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$Problem:$ Every convergent sequence has a unique limit.
$Proof:$ Let's assume that a sequence $\langle a_n\rangle$ converges to two distinct limits $l$ and $l^{'}$.
Let's assume $\epsilon=\frac{1}{2} |l-l^{'}|$.Since, $l\neq l^{'}$,$|l-l^{'}|>0$ so that $\epsilon>0$.
Given $\epsilon>0$, $\exists$ a $+\text{ive}$ integer $m_1$ such that $|a_n-l|<\frac{\epsilon}{2}$ for every $n\geq m_1$.
Similarly, for second limit $l^{'}$
Given $\epsilon>0$, $\exists$ a $+\text{ive}$ integer $m_2$ such that $|a_n-l^{'}|<\frac{\epsilon}{2}$ for every $n\geq m_2$.
Let $m=\max(m_1,m_2)$
Now, $|l-l^{'}|=|(l-a_n)+(a_n-l^{'})|\leq |l-a_n|+|a_n-l^{'}|$
$|l-l^{'}|<\epsilon$ for every $n \geq m$ which contradicts the assumption above. Hence $l = l^{'}$.

My question here is: a) If the limits are different than how can we assume that $\epsilon$ is same for both the limits. Infact, if $l^{'}\gg l$ than $\epsilon^{'}\ll\epsilon$ as n gets larger and larger.
b) How this method to prove theorem is appropriate, if I tend to use $\epsilon$ and $\epsilon^{'}$ rather than common $\epsilon$ at both places.
c) Can I expect to prove it by working out from $x$-axis rather than working on it with the assumption for $\epsilon$ from $y$-axis.
d) In case of $\epsilon=\frac{1}{2} |l-l^{'}|$, what is the $y$ here for which $y+\epsilon$ and $y-\epsilon$ can be considered.

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I would go accept some answers to your other question if you want some answers – Jean-Sébastien Oct 27 '12 at 17:56
6 questions, zero accepted answers. Don't you like the answers you get here? – DonAntonio Oct 27 '12 at 17:57
Also, where are you working? In general topological spaces this result isn't necessarily true. – AsinglePANCAKE Oct 27 '12 at 17:59
i just had this theorem in front of me and I thought of breaking it in parts and tried finding some alternatives.Couldn't find solutions but found questions ! – Rorschach Oct 27 '12 at 18:01
sorry I didn't know that accepting answers matters too..!! – Rorschach Oct 27 '12 at 18:06

There are really two important ideas working in this proof: The first is that we can use $\mathbf{any}$ $\epsilon$ we want by the definition of convergence. The second is the triangle inequality. To see exactly how we take advantage of both of these I'll write the proof slightly differently (no longer by contradiction).

Fix $\mathbf{any}$ $\epsilon >0$ and by the definition of convergence we can find $N$ and $M$ large so that $$\left|l - a_n\right| < \epsilon \text{ for any } n \geq N;$$ $$ \left| l' - a_m\right| < \epsilon \text{ for any } m\geq M.$$So now fix a specific $k\geq \max(N,M)$ then, since $k$ is fixed, we can use the triangle inequality $$\left|l - l'\right| \leq \left| l - a_k \right| + \left| a_k - l'\right| \leq \epsilon + \epsilon = 2 \epsilon.$$

And since we chose $\epsilon > 0$ arbitrary, this means that $\left|l - l'\right| = 0$ and so $l = l'$.

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i was not asking for alternatives..I can find that myself my friend ! – Rorschach Oct 30 '12 at 6:07
@rafiki for you proof: (a) $\epsilon$ is what we control ourself, we chose it so we need not worry about it varying. (b) the proof method you have is using a very specific $\epsilon$, you won't want to use $\epsilon$ and $\epsilon'$ also for the reason in (a). For (c) and (d) it's unclear to me what you mean by $x$-axis or $y$. – Deven Ware Oct 30 '12 at 6:14
if I see this closely this proof is saying that if limit of an infinite sequence is calculated than it's limit will be same even if it will be calculated from two different assumed ends somewhere close to infinity. What I am assuming here is that for the same end can I have two limits. Am I right ? – Rorschach Oct 30 '12 at 9:23
It's probably worth saying that this is identically the proof in baby Rudin. – AnonymousCoward Jan 17 at 1:03

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