How can I prove $[0,1]\cap\operatorname{int}{(A^{c})} = \emptyset$?

If $A \subset [0,1]$ is the union of open intervals $(a_{i}, b_{i})$ such that each rational number in $(0, 1)$ is contained in some $(a_{i}, b_{i})$, show that boundary $\partial A= [0,1] - A$. (Spivak- calculus on manifolds)

If I prove that $[0,1]\cap\operatorname{int}{(A^{c})} = \emptyset$, the proof is complete.

I tried to find a contradiction, but I didn't find one.

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Can $A^c$ in $[0,1]$ contain a non-trivial interval ? –  user18119 Nov 11 '11 at 18:42
i think no and i try to proof. if $A^{c}$ contain such a interval, this interval contain infinitely many rationals. And these rationals belong to both $A$ (from definition) and $A^{c}$. Contradiction! QED? :) thank you and Socrates (for socratic method) –  H. ERGUL Nov 11 '11 at 19:22
@H.ERGUL: The verb is "to prove", not "to proof". So the correct way to write it "to prove", not "to proof". –  Arturo Magidin Nov 11 '11 at 19:40

Since you say you want to prove it by contradiction, here we go:

Suppose that $x\in \mathrm{int}(A^c)\cap [0,1]$. Then there is an open interval $(r,s)$ such that $x\in (r,s)\subseteq A^c$. Every open interval contains infinitely many rational numbers, so there are lots and lots of rationals in $(r,s)$. However, since $A$ contains all rationals in $(0,1)$, then the only rationals that can be in $(r,s)$ are $0$, $1$, and rationals that are either negative or greater than $1$.

Since $x\in [0,1]$, the only possibility is $x=0$ or $x=1$ (there's a small argument to be made here; I'll leave it to you). Why can we not have $x=0$? Well, if $x=0$, then $s\gt 0$, so $(r,s)$ contains $[0,\min\{s,1\})$. Are there any rationals between $0$ an d $1$ that are in $[0,\min\{s,1\})$?

And what happens if $x=1$?

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$0$ and $1$ aren't belong to $\mathrm{int}(A^c)=(- \infty, 0)\cup (1, + \infty)$. therefore $0, 1 \notin \mathrm{int}(A^c) \cap [0,1]$ –  H. ERGUL Nov 11 '11 at 19:40
@H.ERGUL: How do you know that the interior of $A^c$ is $(-\infty,0)\cup(1,+\infty)$? If you already know that, then why are you trying to prove that the interior does not intersect $[0,1]$? –  Arturo Magidin Nov 11 '11 at 19:42
i am wrong. sorry, i hastened. As a consequence of completeness of $\mathbb{R}$, $\exists q\in \mathbb{Q} : 0 < q <\min\{s,1\}$. Then $q$ must be in $A$, but $q$ in $A^{c}$. contradiction! –  H. ERGUL Nov 11 '11 at 19:57

The set $\partial A$ is just the closure of $A$ minus the interior of $A$. As $A$ is a union of open intervals, it's open (and thus equals its interior), and, as $A$ contains all rationals, it's dense and so its closure is $[0,1]$.

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+1.    –  Did Nov 11 '11 at 18:48
lots of thanks. A different approach . –  H. ERGUL Nov 11 '11 at 19:00
@Ergul: sometimes the simplest approach is the best. No contradiction is needed. –  Henno Brandsma Nov 11 '11 at 19:11

All you need is that $A$ contains all the rationals in $[0,1]$. Now suppose, for a contradiction, that the interior of $A^c$ was non-empty -- say that it contained $x$. Then by definition $A^c$ has to contain an open interval around $x$. But every open interval contains a rational number, and by assumption there are no rationals in $A^c$.

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