equivalence classes and real numbers

**yvonnehr** · December 17th 2009, 06:28 PM

Can someone help me with this problem? I wrote up a solution but the professor said that it was not what he wanted. He wants us to use the Cauchy sequence, or something like that. I don't remember exactly but I know he said Cauchy. (2 finals today, brain is fuzzy.)

The problem reads:
Let $\delta$ be the set of all Cauchy sequences $(r_i)$ where $r_i \epsilon \mathbb{Q}$ . Define the relation on $\delta$ as follows

$(r_i)\sim(q_i)\ \Leftrightarrow \lim_{i \to \infty} (r_i-q_i)=0$

Describe the equivalence classes. Show that the set of equivalence classes can be identified with the set of real numbers $\mathbb{R}$ .

Thank you, thank you, thank you.

**Bruno J.** · December 17th 2009, 07:36 PM

The equivalence relation he is talking about is defined on the set $S$ of Cauchy sequences of rational numbers, as follows; for two Cauchy sequences $\{x_n\}$ and $\{y_n\}$ , we have $\{x_n\}\sim \{y_n\}$ iff the sequence $\{x_n-y_n\}$ tends to $0$ .

See what you can do with this and let me know if you need more help!

**Jose27** · December 17th 2009, 07:41 PM

I guess you're dealing with the completion of $\mathbb{Q}$ as a metric space. You most likely were asked to define an equivalence relation like this: If $(x_n),(y_n)$ are Cauchy sequences in $\mathbb{Q}$ we say they're equivalent if $\| x_n-y_n\| \rightarrow 0$ as $n\rightarrow \infty$ . Considering the set of equivalence classes you define a distance as follows if $(x_n) \in P$ and $(y_n)\in Q$ (where $P$ and $Q$ are in said set) $d(P,Q)=d(x_n,y_n)=\lim_{n\rightarrow \infty } \| x_n - y_n\|$ (you need to prove this is well defined and that this last limit actually exists for any two Cauchy sequences).

This generates a metric space in which $\mathbb{Q}$ is dense (we see $\mathbb{Q}$ in this set via the mapping $x\mapsto (x) \in P_x$ (the class that contains the constant sequence) which is an isometry (prove all this)). Finally, this space is unique up to an isometry so by the properties of the reals you get the result.

**yvonnehr** · December 17th 2009, 08:17 PM

Originally Posted by Bruno J.

The equivalence relation he is talking about is defined on the set $S$ of Cauchy sequences of rational numbers, as follows; for two Cauchy sequences $\{x_n\}$ and $\{y_n\}$ , we have $\{x_n\}\sim \{y_n\}$ iff the sequence $\{x_n-y_n\}$ tends to $0$ .

See what you can do with this and let me know if you need more help!

Jose's reply seems way over my head. I wouldn't know how to go about it with that information. Can one of you start me off and give me an outline of how the proof would work? This is my last one...

**Jose27** · December 17th 2009, 08:35 PM

Originally Posted by yvonnehr

Jose's reply seems way over my head. I wouldn't know how to go about it with that information. Can one of you start me off and give me an outline of how the proof would work? This is my last one...

Exactly where are you stuck?

PS. It might help to think the equivalence classes as the set of Cauchy sequences that converge to the same point (this is only to help you visualize what you're trying to do, since obviously there are some Cauchy sequences that will not converge).

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equivalence classes and real numbers

need help--outline maybe

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