Question

1) (Limits and vector space operations) Let V be a vector space with a norm $|| \cdot ||$. Suppose that $(a_n)$ and $(b_n)$ are convergent sequences in V with limits a, b respectively. Prove the following statements: a) For every $k \in \mathbb{R}$ we have $\lim_{n \to \infty} (ka_n) = ka$. b) We have $\lim_{n \to \infty} (a_n + b_n) = a + b$. 

          1) (Limits and vector space operations) Let V be a vector space with a norm $|| \cdot ||$.
Suppose that $(a_n)$ and $(b_n)$ are convergent sequences in V with limits a, b
respectively. Prove the following statements:
a) For every $k \in \mathbb{R}$ we have $\lim_{n \to \infty} (ka_n) = ka$.
b) We have $\lim_{n \to \infty} (a_n + b_n) = a + b$.
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1) (Limits and vector space operations) Let V be a vector space with a norm || · ||. Suppose that (an) and (bn) are convergent sequences in V with limits a, b respectively. Prove the following statements: a) For every k ∈ℝ we have limn →∞ (kan) = ka. b) We have limn →∞ (an + bn) = a + b.

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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(Limits and vector space operations) Let V be a vector space with a norm || -- ||. Suppose that (an) and (bn) are convergent sequences in V with limits a, b respectively. Prove the following statements: a) For every k ∈ R we have limn→∞ (kan) = ka. b) We have limn→∞ (an + bn) = a + b.
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Transcript

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00:01 So here in this question we have to prove that limit this is basically the party of the question limit superior of a of n plus b of n is less than and equals to the limit sup of a n plus limit sup of b n.
00:20 So basically we have to prove this inequality.
00:23 So to prove that we are considering a n and b n are both bounded sequences.
00:37 So we can say that if a n and b n are separately bounded so we can say that a n plus b n 2 is a bounded sequence.
00:51 So here we are considering let limit a of n is equals to l1 and let limit b of n is equals to l2.
01:01 So we can say that limit of a of n plus b of n is equals to p where limit sup of a of n is equals to limit of a n.
01:15 That is we do not limit sup in another way or we can say that here we can choose epsilon where epsilon is greater than 0 since limit of a of n is equals to l1 such that a is a natural number natural number given and we can say that a1 is less than l1 plus epsilon divided by the 2 where n is greater than n equals to given and next we are considering about the limit of b of n that is equals to l2 such that a is a natural number where we are considering a natural number that from here is k2 such that b of n is less than l2 plus epsilon that is divided by 2 where n is greater than n equals to k2.
02:11 Now after that we are considering about the k that is equals to maximum of k1 and k2 then we are considering about a of n which is less than l1 plus epsilon divided by 2 and after that we are considering about the bn which is less than l2 plus epsilon divided by 2 where n is greater than n equals to k.
02:37 So we can say that a of n plus b of n 2 is less than l1 plus l2 plus epsilon where n is greater than n equals to k.
02:46 So it follows that no subsequent limit of the sequence that is a of n plus b of n can be greater than l1 plus l2 plus epsilon.
03:01 So here we are considering about since epsilon is greater than 0 and this is basically a arbitrary.
03:11 So every subsequent limit of the sequence that is a of n plus b of n is less than or equals to l1 plus l2.
03:22 Therefore p is less than n equals to l1 plus l2.
03:26 So we can say that that the limit of a of n plus b of n is less than and equals to limit of a of n plus limit of bn...
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