1. ## Subsequence

If xn converges to zero then every subsequence of xn converges to 0
I posted a proof that we did in class and I wrote *** to the place that I don't understand. Can someone explain me that part?
He suddenly said there exists k such that nk>N . According to what?

Is the logic that we say, ok let's take an increasing sequence nk, that corresponds to the elements of n.
And therefore there exists k such that nk>N
So for all k>K we have nk>nK because it's an increasing sequence and since the sequence nk that we created correspondes to the elements of n we will get |xnk|<epsilon

Is my logic correct?

Note:My analysis exam is coming up so I might ask the questions more than usual, about the things that I don't understand.Therefore I ask for your patience. I will always write my thinking process so if I have something wrong in my thinking I hope you can fix me. Thank you!

2. ## Re: Subsequence

Originally Posted by davidciprut
If xn converges to zero then every subsequence of xn converges to 0
I posted a proof that we did in class and I wrote *** to the place that I don't understand. Can someone explain me that part?
He suddenly said there exists k such that nk>N . According to what?!
If $x_n\to 0$ then almost all of the values of the sequence are "close to" 0.
Any open set $(-\delta,\delta)$ will contain all but a finite collection of the $x$'s.

Now that word finite means that given any $N$ there will be an infinite collection of the $x_n$ where $n>N$.

3. ## Re: Subsequence

Originally Posted by davidciprut
He suddenly said there exists k such that nk>N . According to what?
Actually, it should say, there exists a $K$ such that $N < n_K$.

Originally Posted by davidciprut
Is the logic that we say, ok let's take an increasing sequence nk, that corresponds to the elements of n.
The part "that corresponds to the elements of n" does not make sense to me. For one, $n$ is a variable that ranges over natural number. At most it's a fixed number, so it does not have any elements. You just consider an increasing function $k\mapsto n_k$ that takes and returns natural numbers. The original sequence is also a function $n\mapsto x_n$, but it maps natural numbers to real ones. You can write these functions as $n(k)$ and $x(n)$. Their composition $x_{n_k}=x(n(k))$ is a sequence of reals, i.e., a function that maps natural numbers to reals, and by definition it is a subsequence of $x_n=x(n)$.

Originally Posted by davidciprut
And therefore there exists k such that nk>N
Since $n(k)$ returns natural numbers and is strictly increasing, it is unbounded, so it becomes greater than any given constant such as $N$. Therefore there exists a $K$ such that $x(K)=x_K>N$.

Originally Posted by davidciprut
So for all k>K we have nk>nK because it's an increasing sequence
Yes, since the function $n(k)$ is strictly increasing.

Originally Posted by davidciprut
since the sequence nk that we created correspondes to the elements of n we will get |xnk|<epsilon
Again, I don't understand this part. As the proof says,
$$k>K\implies n_k>n_K>N\implies |x_{n_k}|<\epsilon.$$