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I worked out the following problem

Show that $\{a_n\}$ and $\{b_n\}$ are equivalent Cauchy sequences iff $\{c_n\} = \{a_1,b_1,a_2,b_2, \cdots \}$ is Cauchy.

my proof:

$(\Leftarrow)$ Suppose $a_n \to x$ and $b_n \to y$ and $x \ne y$ while $\{c_n\}$ is Cauchy . Then given any $\epsilon$

$$\exists N \in \Bbb N \text{ s.t }\forall n>N, \\ \begin{align}\quad & d(x,a_n)<\epsilon/3 \\ & d(y,b_n)<\epsilon/3 \\ & d(x,y) \gt \epsilon \\ & d(c_n,c_m) = d(a_n,b_n)<\epsilon/3 .\end{align}$$

So,

$$\begin{align} d(x,y) & \le d(x,a_n)+d(a_n,y) \\ & \le d(x,a_n) + d(a_n,b_n) + d(b_n,y) \\ & \lt \epsilon \end{align}$$

which is a contradiction.

So $\{a_n\}$ and $\{b_n\}$ are equivalent Cauchy sequences.

($\Rightarrow$)

Since $\{a_n\}$ and $\{b_n\}$ have the same limit $x$, given any $\epsilon$

$$\exists N \in \Bbb N \text{ s.t } \forall n > N \\ \begin{align} & d(a_n,x) < \epsilon/2 \\ & d(b_n,x) \epsilon/2 \\ \end{align}$$.

So,

$$\begin{align} d(a_n,b_n) & \le d(a_n,x) + d(b_n,x)\\ & \lt \epsilon \end{align}$$

which shows that $\{c_n\}$ is Cauchy.

QED

Am I on the right track or is there anything I am off ?

share|improve this question
    
does the question explicitly state that the sequences are in $\mathbb{R}$? by your use of $d$ i would assume the question is stated in a general metric space, where the sequences don't necessarily have limits even if they're cauchy –  citedcorpse Jun 14 '13 at 9:02
    
@exitingcorpse: It’s pretty clear from other recent questions that the OP is working through a construction of $\Bbb R$ as equivalence classes of Cauchy sequences. –  Brian M. Scott Jun 14 '13 at 9:11
    
oh woops, i didn't realise it was the same person –  citedcorpse Jun 14 '13 at 9:12
    
At this point are you working in $\Bbb Q$ or in $\Bbb R$? This result seems likely to be part of the build-up to constructing $\Bbb R$, and if you’re working strictly in $\Bbb Q$, you can’t assume that your Cauchy sequences converge. –  Brian M. Scott Jun 14 '13 at 9:13
    
I don't think you need to use any contradiction at all. And then you immediately avoid assuming your sequence(s) converge. –  HSN Jun 14 '13 at 12:05
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