Write a proof for each of the following (a) and (b) a.) (-1/n,1/n) for n=1 and g

Question

Write a proof for each of the following (a) and (b)

a.) (-1/n,1/n) for n=1 and goes to infinity

b.) (-n,n) for n=1 and goes to infinity.

I planned on proving by contradiction. For (a) 1/n Iclaim A(sub)n=0 for n=1 and goes to infinity. By assuming there is an x>0 such that x is an element of A(sub)n for n=1 and goes to infinity. Then x is an element [0,1/n] for all n. Choose N to be any natural number such that N>1/x. This means x>1/N. For N, we have 0<1/N<x. In other words, x is not an element [0,1/N] = A(sub)n. This is a contradiction. Hence A(sub)n = {0} for n=1 and goes to infinity.

I am not sure if this is the same logic I need to follow fornegative 1/nand how to apply the same type of proof to just (-n,n). I also am wanting to verify that I will use all open brackets since the first set is 0,0 and the second is -infinity, infinity.

Thanks

Explanation / Answer

Write a proof for each of the following (a) and (b)

a.) (-1/n,1/n) for n=1 and goes to infinity

b.) (-n,n) for n=1 and goes to infinity.

I planned on proving by contradiction. For (a) 1/n Iclaim A(sub)n=0 for n=1 and goes to infinity. By assuming there is an x>0 such that x is an element of A(sub)n for n=1 and goes to infinity. Then x is an element [0,1/n] for all n. Choose N to be any natural number such that N>1/x. This means x>1/N. For N, we have 0<1/N<x. In other words, x is not an element [0,1/N] = A(sub)n. This is a contradiction. Hence A(sub)n = {0} for n=1 and goes to infinity.

This is a good proof. Nice job.

I am not sure if this is the same logic I need to follow fornegative 1/nand how to apply the same type of proof to just (-n,n). I also am wanting to verify that I will use all open brackets since the first set is 0,0 and the second is -infinity, infinity. Yes, open brackets.

Thanks No problem :)

Since you have a good hang on (a), try replacing each use of 1/x, 1/n, and 1/N with simply x, n, and N. That's it - your proof will be almost exactly identical, as will the contradiction you'll reach.

I'd walk you through it completely, but it honestly seems like you have a very good grasp on this as it is. Good luck! :)

I hope that you found this answer useful towards your studies. It took a considerable amount of thought, time, and effort to compose, and and I'd sincerely appreciate a lifesaver rating! It would really make my day, and will allow me to continue answering your questions :)

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