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The following proof was suggested: suppose [0,1] was the disjoint countable union of closed intervals. Write the intervals as $[a_n,b_n]$. Start by showing the set of endpoints $a_n, b_n$ is closed. At first I thought this was obvious since it seems like the complement is just the union of $(a_n,b_n)$ which is open but then I thought the complement was the infinite intersection of $(0,a_n) \cup (a_n,b_n) \cup (b_n,1)$ which is not necessarily open any more.

This comes from Taylor's proof in Is $[0,1]$ a countable disjoint union of closed sets?

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Your first thought was correct. By hypothesis each $x\in[0,1]$ belongs to exactly one of the intervals $[a_n,b_n]$. If it isn't an endpoint of any of the intervals, it must actually belong to the open interval $(a_n,b_n)$. Thus, the complement of the set of endpoints is indeed $\bigcup_{n\in\Bbb N}(a_n,b_n)$.

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  • $\begingroup$ Thanks Brian! So we have that the set of endpoints is closed. It is countable by construction. It seems we want to get a contradiction by showing it must also be uncountable. Since it's closed, if we can show it's perfect then we are done since it is a perfect subset of a complete space and thus uncountable. However, I am worried that some points are isolated: for example, isn't one of the $a_n=0$ and $b_n=1$? Surely these can be isolated? For the other endpoints it seems more likely that they are not isolated since $a_i\neq 0$ must be "pretty darn close" to some $b_j$ for it to cover $[0,1]$. $\endgroup$ – user224531 Mar 18 '15 at 16:49
  • $\begingroup$ why is you so smart sir? $\endgroup$ – user139708 Mar 19 '15 at 0:41
  • $\begingroup$ @user224531: Taylor Martin was a bit careless in writing up his proof. He was actually working in $(0,1)$, not $[0,1]$, though he did point out why it makes no difference. Yes, $0$ is one of the left endpoints, and $1$ is one of the right endpoints; we can number the intervals so that $0=a_0$ and $1=b_1$. Then the intervals $[a_k,b_k]$ for $k>1$ must be a partition of the open interval $(b_0,a_1)$, and his argument shows that $\{a_k:k>1\}\cup\{b_k:k>1\}$ has no isolated points. $\endgroup$ – Brian M. Scott Mar 19 '15 at 3:01
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Assume that $[0,1]=\bigsqcup_{i=1}^\infty[a_i,b_i]$.

Start from $x_0=b_1$. Assume $b_1<1$. Then $x_0+\epsilon_1\in (0,1)$ lies in some $[a_i,b_i]$, where $\epsilon_1<\frac{1}{2}$. put $x_1=a_i>x_0$. Then $|x_0-x_1|<\frac{1}{2}$. Then $x_1-\epsilon_2\in (0,1)$ lies in some $[a_j,b_j]$, where $\epsilon_2<\frac{1}{4}$, put $x_2=b_j$, then $x_0<x_2<x_1$ and $|x_1-x_2|<\frac{1}{4}$. Continuing this process, we get $$x_0<x_2<\ldots <x_3<x_1\qquad |x_n-x_{n+1}|<\frac{1}{2^n}$$ thus by Leibniz's criteria, $x_n\to x\in (0,1)$ with $$x_0<x_2<\ldots <x<\ldots <x_3<x_1$$ But $x$ cannot lie in any closed interval. For $x\in [a,b]$, if $a<x$, then it is a must that some $a<x_{2i}<x$, but $x_{2i}$ is a right point of some interval, which is contradict to the disjoint property, thus $a= x$. Similarly, $b=x$. The proof is complete.

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