I read, that
$$ \overline{\bigcup^\infty_{n=0} T^{-n}(0)} = I, $$
(where $T$ is Tent map and $I = [0; 1]$) but I don't understand why.
Could you, please, show me the proof of this?
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Sure, that $I$ is the identity here? This does not make any sense. I'd say $I = [0,1]$, the unit interval. – martini Nov 04 '16 at 13:08
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@martini Thank you very much! Of course it's the unit closed interval. – Eenoku Nov 04 '16 at 13:09
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And you want the unit-height tent? – martini Nov 04 '16 at 13:10
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@martini That was not specified in the book, but I suppose so. – Eenoku Nov 04 '16 at 13:10
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We have $T^{-1}(\{0\}) = \{0,1\}$. To compute the higher inverse images, note, that $T^{-1}(\{x\}) = \{\frac x2,1-\frac x2\}$ for $x\in[0,1]$. Therefore $$ T^{-2}(\{0\}) = \left\{0,\frac 12, 1\right\} $$ By induction, we see that $$ T^{-n}(\{0\}) = \left\{ \frac{k}{2^{n-1}} : k = 0,\ldots, 2^{n-1}\right\}$$ This gives us that $\bigcup_n T^{-n}(\{0\})$ consists of all dyadic fractions, which form a dense subset of $[0,1]$.
martini
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So if I understand it well, $\bigcup^\infty_{n=0} T^{-n}(0) = \overline{\bigcup^\infty_{n=0} T^{-n}(0)} = I$? – Eenoku Nov 04 '16 at 14:00
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No, only the closure equals $I$, for example $\frac 13$, not being a dyadic rational, does not belong to any $T^{-n}[{0}]$ – martini Nov 07 '16 at 10:50