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--- math104-s22:notes:lecture_5 [2022/01/31 21:14]
pzhou
+++ math104-s22:notes:lecture_5 [2022/01/31 21:37] (current)
pzhou
@@ Line 34: / Line 34: @@
 notice the inequality is strict, hence we have a contradiction. Thus $A = B$.
+===== Subsequence and Subsequntial Limit =====
+If $n_1 < n_2 < \cdots $ is a strictly increasing sequence in $\N$, and $(a_n)$ is a sequence, then $(a_{n_k})$ is a subsequence of $(a_n)$.
+In this section, we can ask, even if $(a_n)$ itself does not converge to some $a \in \R$, can we cherry-picking some nice subsequence in $(a_n)$ that does converge.
+Example:
+  * $a_n = (-1)^n + 1/n$ has a convergent subsequences, that converges to $+1$; and another subseq convergent to $-1$.
+**Definition**: Let $(a_n)$ be a sequence in $\R$, if $a \in \R$ is the limit of a sequence of $(a_n)$, we say $a$ is a **subsequential limit**.
+** Lemma **: $a$ is a subsequential limit of $(a_n)$ $\Leftrightarrow$ for any $\epsilon>0, N>0$, there exists an $n>N$, with $|a_n - a| < \epsilon$. Equivalently, for any $\epsilon>0$, the set $A_\epsilon = \{n \mid |a_n - a| < \epsilon \}$ is infinite. \\
+Pf: $\Rightarrow$ : let $a_{n_k}$ be a subseq that converges to $n$, then we can find among member within this subsequence.
+$ \Lefgarrow$: it's a good opportunity to introduce the Cantor's diagonal trick. For any positive integer $k$, we know $A_{1/k}$ is infinite, we write it in the $k$-th row, as
+$$ A_{1/k} = n_{k,1} < n_{k,2}< \cdots $$
+As $k=1,2,\cdots$, we have an semi-infinite matrix of indices. We may check that $n_{k, i } \leq n_{k+1, i}$. Thus $n_{k,k} \leq n_{k+1, k} < n_{k+1, k+1}$, hence along the diagonal, we have **strictly** increasing sequence. Consider the subsequence $a_{n_{kk}}$, that will converge to $a$.

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