Question

Prompt: "Prove the set N - N15 is infinite."
Where N is the set of Natural Numbers and the 15 is subscript. I have figured out how to prove N is infinite, but I am unsure how to approach this one. Any tips or assistance would be appreciated, thank you!

Answer 1

\(N_{15}\) ?

Answer 2

Aye, it's N minus that.

Answer 3

oh so it's given that n_15 is finite group is it ?

Answer 4

Aye, it is. The prompt and nothing else on the page mentions having to prove that that part was finite, only that the difference was infinite.

Answer 5

Though I still wondering if I need to use the Lemma that "for every K element N, every subset of N_k is finite" in it though. As said, just not sure how to approach this one. Proofs are not my strong suit aha.

Answer 6

well i need to know what N_k stands for, as i felt earlier N_15 is finite so that lemma helps indeed!


@ganeshie8 any thoughts ?

Answer 7

basically you want to show that removing finite number of elements from an infinite set doesn't change the cardinality

Answer 8

yes @ganeshie8 but what is N_k ?
i mean without that lemma we could never know that N_15 is infinite right ?
we need more info about its definition

Answer 9

I think \(N_{15}\) is a finite set whose elements are first \(15\) natural numbers

Answer 10

\[N_{15}=\{1,2,3,\ldots,15\}\]

@Hlares please confirm if that is correct

Answer 11

yeah that make sense !

Answer 12

Aye, that is correct.

Answer 13

prove \(\mathbb{N}-N_{15}\) is infinite

Answer 14

And thank you. For proving that it will not change the cardinality of N, should I show that N_15 is a subset of N at all?

Answer 15

would you agree that union of two finite sets is finite ?

Answer 16

if you do, then you may try proof by contradiction

Answer 17

(proof by contradiction)


Suppose \(S=\mathbb{N}-N_{15}\) is finite.
Since union of two finite sets is finite, \(S\cup N_{15}\) must be finite.
But \(S\cup N_{15}=\mathbb{N}\) is infinite. Contradiction.

Answer 18

that proof depends on below facts :
1) \(\mathbb{N}\) is infinite
2) union of two finite sets is finite

Answer 19

Ohhh, that is a lot better way to approach it (and easier than the hodgepodge I was thinking of). Aye, I could write out a proof by contradiction like that then to prove it and dig up the appropriate theorems and lemmas in the book. Thank you so much!

Answer 20

may i see your proof for \(\mathbb{N}\) infinite ?

Answer 21

if you have it online already..

Answer 22

You can, just give me a moment to type it up. All of my stuff is written out by hand.

Answer 23

if it is easy please take a screenshot and attach
otherwise it is okay...
i vaguely remember a proof using induction but not so sure of the exact method...

Answer 24

Proof: Suppose N is finite. Then, since N =/= (empty set), there exists k such that N is equivalent to N_k. Thus, as N is equivalent to N_k, there exists a one-to-one correspondence and thus function, f, from N_k onto N. Then let n= {f(1), f(2), f(3),..,f(n), f(n+1)}. Then n =/= f(i) for any i element N_k. Thus, f is not onto N. Therefore, N is an infinite set.

Answer 25

Whoop, forgot the first line (before the "Suppose..." one): This is a proof by contradiction.

Answer 26

Looks good !

Answer 27

thank you :)

Answer 28

Thanks! Took me longer to work on it than it probably should have, but, as said, proofs are not my strong suit, so I'm just happy when I can get them done ahah.

Answer 29

https://youtu.be/mciBPGCvpBk?list=PL04BA7A9EB907EDAF&t=1824

Answer 30

try watching it between 30th minute and 40th minute
it has a similar proof...

Answer 31

I'll watch that right now then. Thank you for finding it!

Answer 32

his \(J_n\) is same as your \(N_k\)

Answer 33

here another thought.
if N is finite then for some k, N=N_k, but there exist m \in N such that
m=k+1 which is a contradiction.