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--- math104-s22:notes:lecture_6 [2022/02/03 06:22]
pzhou created
+++ math104-s22:notes:lecture_6 [2026/02/21 14:41] (current)
@@ 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