- #1
Terrell
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Homework Statement
Let ##x\in\Bbb{R}## such that ##x\neq 0##. Then ##x=LIM_{n\rightarrow\infty}a_n## for some Cauchy sequence ##(a_n)_{n=1}^{\infty}## which is bounded away from zero.
2. Relevant definitions and propositions:
3. The attempt at a proof:
Proof:(by construction)
Let ##x\in\Bbb{R}## such that ##x\neq 0##. Then by definition 5.3.1., ##x=LIM_{n\rightarrow\infty}a_n## such that ##(a_n)_{n=1}^{\infty}## is a Cauchy sequence; That is \begin{align}\forall\epsilon\in\Bbb{Q^+}\exists N\in \Bbb{Z^+}\forall j,k\geq N(\vert a_j-a_k\vert\leq\epsilon)\end{align}
and ##(a_n)_{n=1}^{\infty}## is also bounded. Suppose ##x=LIM_{n\rightarrow\infty}a_n## is not bounded away from zero. Thus, ##\exists j\in\Bbb{N}## such that ##a_j=0##.
Consider ##x'=LIM_{n\rightarrow\infty}a'_n## where ##\forall n\in\Bbb{N}(n\neq j\Leftrightarrow a_n=a'_n)##. Consequently due to definition 5.3.4. and Lemma 5.3.6.
\begin{align}x-x'=LIM_{n\rightarrow \infty}a_n - LIM_{n\rightarrow \infty}a'_n=LIM_{n\rightarrow \infty}(a_n-a'_n)\end{align}
Thus, ##\forall n\in\Bbb{N}(n\neq j\Leftrightarrow a_n-a'_n=0)##. Keep in mind that ##x,x'\in\Bbb{R}## and ##\forall\in\Bbb{N},a_n,a'_n\in\Bbb{Q^+}##. Since ##a'_j\neq a_j=0##, then by trichotomy of rationals ##a'_j>0## or ##a'_j<0##. Without loss of generality, suppose ##a'_j>0## then ##\vert a_j-a'_j\vert =\vert 0-a'_j\vert =\vert -a'_j\vert=-(-a'_j)=a'_j##. Hence, ##LIM_{n\rightarrow\infty}(a_n-a'_n)## where ##(a_n-a'_n)_{n=1}^{\infty}## represents the sequence \begin{align}a_1-a'_1=0,...,a_{j-1}-a'_{j-1}=0,a_j-a'_j=a'_j,a_{j+1}-a'_{j+1}=0,0,...\end{align}
Clearly, for all ##\epsilon'\in\Bbb{Q^+}## there exists ##\overline{n}\in\Bbb{Z^+}## such that ##\forall k\geq\overline{n},\vert a_k-a'_k\vert\leq\epsilon'##. Since ##x## is equivalent to ##x'## and the sequence ##(a'_n)_{n=1}^{\infty}## representing ##x'## is bounded away from zero, then ##(a'_n)_{n=1}^{\infty}## is a cauchy sequence representing ##x## that is bounded away from zero because by definition 5.3.1. equivalent cauchy sequences represent the same real number.
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