1
$\begingroup$

$1)$ If $(x_n)$ is Cauchy and $x_n \in \mathbb{Z} \,\,\forall n \in \mathbb{N}$ then $x_n$ is eventually constant (for $ n \geq N, x_n = x_{n+1})$

Attempt: If $(x_n) $ is Cauchy then for all $\epsilon > 0,\,\, \exists N \in \mathbb{N} $ such that $|x_n - x_m| < \epsilon$ for $n,m \geq N$.
Hence pick $\epsilon = 1$. Then we are done already since if $|x_n - x_m| < 1 $, implies that the difference between two integers (absolute value) is strictly less than 1. Since the statement $|x_n - x_m| < 1$ holds for all $n,m$ take two consecutive integers. Their difference is exactly 1. So the only way for the inequality to hold is if $x_n = x_m\,\forall n,m, $so in particular $x_n = x_{n+1}$.

Is it okay?

2)Formulate a definition for the following: $\operatorname{lim}_{x \rightarrow a^+} f(x) = - \infty$

Attempt: $f(x) \rightarrow -\infty$ as $x \rightarrow a^+$ if for every $N > 0$ such that if $ a < x < a + \delta$ then $f(x) < -N$

Is it okay?

Many thanks.

$\endgroup$
1
$\begingroup$

For (1):

If $(x_{n})$ is Cauchy, then for any $\epsilon > 0$ we have $| x_{n} - x_{m}| < \epsilon $ for $n,m > N \in \mathbb{N}$. Since $(x_{n}) \in \mathbb{Z}$, $x_{n} = x_{m}$ for all $n,m > N $.

$\endgroup$
0
$\begingroup$

Yes, the first one is correct.

For 2), you maybe miss a quantifier: for all $N>0$ there exists $\delta$...

$\endgroup$
1
  • $\begingroup$ Do you mean for all $N > 0,$ there exists $\delta$ such that if $a < x< x + \delta$ then $f(x) < -N$? $\endgroup$ – CAF Mar 20 '13 at 21:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.