My textbook has not been very clear (at least to me) with respect to closed sets.
I have the following understanding. These two definitions are equivalent with respect to closed sets:
(1) A closed set is any set that contains all of its limit points.
(2) A closed set is any set that contains all of its boundary points.
Does (1) imply (2), and vice versa, or is the implication only from (1) to (2)?
Are there any other definitions of a closed set?
It is not true that simply because a set contains a point $p$ such that all neighborhoods $N_rp$ contain points in the set and points not in the set that the set is closed. In other words, simply because a set contains some of its boundary points does not mean it is closed.
So, I was thinking about the $\mathbb{N}$atural numbers and was confused when considering any singleton $\{a\} \subset \mathbb{N}.$
First, can you even talk about neighborhoods of non-integer radii when talking about the natural numbers? Isn't there something unnatural, or "unfair" about considering a radius $r$ such that $0 < r < 1?$ Does the radius have to be an element of the metric space you are considering?
Assuming you can, it is easily apparent that the singleton is open, for there exists a neighborhood $N_ra$ of $a$ that contains only points in $\{a\},$ namely any neighborhood that has a positive radius less than 1.
By definition (1), the singleton is closed, as we can construct the constant sequence $(x_n)$ such that $x_n = a$ for all $n \in \mathbb{N}.$ This is the only singleton that may defined in $\{a\},$ it it converges to $a,$ which, of course, is in the set.
How would you consider the singleton under the second definition? I suppose it doesn't have ANY boundary points, as there are neighborhoods of $a$ that do not contain points in $\mathbb{N} \setminus \{a\}.$ If we consider any singleton set in $\mathbb{R},$ the set is closed, but not open by the same understanding.