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Let $K$ be an ordered field. Then $K$ contains a unique copy of the ordered field $\mathbb{Q}$. Also, we can define the absolute value $|\cdot|$ and sequence convergence in exactly the same way as in $\mathbb{R}$. In particular, a sequence $(a_n)\subset K$ is said to converge to $l\in K$ if $$ \forall\epsilon\in K(\epsilon>0\to\exists N\in\mathbb{N}(\forall n\in\mathbb{N}(n\geq N\to|a_n-l|<\epsilon))) $$

Question: In general, will $l$ be unique? If not, what is wrong with the usual proof given below?

I'm asking since it was remarked here that it will not in general. The example of the hyperreals in which every other point infinitesimally close to $l$ will also be a limit of $(a_n)$ was given.

Proof (possibly flawed): Suppose $l,l'$ are two limits of $(a_n)$. WLOG, suppose that $l'<l$. Let $\epsilon:=\frac{l'-l}{2}>0$, so that there exists $N_1\in\mathbb{N}$ such that if $n\geq n_1$ then $$ |a_n-l|<\frac{l'-l}{2} $$ Also, there exists $N_2\in\mathbb{N}$ such that if $n\geq N_2$ then $$ |a_n-l'|<\frac{l'-l}{2} $$ If $N:=\max\{N_1,N_2\}$, then $$ a_N-l<\frac{l'-l}{2}\text{ and }-\frac{l'-l}{2}<a_N-l' $$ so $$ \frac{l'+l}{2}<a_N<\frac{l'+l}{2} $$ which is impossible.

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The proof of uniqueness works fine. The problem (if one thinks of it as a problem) is that for example the sequence $(1/n)$ need not have a limit.

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  • $\begingroup$ Then should I conclude that the affirmation "In [the ordered field of the hyperreals] the limit of a Cauchy sequence is not unique. It converges to every point which is infinitesimally close to any other limit point." is false? $\endgroup$ – Guest Dec 18 '15 at 5:55
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    $\begingroup$ I have only checked the correctness of your proof, given your definition of limit. If we change the definition, by saying that for every positive integer $k$, there is an $N$ such that if $n\gt N$ then $|a_n-l|\lt 1/k$, then it is true that limit need not be unique. But you asked instead that for every $\epsilon\gt 0$, $\dots$. That is a much stronger condition. As to Cauchy sequences, if a Cauchy sequence has a limit (using your definition of limit), yes, it will be unique. $\endgroup$ – André Nicolas Dec 18 '15 at 6:11
  • $\begingroup$ @Guest In hyperreals the sequence $1/n$ is not Cauchy - for any infinitesimal $\varepsilon$, we always have $|1/n-1/m|>\varepsilon$. $\endgroup$ – Wojowu Dec 18 '15 at 6:22

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