adherent point contained by a open ball

Question

In any metric space $(X,d_X)$, $x$ is an adherent point of $A \subseteq X$ $\iff$ for every $r>0$, there exists $y \in A$ such that $y \in B(x;r)$. I sort of understand the main idea behind this theorem. My question is why does it have to be $y \in B(x;r)$ and not $y \in B[x;r]$.

Well, I'm sure it has to be an open ball to meet the definition of convergence of a sequence but I cannot think of any counter example of using the close ball. Intuitively, I think a close ball should also do the job.

Could anybody be kind enough to present any counter-example which shows that using a close ball invalidates the theorem?

Answer 1

The way you've stated this, it looks like a definition rather than a theorem (so it is open not closed because that is the way it is defined). If this is indeed a theorem, you should provide us with the definition you are working from. To figure out why it is defined that way, you should look at some other theorems that use the concept of adherent points.

From Wikipedia, here are some equivalent characterizations of adherent points: $x$ is an adherent point of $A$ if and only if every open set containing $x$ contains an element of $A$. Or, $x$ is an adherent point if and only if $x$ is in the closure of $A$. I leave the proof of these to you, as they are a good exercise. Feel free to ask for hints though.

Answer 2

Technically it need not be an open ball! The concept of an adherent point to a set can be defined in topological spaces, which are more general than metric spaces in the sense that it can be shown that every metric space is a topological space.

In a topological space $X$, an adherent point to a set $A \subset X$ is defined as a point $x \in X$ such that every neighborhood of $x$ (every subset of $X$ whose interior contains $x$) contains at least one point of $A$. Some authors would require at first that a neighborhood of a subset of $X$ to be an open set. But in fact this is more or less just a personal preference thing; we can also require a neighborhood of a subset of $X$ to be just a subset of $X$ whose interior includes the given subset.

Point set topology in metric spaces customarily employ open balls as a device instead of neighborhoods in general. This is a matter of convenience.

Answer 3

It is fine to use a closed balls, because every closed ball contains an open ball.

Adherent points of $A$ are points that are intrinsically "close" to $A$ -- either because they already belong to $A$, or because they are accumulation points of $A$ (which means there are other points of $A$ which are as close to the point as you like).

If a point is an adherent point of $A$, but is not actually in $A$ (so, it is an accumulation point of $A$), then this means that the point fills in a sort of "hole" in $A$. For example, the point $1$ in the case $A=[0,1)\cup(1,2]\cup \{3\}$. You can literally "see" the hole at $1$.

Note in this example $3$ is also an adherent point of $A$, but it is not an accumulation point of $A$, since there are no other points of $A$ close to $3$. On the other hand, $0$ is an adherent point of $A$ as well (it is in $A$, and it also happens to be an accumulation point of $A$ since there are other points of $A$ close to it -- like $\frac12,\frac13,\frac14,$ etc as close as you like).

In the end, using closed balls doesn't buy you anything more, and open balls are more natural as they are what make it work. You can use closed balls only because they lead you to open balls.