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I am trying to prove that if $\{x_n\}$ is a Cauchy sequence in $X$ then the sequence of norms $\{\|x_n\|\}$ is a Cauchy sequence of real numbers.

This is the proof:

Since $\|u_m\| - \|u_n\| \leq \|u_m - u_n\|$ the sequence is a Cauchy sequence in $\mathbb{R}$

I am not sure what the question is trying to ask, surely the question is trivial?

I understand what a Cauchy sequence is but why does the question specify "real numbers"?

How does the proof answer the question?

Michael Hardy
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Al jabra
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  • Some times it may ask you to use $\epsilon$. :) – Bumblebee Apr 22 '16 at 09:44
  • If ${x_n } $ is Cauchy seqence then for any $\varepsilon >0$ there exists $N$ such that $||x_k -x_j ||\leq \varepsilon $ for $k,j \geq N.$ Thus if you take any $\varepsilon >0$ then $$|||x_k || -||x_j |||\leq ||x_k - x_j ||\leq \varepsilon$$ for $k,j \geq N.$ hence ${||x_n ||} $ is Cauchy sequence. –  Apr 22 '16 at 09:45
  • @motylanogatomkamazura I understand that but why does the question specify real numbers? – Al jabra Apr 22 '16 at 09:46
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    Because $||x||$ is a real number. –  Apr 22 '16 at 09:47
  • Could the three of you please write $|x|$ instead of $||x||$? I corrected this in the question itself. Notice how conspicuous this looks in the following context: $||a|| ||b||$ versus $|a||b|$ (as if it weren't already conspicuous before that). $\qquad$ – Michael Hardy Apr 22 '16 at 09:57

2 Answers2

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The hypothesis that $\langle x_n:n\in\Bbb N\rangle$ is a Cauchy sequence in $X$ means that

for each $\epsilon>0$ there is an $m\in\Bbb N$ such that $\|x_k-x_\ell\|<\epsilon$ whenever $k,\ell\ge m$.

The desired conclusion, that $\langle\|x_n\|:n\in\Bbb N\rangle$ is a Cauchy sequence in $\Bbb R$, means that

for each $\epsilon>0$ there is an $m\in\Bbb N$ such that $\big|\|x_k\|-\|x_\ell\|\big|<\epsilon$ whenever $k,\ell\ge m$.

As you can see, these are not the same statement: the first is a statement about a sequence of vectors in $X$, and the second is a statement about the sequence of the norms of those vectors, which is a sequence of real numbers. Thus, there is definitely something to be proved.

To prove it, suppose that $\langle x_n:n\in\Bbb N\rangle$ is Cauchy in $X$, and let $\epsilon>0$ be arbitrary. By hypothesis there is an $m\in\Bbb N$ such that $\|x_k-x_\ell\|<\epsilon$ whenever $k,\ell\ge m$. Now use the fact, mentioned in your question, that

$$\|u\|-\|v\|\le\|u-v\|\tag{1}$$

for any $u,v\in X$: since $(1)$ holds for all $u$ and $v$, it’s equally true that

$$-(\|u\|-\|v\|)=\|v\|-\|u\|\le\|v-u\|=\|u-v\|\;.\tag{2}$$

$(1)$ and $(2)$ together imply that

$$\big|\|u\|-\|v\|\big|\le\|u-v\|\;.$$

Now apply this to any $x_k$ and $x_\ell$ with $k,\ell\ge m$:

$$\big|\|x_k\|-\|x_\ell\|\big|\le\|x_k-x_\ell\|<\epsilon\;.$$

Thus, $\big|\|x_k\|-\|x_\ell\|\big|<\epsilon$ whenever $k,\ell\ge m$, and $\langle\|x_n\|:n\in\Bbb N\rangle$ is indeed a Cauchy sequence in $\Bbb R$.

Brian M. Scott
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$$ \|u\| = \overbrace{\|(u-v)+v\| \le \|u-v\| + \|v\|}^\text{This is a basic property of norms.}; $$ therefore $$ \|u\|-\|v\| \le \|u-v\|. \tag 1 $$ But the same reasoning can show that $$ \|v\|-\|u\| \le \|v-u\|. \tag 2 $$ To say that $(1)$ and $(2)$ are both true is to say that $$ \left| \|u\| - \|v\| \right| \le \|u-v\|. $$ Consequently if you've shown that $\|u-v\| \le \varepsilon$, you can conclude that $\left| \|u\|-\|v\| \right| \le \varepsilon$.

Michael Hardy
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