Closed sets: definition(s) and applications

Question

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.

Answer 1

I'm assuming that you're familiar with metric spaces based on the depth of your question. In this answer let the metric space be $(X,d)$.

"Does (1) imply (2), and vice versa, or is the implication only from (1) to (2)?

Yes.

A boundary point is a point which is neither interior nor exterior to the set. A point, $p \in X$ is an interior point of a set $A$ if there exists an open neighborhood $N$ of $p$ such that $N \subset A$. A point $q \in X$ is an exterior point of $A$ if $q$ is an interior point of $A^c$. That is, a point $r \in X$ is a boundary point of $A$ if it is interior to neither $A$ nor $A^c$.

I suggest using the neighborhood definition of a limit point, that is that $p \in X$ is a limit point of $A$ if $\forall \epsilon>0$, the ball $N_\epsilon(p)$ contains an $x \in A$ with $x \neq p$.

If $p \in X$ is a boundary point of $A$ then no neighborhood, $N_\epsilon(p)$, of $A^c$ is contained in $A^c$ (else it would be interior to $A^c$). Therefore $\forall \epsilon>0: N_\epsilon(p) \cap A \neq \emptyset$.

Case 1: $p \in A^c$. Then if there is no $x \neq p$ in $A \cap N_\epsilon(p)$, we get a contradiction. Thus there must exist such an $x \in N_\epsilon(p)$ with $x \neq p$.

Case 2: $p \in A$. (I'll let you handle this one. You can show that there exists an $x \in N_\epsilon(p)$ with $x \neq p$)*.

Thus every $\epsilon$ neighborhood of $p$ contains an $ x \neq p$.*

I'll let you handle the converse proof.

Are there any other definitions of a closed set?"

Yes, the most important being that a set is closed if and only if its complement is open. This is equivalent to the other definitions.

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?

Radii are always taken to be non-negative real numbers. In particular, if the radius had to be an element of the metric space then we would always need some kind of ordering on our metric spaces, which is undesirable. Is it unfair to use real radii in the metric space $X=\mathbb{N}$? I don't see why it's "unfair". When you study topology you will find that there are more general ways to imbue sets like $\mathbb{N}$ with the structure of a "space" other than as a metric space.

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.

I think you meant to say "sequence" instead of "singleton" in that last sentence. Be careful with your wording. Your general idea is right that there is only one possible sequence in $\{a\}$, and the conclusion that the limit is $a \in \{a \}$ is correct, but don't "construct" a sequence, let $\{x_n\}$ be any sequence in $\{a\}$ and show its limit is $a$.

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 $\mathbb{N} \backslash \{a\}$.

You're correct that $\{a\}$ (in the metric $X=\mathbb{N}$) contains no boundary points. Can you prove this using the definition of boundary points I gave above? If you want to understand why this means it "contains all its bounday points", look up "vacuous truth". The reasoning is similar as to why the empty set is closed under either of your definitions of closedness.

Answer 2

Another definition of a closed set is ;

A set is closed iff its complement is open.

This way you can figure out that the set of natural numbers or any subset of it is closed in the standard topology of real line.

Singletons are also closed with this definition.