- #1

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**Problem:**If sequence ## (a_n) ## has ##10-10## as partial limits and in addition ##\forall n \in \mathbb{N}.|a_{n+1} − a_{n} |≤ \frac{1}{n} ##, then 0 is a partial limit of ## (a_n) ##.

**Proof :**Suppose that ## 0 ## isn't a partial limit of ## (a_n) ##. Then there exists ## \epsilon_0 > 0 ## and ## N \in \mathbb{N} ## s.t. ## \forall n \geq N. | a_n - 0 | \geq \epsilon_0 ## ( the defintion for a partial limit ## L ## of some sequence sequence ## (b_n) ## is ## \forall \epsilon>0. \forall N \in \mathbb{N} .\exists n \geq N. | b_n - L | < \epsilon ##, so we just used the negation of it in our proof ).

Hence ## \forall n \geq N. a_n \leq -\epsilon_0 \lor a_n \geq \epsilon_0 ##. There exists ## N^* \in \mathbb{N}## s.t. ## N^* \geq max \{ N,\frac{1}{2\epsilon_0} \} ##.

Instantiatng ## N^* ## we get that ## a_{N^*} \leq -\epsilon_0 \lor a_{N^*} \geq \epsilon_0 ##. Suppose without lose of generality that ## a_{N^*} \leq -\epsilon_0 ##.

We define the set ## A = \{ n \in \mathbb{N} : n>N^* , a_n \geq \epsilon_0 \} ##, since ## +10 ## is a partial limit of ## (a_n) ## then ##A \neq \emptyset## , Hence ## A ## has a minimum ( since ## A ## is a set of naturals, it has minimum [ we use well-ordering principle here ] ) which we designate as ## M = min A ##.

Hence ## a_M \geq \epsilon_0 ## and ## a_{M-1} \leq -\epsilon_0 ##, Hence ## a_M - a_{M-1} \geq 2 \epsilon_0 > \frac{1}{N^*} \geq \frac{1}{M-1} ## ( Note: since we have ## M > N^* ## then ## M -1 \geq N^* ## hence ## \frac{1}{N^*} \geq \frac{1}{M -1} ## ), so ## |a_M - a_{M-1} | \geq 2 \epsilon_0 > \frac{1}{N^*} \geq \frac{1}{M-1} ##. But after instantiation ( of ## M-1 ## in ##\forall n \in \mathbb{N}.|a_{n+1} − a_{n} |≤ \frac{1}{n} ## ) we get ## |a_M - a_{M-1} | \leq \frac{1}{M-1} ## which is a contradiction.

**My question:**I didn't fully understand why the set ## A = \{ n \in \mathbb{N} : n>N^* , a_n \geq \epsilon_0 \} ## is non-empty, how does the fact that ## +10 ## is a partial limit help me understand this? ( since ##+10 ## is a partial limit we have ## \forall \epsilon>0. \forall N \in \mathbb{N} .\exists n \geq N. | a_n - 10 | < \epsilon ## , but I have no idea how instantiating here with ## \epsilon_0 ## or ## N^* ## helps me)