Consider the continuous map $ f: \Sigma_{2}^{+} \rightarrow \Sigma_{2}^{+} $ , which is shift map also by

Question

Define $ \Sigma_{2}^{+}=\{x=\{x_n\}: x_n=0 \ or \ 1 \} $ is infinite string of $ \ 0 's \ \ and \ \ 1 's $. Consider the continuous map $ f: \Sigma_{2}^{+} \rightarrow \Sigma_{2}^{+} $ , which is shift map also by

$ f(x_0,x_1,x_2,..........)=(x_1,x_2,x_3,...............) $.

For $ x \in \Sigma_{2}^{+} \ \ and \ \delta >0 $ , define an open ball centered at $ x $ by $ B_{\delta}(x)=\{ z \in \Sigma_{2}^{+} : d(x,z ) < \delta \} $ .

Let $ \bar{0}=(0,0,0,..........) \ \ and \ \ \bar{1}=(1,1,1,.........) $ are elements of $ \Sigma_{2}^{+} $.

Let $ 0 < \delta <1 $.

Then prove that there exists $ z \in B_{\delta}(\bar{0}) $ such that $ f^n (z)=\bar{1} $ , for some $ n \in \mathbb{N} $.

Answer:

Consider the point $ z=(0,0,0,........,0,\bar{1},..........) , $ where first $ n $ terms are $ 0 \ \ and \ \ then \ all \ \ 1's $.

Clearly , $ d(\bar{0} , z) < 1 \ \ and \ \ hence \ \ z \in B_{\delta}(\bar{0}) $.

Therefore the $ n^{th} $ iteration of the shift map $ f : \Sigma_{2}^{+} \rightarrow \Sigma_{2}^{+} $ give rise the desired result,

i.e.; $ f^n(z)=\bar{1} $.

I need confirmation about my work. Is there any help?

Answer 1

With your $z$ from above, we have $d(\bar{0} , z)=2^{-n}$. Now choose $n$ so that $2^{-n} < \delta$. Then $z \in B_{\delta}(\bar{0})$ and $f^n(z)=\bar{1}$.

Answer 2

Your idea is essentially correct but you have not specified the choice of $n$. For any $n>1$, $d(\overline{0}, z)<1$ but this does not guarantee that $d(\overline{0}, z)<\delta$ which is required to conclude that $z\in B_\delta(\overline{0})$. For this you will have to select $n$ large enough so that $$d(\overline{0}, z)=\frac{1}{2^n}+\frac{1}{2^{n+1}}+\cdots=\frac{1}{2^{n-1}}<\delta.$$