Given $X_n$ is Cauchy in $\mathbb{R}$. Is $\frac{1}{X_n}$ Cauchy?

Question

Given ${x_n}$ is Cauchy sequence in $\mathbb{R}$. I want to proof (or we can add the condition) so that $\frac{1}{x_n}$ is Cauchy sequence too.

Definition: $X_n$ Cauchy sequence iff $\forall \epsilon>0,$ there is $N \in \mathbb{N}$ such that $$m,n > N \Longrightarrow |X_m - X_n| < \epsilon.$$

Theorem: If $X_n$ is Cauchy then $X_n$ is bounded.

From the theorem above, we know $X_n < K$ for some $K \in \mathbb{R}$.

Question: I want to proof so that $\frac{1}{x_n}$ is Cauchy sequence, that is $\forall \epsilon > 0$ choose $N \in \mathbb{N}$ such that $$m,n > N \Longrightarrow \left|\frac{1}{X_m}-\frac{1}{X_n}\right| < \epsilon.$$

My Idea First we observe that $$\left|\frac{1}{X_m}-\frac{1}{X_n}\right| = \left| \frac{X_n-X_m}{X_nX_m}.\right|$$ Because of boundedness of Cauchy sequence we can write $\frac{1}{X_m} < K_1$ and $\frac{1}{X_n} < K_2$

Because $X_n$ is Cauchy, so $$|X_n - X_m| < \frac{\epsilon}{K_1K_2}.$$

So $$\left| \frac{X_n-X_m}{X_nX_m}.\right| < \frac{\epsilon}{K_1K_2} K_1K_2 = \epsilon. $$

Is it right? I am doubt about my proof especially when I am using boundedness of Cauchy seq. Thank you so much.

Answer 1

The sequence might converge to $0$, therefore, it is necessary for the sequence to converge to a number which is not $0$.

The following is a well known theorem from the theory of sequences in $\mathbb{R}$: Let $(a_n)_{n=1}^{\infty}$ be a Cauchy sequence in $\mathbb{R}$. Then $(a_n^{-1})_{n=1}^{\infty}$ is a Cauchy sequence if and only if $\lim_{n \to \infty} a_n \neq 0$.

$\textit{Hint:}$ Prove that if for $\epsilon >0$, $y_0 \neq 0$ and $y \in \mathbb{R}$,

$|y - y_0| < \min \{ |y_0|/2, \epsilon |y_0|^2/2 \} $,

then $y \neq 0$ and

$|\frac{1}{y} - \frac{1}{y_0}| < \epsilon$

Answer 2

Your argument would work if we knew that $\left\{\frac{1}{x_n}\right\}_{n\in\mathbb{N}}$ was bounded, however with the given assumptions that is not necessarily true, and neither is the conclusion. Indeed consider $x_n=\frac{1}{n+1}$. Then $\frac{1}{x_n}=n+1$, and this is clearly not bounded nor Cauchy. To fix the problem, we need to add the assumption that $\{x_n\}_{n\in\mathbb{N}}$ does not converge to $0$. Then, as $\mathbb{R}$ is complete, it converges to some $x\in\mathbb{R}\setminus\{0\}$. Then just choose $N\in\mathbb{N}$ so that

$$\lvert x_n-x\rvert<\frac{\lvert x\rvert}{2}$$

for all $n\geq N$. Then in particular,

$$\lvert x_n\rvert \geq \frac{\lvert x\rvert}{2}>0,$$

and so

$$\frac{1}{\lvert x_n\rvert}\leq\frac{2}{\lvert x\rvert}.$$

Thus, by adding the assumption that $\{x_n\}_{n\in\mathbb{N}}$ does not converge to $0$, we can bound $\left\{\frac{1}{x_n}\right\}_{n\in\mathbb{N}}$, which fixes your argument.