$X = \{0 ,1\}^{\mathbb N}$ be the metric space. Can anyone please tell me how to define a continuous injective function from $X = \{0 ,1\}^{\mathbb N}$ to the cantor set ?
Can anyone please give an idea ?
$X = \{0 ,1\}^{\mathbb N}$ be the metric space. How to define a continuous injective function from $X = \{0 ,1\}^{\mathbb N}$ to the cantor set
general-topologyreal-analysis
Related Solutions
Yes (assuming it's nonempty, of course). Moreover if you google "continuous image of the Cantor set", the first hit takes you to
http://en.wikipedia.org/wiki/Cantor_set
where you can read that this theorem is true and a reference is given to Willard's General Topology. (The article does not say this and perhaps it should, but it is specifically Theorem 30.7 on p. 217 of the Dover edition.)
$(i)$ Take a Cauchy sequence $(f_{n})_{n=1}^{\infty}\subseteq C_{b}(X;\mathbb{R})$ in the sup-norm $\|\cdot\|_{\infty}$ and $\varepsilon>0$. Hence there exists $n_{\varepsilon}\in\mathbb{N}$ so that \begin{equation*} |f_{n}(x)-f_{m}(x)|\leq \|f_{n}-f_{m}\|_{\infty}<\frac{\varepsilon}{3} \end{equation*} for all $n,m\geq n_{\varepsilon}$, which shows that $(f_{n}(x))_{n=1}^{\infty}$ is a Cauchy sequence in $\mathbb{R}$ for every $x\in X$. Since $\mathbb{R}$ is complete, for every $x\in X$ there exists $f(x)\in \mathbb{R}$ so that $f_{n}(x)\to f(x)$. Since limits are unique in metric spaces, you may define a function $f:X\to\mathbb{R}$ so that $x\mapsto f(x)$. Now $f$ is the point-wise limit of $f_{n}$. We show that $\|f_{n}-f \|_{\infty}\to 0$ and $f\in C_{b}(X;\mathbb{R})$. Fix $x\in X$. Now there exists $n_{0}\in \mathbb{N}$ so that $|f_{n}(x)-f(x)|<\frac{\varepsilon}{3}$ for all $n\geq n_{0}$.. Thus for every $n\geq n_{1}:=\max\{n_{\varepsilon},n_{0}\}$ we have \begin{equation*} |f_{n}(x)-f(x)|\leq |f_{n}(x)-f_{n_{0}}(x)|+|f_{n_{0}}(x)-f(x)|<\frac{\varepsilon}{3}+\frac{\varepsilon}{3}=\frac{2\varepsilon}{3}. \end{equation*} Hence by taking supremum over all $x\in X$, we have $\|f_{n}-f\|_{\infty}\leq \frac{2\varepsilon}{3}<\varepsilon$ for all $n\geq n_{1}$. Hence $\|f_{n}-f\|_{\infty}\to 0$. You may now either conclude that $f\in C_{b}(X;\mathbb{R})$ since $f$ is the uniform limit of continuous bounded functions, or prove it as follows. Let $x\in X$ and everything else as above remain fixed. Since $f_{n_{1}}$ is continuous, there exists $\delta>0$ so that $f_{n_{1}}B(x,\delta)\subseteq B(f_{n_{1}}(x),\frac{\varepsilon}{3})$. Now for all $y\in B(x,\delta)$ we have \begin{align*} |f(x)-f(y)| &\leq |f(x)-f_{n_{1}}(x)|+|f_{n_{1}}(x)-f_{n_{1}}(y)|+|f_{n_{1}}(y)-f(y)| \\ &\leq 2\|f_{n_{1}}-f\|_{\infty}+|f_{n_{1}}(x)-f_{n_{1}}(y)| \\ &<3\frac{\varepsilon}{3}=\varepsilon, \end{align*} and thus $fB(x,\delta)\subseteq B(f(x),\varepsilon)$. Hence $f$ is continuous. Finally, since \begin{equation*} \|f\|_{\infty}\leq \|f-f_{n_{1}}\|_{\infty}+\|f_{n_{1}}\|_{\infty}<\frac{\varepsilon}{3}+\|f_{n_{1}}\|_{\infty}<\infty, \end{equation*} then $f$ is bounded. Hence $f\in C_{b}(X;\mathbb{R})$, and since $\|f_{n}-f\|_{\infty}\to 0$, we have proven that $C_{b}(X;\mathbb{R})$ is complete.
$(ii)$ Note that $|O(x)(y)|=|d(y,x)-d(y,x_{0})|\leq d(x,x_{0})$ for all $y\in X$ by the reverse triangle-inequality. Hence $\|O(x)\|_{\infty}\leq d(x,x_{0})$, which shows that $O(x)$ is bounded for every $x\in X$. The functions $y\mapsto d(y,x)$ and $y\mapsto -d(y,x_{0})$ are continuous, so each $O(x)$ is continuous as a sum of two continuous functions. Hence $O(x)\in C_{b}(X;\mathbb{R})$ for every $x\in X$.
$(iii)$ Note at first, that again by the reverse triangle-inequality it follows that for all $x,y\in X$: \begin{align*} \|O(x)-O(y)\|_{\infty}&= \sup_{z\in X}\|O(x)(z)-O(y)(z)\|_{\infty} \\ &=\sup_{z\in X}\|d(x,z)-d(z,x_{0})-d(z,y)+d(z,x_{0})\|_{\infty} \\ &=\sup_{z\in X}\|d(x,z)-d(z,y)\|_{\infty} \\ &\leq d(x,y). \end{align*} On the other hand, \begin{align*} 0\leq d(x,y)&=d(x,y)-d(x,x_{0})+d(x,x_{0})-d(x,x) \\ &=O(y)(x)-O(x)(x) \\ &= |O(y)(x)-O(x)(x)| \\ &\leq \|O(y)-O(x)\|_{\infty}. \end{align*} Hence $\|O(y)-O(x)\|_{\infty}=d(x,y)$ for all $x,y\in X$, which shows that $x\mapsto O(x)$ is an isometric embedding $X\to C_{b}(X;\mathbb{R})$.
Best Answer
For each $x\in X$, we have a sequence of zeros and ones. Meanwhile the Cantor set is the set of all real numbers in the unit interval whose ternary expansion contains no $1$'s. So the natural map would be to send a given sequence $x=(x_n)$ of zeros and ones to the number in the Cantor set whose ternary expression is $\sum a_n/3^n$, where $a_n=\begin{cases} 0 , x_n=0\\2, x_n=1\end{cases}$.
The map is automatically continuous because the $X$ is discrete.
It remains to prove injectivity. But that's straight forward, because different sequences result in numbers with different ternary representations.