Question

Prove that the set {n|n=1 or n in N and n-1 in N} is inductive.

Answer 1

N as in natural numbers

Answer 2

Is it not that by definition of inductive set, we need 2 things:
1) \(1\in S\)
2)\(\forall n\in S, n+1\in S\),
And we have 1 ) already,
2) is easy by inductive rule :
\[\dfrac{e_1 \in S ~~e_2\in S}{(e_1+e_2)\in S}\]
where \(e_1=1, e_2=n ~~or ~~e_2=n-1\)
Or I underestimate the problem??

Answer 3

that is a bit contradicting. I'm having a tough time because I thought 1 is excluded.

Answer 4

suppose n=1, then n-1=1-1=0 which means n-1 is not a natural number. Am I just overthinking this?

Answer 5

\(S={n|n=1 "or" n\in N "and" n-1\in N}\)
To me, it means
n=1 and n in N
or n=1 and n-1 in N
both works

Answer 6

@linyu since you post at College level, I assume that the problem is from Set theory course. Am I right?

Answer 7

Yes. It is advanced calculus one variable

Answer 8

oh, advance calculus, which book do you use? which chapter?

Answer 9

@Loser66 It's called "Advanced Calculus" 2ed by M. Fitzpatrick

Answer 10

section 1.1 problem 7

Answer 11

oh, my book is Elementary Classical Analysis by Marsden.

Answer 12

I think they teach the same techniques.

Answer 13

And I don't have inductive set

Answer 14

Good luck, my friend. It killed me when I took the course. I got a C

Answer 15

I'm new to classical analysis so..

Answer 16

@Loser66 I appreciate your help haha

Answer 17

ok, at the beginning, it was easy, but later on, it was harder and harder. I struggled with homework. I was always around 80!! hehehe...

Answer 18

@Loser66 pray for me oh god...

Answer 19

Yes, I will. hehehe... find some classmates to study together, knock on your professor's door when you need help.

Answer 20

@Loser66 Just as you said the second assumption of this question is easy. I just don't know how to bypass the first 1-1=0 is just wrong

Answer 21

Don't trust anybody else but the prof.

Answer 22

true that haha

Answer 23

The set S has more than 1 element. either it has 1 and n, or 1 and n-1

Answer 24

and at very first step, you need say: 1 in S, hence S is not empty.

Answer 25

Does it have something to do with the completeness axiom?

Answer 26

The completeness property is about monotone sequence and N is completeness set.

Answer 27

but I don't think we need it on this problem

Answer 28

@Loser66 take my class for me please *sob*

Answer 29

hehehe... good luck my friend. You can do it. trust me,