0
$\begingroup$

I need some help with this proof. A subsequential limit is a limit of a subsequence.

Suppose $a_n$ is a sequence with $\{ L_1, L_2, ...\}$ subsequential limits. Suppose $L_n \to L$. Prove that $L$ is a subsequential limit of $a_n$.

Proof:

We know (1): $\forall_{\epsilon > 0} \exists_{N_0} s.t \forall_{n>N_0} \implies |L_n - L| < \epsilon$

(2) For i = 1 to ... $\forall_{\epsilon_i > 0} \exists_{N_i} s.t \forall_{n_i > N_i} \implies |a_{n_i} - L_i| < \epsilon$

  1. Can I just take my subsequence terms from the interval $|L_n - L| < \epsilon$ and then conclude by the definition of the limit that $L$ is subsequential limit?
  2. Other than that I am unsure of a general strategy of how to construct proofs with subsequences
$\endgroup$
3
  • $\begingroup$ What is a subsequential limit? Can you rephrase in terms of limits? For example are you asking: Suppose a_n has limit L and L_n is a subsequence of a_n with limit L, and for each n x_n,j is a sequence with limit L_n, can we conclude that the sequence x_n,j approaches L? If so use triangle inequality. $\endgroup$
    – coffeemath
    Oct 13, 2012 at 23:51
  • $\begingroup$ A subsequential limit is the limit of a subsequence. $\endgroup$ Oct 13, 2012 at 23:54
  • $\begingroup$ So $L_n$ is the sequence of the subsequential limits $\endgroup$ Oct 13, 2012 at 23:55

3 Answers 3

2
$\begingroup$

You can't use the sequence of $L_n$ as your sequence of $a$'s which converge to $L$, because the $L_n$ would typically not be terms in the a sequence. But you can take any $\varepsilon>0$ and first find an $L_n$ within $\varepsilon/2$ of $L$, and after that look at your sequence of $a$'s which approach $L_n$ (because $L_n$ is a subsequential limit of the $a$'s), and select some $a_n$ in the sequence of $a$'s approaching $L_n$ which is within $\varepsilon/2$ of $L_n$.

Now you can use that $a_n-L=(a_n-L_n)+(L_n-L)$, apply the triangle inequality, and get

$$|a_n-L|\le|a_n-L_n|+|L_n-L|<\varepsilon/2+\varepsilon/2=\varepsilon.$$

This gives a sequence of $a_n$'s which approaches $L$, so in the subsequential limit terminology, we can say that $L$ is a subsequential limit of the sequence $a_n$.

$\endgroup$
0
0
$\begingroup$

For each $n\geqslant1$, there exists a sequence $(a_{\varphi_n(k)})_{k\geqslant1}$ such that $a_{\varphi_n(k)}\to L_n$ when $k\to\infty$. Choose $k_n$ such that $|a_{\varphi_n(k_n)}-L_n|\leqslant1/n^2$. For every $n\geqslant1$, let $\psi(n)=\varphi_n(k_n)$. Then $(a_{\psi(n)})_{n\geqslant1}$ is a subsequence of $(a_n)_{n\geqslant1}$ and $a_{\psi(n)}\to L$ when $n\to\infty$.

$\endgroup$
0
$\begingroup$

Hint: Let $\epsilon>0$. You want to find $k\in \mathbb N$ such that $a_k \in (L-\epsilon,L+\epsilon)$. You know that exist $n\in \mathbb N$ s.t. $L_n \in (L-\frac{\epsilon}{2},L+\frac{\epsilon}{2})$ and $m\in \mathbb N$ s.t. $a_m \in (L_n-\frac{\epsilon}{2},L_n+\frac{\epsilon}{2})$. Try $k=m$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .