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--- math104-s22:notes:lecture_6 [2022/02/02 22:22]
pzhou created
+++ math104-s22:notes:lecture_6 [2022/02/02 22:29] (current)
pzhou
@@ Line 7: / Line 7: @@
   * Every bounded sequence has convergent subsequence. (just take a monotone sequence, then it will be convergent)
   *  $\limsup$  and  $\liminf$  can be realized as subseq limits.
-  * Let  $(s_n)$  be a seq, and  $S$  denote the set of all subseq limits. Then,  $S$  is non-empty.  $\sup S = \limsup s_n$  and  $\inf S = \liminf s_n$ .
+  * Let  $(s_n)$  be a seq, and  $S$  denote the set of all subseq limits. Then,  $S$  is non-empty.  $\sup S = \limsup s_n$  and $\inf S = \liminf s_n $.$ lim s_n $exists if and only if$ S$ contains only one element. (All these are just summary of previously proven results)
+  *  $S$  is closed under taking limits. (i.e.  $S$  is a closed set)
+    * Proof of this is fun. Suppose one has a sequence of  $t_n$  in  $S$ , and  $t_n \to t$ . Can we show that  $t$  is in  $S$  as well? Well, we need to show that for any $\epsilon>0$, there are infinitely many  $s_n$  in $(t-\epsilon, t+\epsilon $. First, we find a$ t_n $, such that$ |t_n - t| < \epsilon / 2 $, then we know there are infinitely many$ s_m $with$ |t_n - s_m| < \epsilon/2$, thus these same set of  $s_m$  will be  $\epsilon$ -close to $t$.
+Discussion time: Ross 11.2, 11.3, 11.5.

Lecture Notes