For $a$ is a real number, suppose $(x_n)$ is a sequence such that for all $n$, $|x_n - a| < 1/n$. Prove $lim x_n = a$?

Question

This problem is two parts, and I'm not really sure how to do either part since I need the first part to get the second part. I kind of get limit laws and all that, but I don't know how to do this problem with these other variables to consider. Can someone help explain the proof for this pretty thoroughly?

Answer 1

For part $(a)$, show that given $\epsilon > 0$, setting $N > \frac{1}{\epsilon}$ will make $|x_n - a| < \epsilon$ for all $n\ge N$. For $(b)$, note that since $S$ is dense in $\Bbb R$ and $a\in \Bbb R$, then there exists an $x_1\in S$ such that $a < x_1 < a + 1$, i.e., $x_1 - 1 < a < x_1$. Then choose $x_2\in S$ such that $a < x_2 < a + 1/2$. Continue the process to create a sequence $(x_n)$ in $S$ such that $a < x_n < a + 1/n$ for all $n \ge 1$. From this it follows that $|x_n - a| = x_n - a < 1/n$ for all $n$. Now you can use part $(a)$ to deduce that $\lim x_n = a$.

Answer 2

Look at $y_n = |x_n-a|$. It's a monotone decreasing sequence bounded above by $\frac{1}{n} \to_n 0$. It's also positive, so it's limit is 0. Hence...

Answer 3

Let $\epsilon > 0$. We have that the sequence $\{\frac1n\}_{n \ge 1}$ is convergent to $0$. So:

$$\exists \ n_0 \in \Bbb N \ such \ that: \ \forall \ n \ge n_0, \ \frac1n < \epsilon \ \ (*)$$

Let $n \ge n_0$, then look at:

$$|x_n - a| < \frac1n < \epsilon, \ \ using \ \ (*)$$

This shows that $x_n \rightarrow a$.