# If $(x_{k})\to L$ and $\forall x_{i}\in (x_{k})$, $x_{i}$ is a subsequential limit of $a_{n}$ then

I want to prove that:

If $(x_{k})\to L$ and $\forall x_{i}\in (x_{k})$, $x_{i}$ is a subsequential limit of $a_{n}$ then $L$ is also a subsequential limit of $a_{n}$.

I came up with the following: Let $\epsilon\gt0$; if $(x_{k})\to L$ then we simply pick $x_{i}\in(L-\epsilon, L+\epsilon)$ and because $x_{i}$ is a subsequential limit of $a_{n}$ we know that in every neighborhood of $L$ there are infinite elements of $a_{n}$ and we conclude that $L$ is also a subsequential limit of $a_{n}$.

This seems a bit clumsy, is there a better way to show this? Perhaps with Bolzano-Weistrass?

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Please include full text in the body of your post; in particular, don't use the Title as half a sentence, and continue immediately on the body. And perhaps try to make the title informative in terms of what the question is about, rather than half a sentence? –  Arturo Magidin Nov 12 '10 at 17:54
I don't think that is clumsy at all. Except, perhaps you would want to make explicit that you are using the fact that $(L-\epsilon,L+\epsilon)$ is a neighborhood of $x_i$. –  Jonas Meyer Nov 12 '10 at 17:57
You mean taking an explicit $\epsilon_{0}$ such that $N_{{\epsilon}_{0}}(x_{i})\subseteq N_{\epsilon}(L)$, like $\epsilon_{0}=\min(|x_{i}-L|, |x_{i}-L-\epsilon|)$? –  daniel.jackson Nov 16 '10 at 19:51

I'm as mystified as Jonas Meyer on why you think this is "clumsy". It follows exactly along the intuition: I can get arbitrarily close to $L$ using the $x_i$, and I can find subsequence of $(a_i)$ that gets arbitrarily close to the $x_i$, so I can find subsequences that get arbitrarity close to things that get arbitrarily close.

But perhaps what you want is some idea of which subsequence that might be? Well, we can get it done as follows:

There is an $N_1$ such that if $k\geq N_1$, then $|x_k-L|\lt 1$. And since $x_k$ is the limit of a subsequence of $(a_n)$, there is an $n_1$ such that $|a_{n_1}-x_{N_1}|\lt 1$. In particular, $|a_{n_1}-L|\lt 2$.

Now, there is an $N_2\gt N_1$ such that for all $k\geq N_2$, $|x_k-L|\lt\frac{1}{2}$. Since $x_{N_2}$ is the limit of a subsequence of $(a_n)$, there is an $n_2$, $n_2\gt n_1$, such that $|a_{n_2}-x_{N_2}|\lt \frac{1}{2}$; in particular, $|a_{n_2}-L|\lt 1$.

Continue this way; assume that we have found $N_k$, $N_k\gt\cdots\gt N_1$ such that $|x_{N_i}-L|\lt \frac{1}{2^{i-1}}$, and $n_1\lt n_2\lt\cdots\lt n_k$ with $|x_{n_i}-x_{N_i}|\lt \frac{1}{2^{i-1}}$, so $|x_{n_i}-L|\lt \frac{1}{2^{i-2}}$.

Then there is an $N_{k+1}\gt N_k$ such that for all $j\geq N_{k+1}$, $|x_{j}-L|\lt \frac{1}{2^k}$. Since $x_{N_{k+1}}$ is the limit of a subsequence of $(a_n)$, there is an $n_{k+1}\gt n_k$ such that $|a_{n_{k+1}}-x_{N_{k+1}}|\lt \frac{1}{2^k}$, and in particular $|a_{n_{k+1}}-L|\lt \frac{1}{2^{k-1}}$.

Inductively, we get a subsequence $(a_{n_k})$ of $(a_n)$. I claim this subsequence converges to $L$. Let $\epsilon\gt 0$; find $k$ such that $0\lt \left(\frac{1}{2}\right)^{k-2}\lt \epsilon$. Then for all $\ell\geq k$ we have \begin{equation*} |a_{n_{\ell}} - L|\lt \frac{1}{2^{\ell-2}} \lt \frac{1}{2^{k-2}}\lt \epsilon. \end{equation*} Thus, the sequence converges to $L$, as claimed. QED

Personally, I don't think this is particularly "elegant", but I don't think it is clumsy either. It is exactly the intuition: get very close to $L$ using the $x_i$, then get very close to $x_i$ using some $a_j$, and this gives you an $a_j$ that is very close to $L$. Just keep doing it and you get a subsequence converging to $L$.

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