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$\newcommand{\and}{\text{ and }}$ $\newcommand{\set}[1]{\{#1\}}$ $\newcommand{\Or}{\text{ Or }}$ We seek to prove that $f$ is one to one, that is, for each $x, y \in X$, if $f(x) = f(y)$ then $y = x$. We will instead show the equivalent proposition, if $x \neq y$ then $f(x) \neq f(y)$. Let $$f_i(x) = \sum_{n = 1}^i \frac{x_n}{3^n}$$ and assume $x \neq y$. We will prove by induction on the number of components of $x$ and $y$ which are not equal, that $\lim_{i \to \infty} f_i(x) \neq \lim_{i \to \infty} f_i(y)$ \begin{enumerate} \item First the base case. Assume there is exactly one $j$ such that $x_j \neq y_j$. Then, for $i \geq j$, we have $$f_i(x) = \frac{x_j}{3^j} + \sum_{n \neq j}^i \frac{x_n}{3^n}$$ Since $x_j$ and $y_j$ are either 0 or 2, and $x_j \neq y_j$, we have that $x_j = 0 \and y_j = 2$ or $x_j = 2 \and y_j = 0$ Assume the first case, $x_j = 0 \and y_j = 2$ $f_i(x) = \sum_{n \neq j}^i \frac{x_n}{3^n}$ and $f_i(y) = \frac{2}{3^j} + \sum_{n \neq j}^i \frac{x_n}{3^n} = \frac{2}{3^j} + f_i(x)$. Therefore, $\lim_{i \to \infty} f_i(y) = \frac{2}{3^j} + \lim_{i \to \infty} f_i(x)$ So $\lim_{i \to \infty} f_i(x) \neq \lim_{i \to \infty} f_i(y)$ as required. \item Let $I = \set{j_1, j_2, \dots, j_k}$, Assume that, if it is true that $x_\alpha \neq y_\alpha$ if and only if $\alpha \in I$, then $\lim_{i \to \infty} f_i(x) \neq \lim_{i \to \infty} f_i(y)$. Now let $I' = \set{j_1, j_2, \dots, j_k, j_{k + 1}}$ and let $x', y'$ be in $X$ and $x'_\alpha \neq y'_\alpha$ if and only if $\alpha \in I'$. Now let $I \subset I'$ = $\set{j_1, j_2, \dots, j_k}$ There exists an $x$ and $y$ such that $x_\alpha = x'_\alpha$ and $y_\alpha = y'_\alpha$ if $\alpha \neq j_{k + 1}$, and $x_{j_{k+1}} = y_{j_{k+1}} = x'_{j_{k+1}}$ otherwise. Then $$f_i(x) = f_i(x')$$ and $$f_i(y) - \frac{2}{3^{j_{k + 1}}} = f_i(y') \Or f_i(y) + \frac{2}{3^{j_{k + 1}}} = f_i(y') $$ This implies that $$\lim_{i \to \infty}f_i(x) = \lim_{i \to \infty}f_i(x')$$ and $$|\lim_{i \to \infty} f_i(y) - \lim_{i \to \infty} f_i(y')| = \frac{2}{3^{j_{k + 1}}}$$ By the induction hypothesis $$f_i(x) - f_i(y) = \sum_{n \in I} \frac{x_n}{3^n} - \sum_{n \in I} \frac{y_n}{3^n}$$ and so $|\lim_{i\to\infty}f_i(x') - \lim_{i\to\infty}f_i(y)| = |\sum_{n \in I} (\frac{x'_n}{3^n} - \frac{y_n}{3^n})| = |\sum_{n \in I} \pm \frac{2}{3^n}|$ $|\lim_{i\to\infty}f_i(x') - \lim_{i\to\infty}f_i(y')| = |(\sum_{n \in I} \pm \frac{2}{3^n}) \pm \frac{2}{3^{j_{k+1}}}| = |(\sum_{n \in I'} \pm \frac{2}{3^n})|$ \end{enumerate}
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Sun, 15 Sep 2019 17:49 GMT