If a sequence $\{a_n\}$ converges to both $m$ and $n$, prove that $m = n$

Question

anonymous · Answer

Assume $m\not=n$. If $\{a_n\}$ converges to both of these values, then there exist $N_1$ and $N_2$ such that $m\ge N_1$ and $n\ge N_2$ imply, respectively,
$$|a_n-m|<\epsilon_1=\frac{\epsilon}{2}\quad\text{and}\quad|a_n-n|<\epsilon_2=\frac{\epsilon}{2}$$
for a given $\epsilon>0$. (The fact that $\epsilon_1=\epsilon_2$ is due to $\epsilon$ being arbitrary.)

What does this tell you?

rsadhvika · Answer

that tells $N_1$ is with in $\epsilon/2$ of$m$  and $N_2$ is with in $\epsilon/2$ of $m$ ?

rsadhvika · Answer

and sorry for mixing index variable with the limit, they are supposed to be different
shall i change the index variable to i ?

anonymous · Answer

It doesn't matter to me :P

Do you see where this proof is going? We assume there are two distinct limits, but we want to show that this leads to a contradiction.

rsadhvika · Answer

it would be contradiction for a sequence to converge to two limits, im struggling with putting steps

rsadhvika · Answer

** that tells $N_1$ is with in $\epsilon/2$ of $m$ and   $N_2$ is with in $\epsilon/2$ of $n$

anonymous · Answer

The contradiction here would be to show that if $m\not=n$, we still arrive at $m=n$.

To do this, we could consider it in terms of the distance between these "distinct" limits. Since $\epsilon>0$ can be arbitrarily small, if you can show that $|m-n|<\epsilon$, then you are finished.

rsadhvika · Answer

Okay I see  $$|a_n-m|< \frac{\epsilon}{2}\quad\text{and}\quad|a_n-n|<\frac{\epsilon}{2}$$

rsadhvika · Answer

what would be my next step ?

anonymous · Answer

How can you combine the absolute value terms?

anonymous · Answer

If we add corresponding sides together, we'll arrive at something close to what we want:
$$|a_n-m|+|a_n-n|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$$

rsadhvika · Answer

the indexes are different right ?

rsadhvika · Answer

i mean the first absolute value inequality holds only when n > N1
and the second absolute value inrquality holds only when n > N2

anonymous · Answer

Sure, but if we only consider the greatest of the two, i.e. $N=\max\{N_1,N_2\}$, then we don't have to worry about that.

rsadhvika · Answer

Ohk that makes sense :  n >  max(N1, N2)

rsadhvika · Answer

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