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Show that if $(x_{n})_{n}$ is a Cauchy sequence in X and $\lambda \in \mathbb{R}$, then the sequence $(\lambda x_{n})_{n}$, is also Cauchy in X.

We know that for $(x_{n})_{n}$, we have $$\forall \epsilon >0:\exists N\in \mathbb{N} : n,m\ge N\implies ||x_{n}-x_{m}||\le \epsilon$$

We can also assume that $$||\lambda (x_{n}-x_{m})||\le \epsilon$$

So to prove this, we can say that:
$$||\lambda (x_{n}-x_{m})||\le |\lambda |\cdot||x_{n}-x_{m}|| \le |\lambda|\epsilon$$

But I can't help but feel dubious about having the $\lambda$ at the end. Any tips?

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I would consider having the $|\lambda|\epsilon$ more unfashionable than wrong. But the fashionable thing to do is to say that by the definition of Cauchy, there is an $N$ such that if $m, n \ge N$, then $\dots <\frac{\epsilon}{\lambda}$. Note that we must separate out for special treatment the case $\lambda=0$. – André Nicolas Feb 29 '12 at 23:06
up vote 0 down vote accepted

Given any $\epsilon$ we can find $N_\epsilon$ such that for $n,m\geq N_\epsilon$, $||x_n-x_m|| \leq \epsilon$

We want to show that given any $\epsilon '$, we can find an $\tilde{N}_{\epsilon '}$ such that $n,m\geq \tilde{N}_{\epsilon '}$, $||\lambda x_n - \lambda x_m || \leq \epsilon$.

Well, start with your given $\epsilon$. Using the first sentence in the answer, find $N_{\epsilon/\lambda}$ That means for $n,m \geq N_{\epsilon/\lambda}$, $$||x_n-x_m||\leq \epsilon/\lambda$$ i.e. $$||\lambda x_n -\lambda x_m||\leq \epsilon$$

Specifically $\tilde{N}_\epsilon=N_{\epsilon/\lambda}$

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The $\lambda$ at the end doesn't give you any problems. You could have started with some $\epsilon'$ in the definition instead of $\epsilon$ itself, and then set $\epsilon = |\lambda| \epsilon'$.

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Oh, that rings a bell, I remember reading that a coefficient doesn't matter and we like to have just $\epsilon$ "for cosmetic reasons". Thanks. – dplanet Feb 29 '12 at 22:58

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