Closed/Open Sets and Natural Numbers

[emergentecon]

Homework Statement

I am trying to understand why ℕ the set of natural numbers is considered a Closed Set.

2. Relevant definition

A Set S in R^m is closed iff its complement, S^c = R^m - S is open.

The Attempt at a Solution

I believe I understand why it is not an Open Set:
Given that it includes 0 as a boundary point, it cannot be an open set.

As to being Closed. What I am thinking is this:
S^c = R^m - ℕ is open because in so doing, we have removed the only boundary point of 0
Therefore, per definition, ℕ or in this case, S is closed.

Am I close?

 

[Orodruin]

Yes, although you probably want to use ##\mathbb R## rather than ##\mathbb R^m## to embed your natural numbers into.

How deeply you wish to answer this question also depends on the level you are at and what topology you put on ##\mathbb R##. (Disregard the last part if you are not taking a topology class in which case the standard topology is probably inferred)

 

[emergentecon]

Ok perfect thank you very much!
Not taking a topology class, that was simply the notation used in my textbook.

 

[HallsofIvy]

It is not enough to deal with "notation"- you need to know the definitions as well. And you clearly are mistaken about the definition of "boundary point". "0" not only is NOT the "only"" boundary point of the set of natural numbers (as a subset of R with the "standard topology") it is not a boundary point at all because the open interval (-1/4, 1/4) which contains 0, does NOT contain any natural numbers. A set is said to be "closed" if and only if it contains all of its boundary points. You appear to know that. A point is a "boundary point" of a given set if and only if every open interval containing that point contains some points in the set and some points not in the set. Let "n" be any natural number (positive integer). Then any open interval about it is of the form (n- a, n+ a) for some positive real number, a. That contains n so clearly contains a member of the set of natural numbers. Does it contain any points that are NOT in the set of natural numbers?

 

[pasmith]

It is not enough to deal with "notation"- you need to know the definitions as well. And you clearly are mistaken about the definition of "boundary point". "0" not only is NOT the "only"" boundary point of the set of natural numbers (as a subset of R with the "standard topology") it is not a boundary point at all because the open interval (-1/4, 1/4) which contains 0, does NOT contain any natural numbers.

That depends on whether you regard 0 (the cardinality of the empty set) as a natural number.

 

[HallsofIvy]

First, the usual definition of "natural numbers" does NOT "regard 0 as a natural number". Second, even if the question were about the "whole numbers" (the natural numbers together with 0), the same argument applies, using any natural number rather than "0". The boundary points for either of those sets are exactly the points in the set. Since they contain all of their boundary points, both sets are closed.

 

[mfb]

First, the usual definition of "natural numbers" does NOT "regard 0 as a natural number".

Both definitions, with and without 0, are common.
Anyway, 0 is not special here, so it does not matter for the original question.

 

[PeterDonis]

Both definitions, with and without 0, are common.

For what it's worth, the Peano axioms require 0 to be included, as does the set theoretic definition of natural numbers. So there are good reasons why the definition that includes 0 is used.