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We can construct reals by defining a real number as a equivalence class of rational Cauchy sequences. The set of all such equivalence classes have a cardinality strictly larger than $\mathbb Q$.

Now consider, in the same fashion, Cauchy sequences of real numbers. Consider two such sequences $\left\{a_k\right\}_{k=1}^\infty$ and $\left\{b_k\right\}_{k=1}^\infty$. Define an equivalence relation, in the same way, by $$(a\sim b) \iff \displaystyle\lim_{k\to\infty} \left|a_k - b_k\right| =0$$

By similar proofs as in the rational case, this relation is really an equivalence relation. Now, consider the set of all equivalence classes with respect to the relation $\sim$ defined above. Will this give us a "new type of numbers" or will we just get $\mathbb R$ all over again?

If the cardinality of the set of all equivalence classes of real Cauchy sequences is $|\mathbb R|$, how would one go on and prove it?

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  • $\begingroup$ A metric space is said to be complete if every Cauchy sequence in that space converges to a point in that space. The real numbers are complete, hence every Cauchy sequence of real numbers converges to a real number. To see this, recall that if you have a sequence of real numbers, each of those can be represented by a series of Cauchy sequences of rational numbers. The limiting object can then be constructed by extracting an appropriate sequence of rational numbers from this sequence of sequences. $\endgroup$ Commented Apr 8, 2018 at 17:33
  • $\begingroup$ We get $\Bbb R.$ The axiomatic definition of $\Bbb R$ implies that a Cauchy sequence of members of $\Bbb R$ converges to a member of $\Bbb R$. If you are going to study sequences, limits, and calculus, you need to be familiar with the logical foundation of $\Bbb R$ and its basic consequences. $\endgroup$ Commented Apr 9, 2018 at 3:40

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Each Cauchy sequence $(a_n)$ of reals tends to a real limit $a^*$. Under your definition $(a_n)\sim (b_n)$ iff $a^*=b^*$. So, you'll just get the real numbers again.

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    $\begingroup$ So the bijection would simply be between the equivalence class and the limit of the sequences in that equivalence class? $\endgroup$ Commented Apr 8, 2018 at 17:35
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    $\begingroup$ Indeed, it would. @MarkusKlyver $\endgroup$ Commented Apr 8, 2018 at 17:36

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