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--- math104-s22:notes:lecture_5 [2022/02/01 00:41]
pzhou created
+++ math104-s22:notes:lecture_5 [2026/02/21 14:41] (current)
@@ Line 18: / Line 18: @@
 ** Lemma **: Let $(a_n)$ be a bounded sequence. $(a_n)$ is convergent if and only if $\limsup a_n = \liminf a_n$. \\
-Pf: Let $A_n = \sup_{m \geq n} a_m, B_n = \inf_{m \geq n} a_m$, then $A_n \geq a_n \geq B_n$, we know $A_n, B_n, a_n$ all convergent, hence
+Pf: Assume $\limsup a_n = \liminf a_n = a$.  Let $A_n = \sup_{m \geq n} a_m, B_n = \inf_{m \geq n} a_m$, then $A_n \geq a_n \geq B_n$. By preview lemma, we know $a_n \to a$.
-$$ \lim A_n \geq \lim a_n \geq \lim B_n. $$
-Suppose $(a_n)$ is convergent to $a$, then
+Assume $(a_n)$ is convergent to $a$. Then, for any $\epsilon > 0$, there is an $N>0$, that for all $n > N$, $|a_n - a|<\epsilon$. In particular, we know $\limsup a_n \leq a+\epsilon$. Since this is true for all $\epsilon>0$, we have $\limsup a_n \leq a$. Similarly, $\liminf a_n \geq a$. On the other hand, since $A_n \geq B_n$, we have $\limsup a_n = \lim A_n \geq \lim B_n = \liminf a_n$. Hence, $\limsup a_n = \liminf a_n = a$.
+** Lemma: ** Cauchy sequence is bounded. (Exercise)
+** Lemma: ** If $\limsup a_n = A$, then for any $\epsilon > 0$, and any $N > 0$, there exists $n>N$ with $|a_n - A| <  \epsilon$. Similarly, if $\limsup a_n = B$, then for any $\epsilon > 0$, and any $N > 0$, there exists $n>N$ with $|a_n - B| <  \epsilon$.\\
+Pf: Since $A_n = \sup_{m>n} a_m$ is a monotone decreasing sequence with limit $A$, then for any $\epsilon>0$ and $N>0$, we can find $M > N$ such that  $A+\epsilon/2 > A_{M} \geq A$. By definition of $A_{M}$, there exists some $a_n$ with $n \geq M$, that $A_{M} \geq a_n > A_{M} - \epsilon/2$. Now, we have
+$$ |a_n - A| \leq |a_n - A_M| + |A_M - A| \leq \epsilon/2 + \epsilon/2 = \epsilon $$
+The result is proven.
+** Theorem **: Cauchy sequence in $\R$ is convergent. \\
+Proof: Let $(a_n)$ be a Cauchy sequence. By previous lemma, it is bounded. Let $A = \limsup a_n$ and $B = \liminf a_n$, we only need to show that $A = B$ to show $\lim a_n$ exists. We know $A \geq B$ for all bounded sequence $a_n$, suppose $A > B$, and let $\epsilon = (A-B)/3$. Then, by Cauchyness of $(a_n)$, we there is an $N>0$, such that for all $n,m> N$, we have $|a_n - a_m| < \epsilon$. By previous lemma, there exists $n>N$, with $|a_n - A|<\epsilon$, and $m>N$ with $|a_m - B| < \epsilon$. Hence
+$$ |A - B| \leq |A-a_n|+|a_n - a_m| + |a_m - B| < 3 \epsilon = |A-B|, $$
+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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