# Thread: Help with Intro to Real Analysis proof with sequence and limit.

1. ## Help with Intro to Real Analysis proof with sequence and limit.

Problem states:
Let be a sequence in with = a.
If a > 0, prove (the set of Natural numbers) such that .

My work so far:
I don't have a proof written up for this problem yet, but I have jotted down some ideas that I think might help me:

I know that the sequence { $a_n$} converges to a and a > 0.
So by definition of convergence, I know that there exists an $n_0$ in the Naturals such that
| $a_n$ - a| < ε for all n ≥ $n_0$.

There is also the theorem that every convergent sequence is bounded, so by definition of bounded: there exists an M > 0 such that
| $a_n$| ≤ M for all n in the Naturals.

So for this proof do I need cases? Would I do proof by contradiction?
I worked with a study group on this, and my classmate drew a picture and talked us through the thinking. It kind of made sense to me that if a sequence converges to zero, then eventually the sequence has to be greater than zero at some point.

Any help, suggestions, and/or tips are greatly appreciated.
Thank you for your time! =)

2. From the fact that $\left( {a_n } \right) \to a>0$ you use the definition of sequence convergence.
But in place of just $\varepsilon >0$ you are free to use $\frac{a}{2} >0$.
$\begin{gathered} \left| {a_n - a} \right| < \frac{a}{2} \hfill \\
- \frac{a}{2} < a_n - a < \frac{a}{2} \hfill \\ 0 < \frac{a}{2} < a_n < \frac{{3a}}{2} \hfill \\ \end{gathered}$

3. Okay, so can I just choose any value for epsilon? Like could I have chosen a/3 or a/7 > 0?

Here's my proof so far:

QED.

is that okay?

4. You could do this proof by contradiction. Assume that there does not exist a $N$ such that $N\leqslant n\implies a_n>0$

By definition of $\lim_{n\to\infty}a_n=a$ we must have that for every $\varepsilon>0$ there exists a corresponding $N$ such that $N\leqslant n\implies|a_n-a|<\varepsilon$. So if we take $\varepsilon=\frac{a}{2}$ we must be able to find a $N$ such that $|a_n-a|<\frac{a}{2}$, but now by assumption $a\leqslant|a_n-a|$ which gives a contradiction.

5. Originally Posted by Mathstud28
You could do this proof by contradiction. Assume that there does not exist a $N$ such that $N\leqslant n\implies a_n>0$

By definition of $\lim_{n\to\infty}a_n=a$ we must have that for every $\varepsilon>0$ there exists a corresponding $N$ such that $N\leqslant n\implies|a_n-a|<\varepsilon$. So if we take $\varepsilon=\frac{a}{2}$ we must be able to find a $N$ such that $|a_n-a|<\frac{a}{2}$, but now by assumption $a\leqslant|a_n-a|$ which gives a contradiction.
Confused, it can be done this way (though I can't see why one would want to) and this is probably OK when all the missing detail is put in but in this form it is near incomprehensible (and the last $n$ is not the same as the $n$ introduced earlier).

CB

6. Originally Posted by CaptainBlack
Confused, it can be done this way (though I can't see why one would want to) and this is probably OK when all the missing detail is put in but in this form it is near incomprehensible (and the last $n$ is not the same as the $n$ introduced earlier).

CB
Yes you are right, that is unclear. Let me fix it up.

Suppositions:
1. $a>0$
2. $\lim a_n=a$
3. There exists a $N$ such that $N\leqslant n\implies a_n\leqslant 0$

A fact we will use, if $x>0$ and $y<0$ then $x\leqslant|x-y|$

Now by the definition of a sequence $\left\{a_n\right\}$ converging to $a$ we must have that for every $\varepsilon>0$ there exists a corresponding $N_1$ such that $N_1\leqslant n\implies |a-a_n|<\frac{a}{2}$. So now let $\varepsilon=\frac{a}{2}$, so we must be able to find a $N_1$ such that $N_1. So it must be true that if we take $N_2=\max\left\{N,N_1\right\}$ that $N_2\leqslant n\implies |a_n-a|<\frac{a}{2}$. But here we arrive at a contradiction. The reason being that because of supposition three we must have that $N_2\leqslant n\implies |a-a_n|\geqslant a\implies |a_n-a|\not<\frac{a}{2}$