated points except perhaps $0$ and $1$. Any other isolated point would be an endpoint of two (this why we need two) of the components and that is impossible if the closed intervals are disjoint.

Perfect sets are uncountable and that violates the assumptions. (Somebody please check and provide a commonly used reference for why perfect sets are not countable.)

B. S. Thomson
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