# Thread: Cauchy sequence, help plz

1. ## Cauchy sequence, help plz

If anyone can help me with this problem, would be greatly appreciated.

Let b be $\displaystyle \mathbb R$ with b > 0 . Show that there exists a Cauchy sequence of Rational numbers {$\displaystyle \ a_n$} with $\displaystyle \ n \geq 1$ such that $\displaystyle \ a_n$ > 0 for all n of N and that $\displaystyle \ lim_{n -> \inf} a_n = b$

2. Originally Posted by Cato
If anyone can help me with this problem, would be greatly appreciated.

Let b be $\displaystyle \mathbb R$ with b > 0 . Show that there exists a Cauchy sequence of Rational numbers {$\displaystyle \ a_n$} with $\displaystyle \ n \geq 1$ such that $\displaystyle \ a_n$ > 0 for all n of N and that $\displaystyle \ lim_{n -> \inf} a_n = b$
If $\displaystyle b$ is rational then there is nothing to prove - why?

Thus, it is safe to assume $\displaystyle b$ is irrational. The for any $\displaystyle n\geq 1$ there exists a rational number $\displaystyle a_n$ so that $\displaystyle -\tfrac{1}{n} < b - a_n < \tfrac{1}{n}$. Define a sequence $\displaystyle \{a_n\}$. Then it turns out this sequence is Cauchy with $\displaystyle \lim a_n = b$.

3. But it says b is a real number, not rational.

4. Originally Posted by Cato
But it says b is a real number, not rational.
So what? There are two possibilities. Either it is rational or it is irrational. If rational there is nothing to prove. If irrational then we use the fact that between any two irrational numbers there is a rational number - that is the second case.

5. Between any two numbers there is a rational number.
$\displaystyle \left( {\forall n \in \mathbb{Z}^ + } \right)\left( {\exists r_n \in \mathbb{Q}} \right)\left[ {b < r_n < b + \frac{1}{n}} \right]$.
Now it is easy to see that $\displaystyle \left( {r_n } \right) \to b$.
Any convergent sequence is a Cauchy sequence.

6. Thank you both for the quick replies