Cauchy

**taypez** · September 25th 2007, 05:17 PM

Prove that the sum of two Cauchy sequences is Cauchy without using:

Let {x_n} be a sequence of real numbers. Then {x_n} ic Cauchy iff {x_n} converges.

**Jhevon** · September 25th 2007, 05:29 PM

Originally Posted by taypez

Prove that the sum of two Cauchy sequences is Cauchy without using:

Let {x_n} be a sequence of real numbers. Then {x_n} ic Cauchy iff {x_n} converges.

define $\{ z_n \} = \{ x_n \} + \{ y_n \}$ . we need to show that: for all $\epsilon > 0$ , there exists a $N \in \mathbb {N}$ , such that $m,n > N$ implies $|z_n - z_m| < \epsilon$

hint: you will need the triangle inequality here

can you continue?

**taypez** · September 25th 2007, 05:45 PM

That would lead me to use the Cauchy Theorem which it said not to and since the Cauchy Thm uses Bolzano-Weierstrass, I'm assuming I can't use that either.

**Jhevon** · September 25th 2007, 05:54 PM

Originally Posted by taypez

That would lead me to use the Cauchy Theorem which it said not to and since the Cauchy Thm uses Bolzano-Weierstrass, I'm assuming I can't use that either.

no it would not. did you try it? all you need is the definition of Cauchy and the triangle inequality, nothing else. and that is allowed for this question

**ThePerfectHacker** · September 25th 2007, 06:00 PM

Originally Posted by taypez

Prove that the sum of two Cauchy sequences is Cauchy without using:

Let {x_n} be a sequence of real numbers. Then {x_n} ic Cauchy iff {x_n} converges.

It is a trivial problem.

$|x_n+y_n - x_m - y_m|\leq |x_n-x_m|+|y_n-y_m| < \frac{\epsilon}{2}+\frac{\epsilon}{2} = \epsilon$ .

**Jhevon** · September 25th 2007, 08:25 PM

Bored.

Let $\{ x_n \}$ and $\{ y_n \}$ be Cauchy sequences.

Define $\{ z_n \} = \{ x_n \} + \{ y_n \}$ . We show that $\{ z_n \}$ is Cauchy.

Since $\{ x_n \}$ is Cauchy, for all $\epsilon > 0$ , there exists an $N_1 \in \mathbb {N}$ such that $m,n > N_1$ implies $|x_n - x_m|< \frac {\epsilon}2$ .

Similarly, since $\{ y_n \}$ is Cauchy, we can find an $N_2 \in \mathbb{N}$ such that $m,n > N_2$ implies $|y_n - y_m|< \frac {\epsilon}2$

Now, fix such an $\epsilon > 0$ , and choose $N = \mbox {max} \{ N_1, N_2 \}$ . Then $m,n > N$ implies that:

$|z_n - z_m| = |(x_n + y_n) - (x_m + y_m)| = |(x_n - x_m) + (y_n - y_m)|$ $\le |x_n - x_m| + |y_n - y_m| < \frac {\epsilon}2 + \frac {\epsilon}2 = \epsilon$

Thus, $\{ z_n \}$ is a Cauchy sequence

QED.

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