Is accumulation point = adherent point in a closed set?

[kadas]

A set S is closed iff it contains all its adherent points iff it contains all its accumulation point?

From what I know, in general accumulation point is a subset of adherent point, but if supposed I have a closed set, then the "if and only if" forces me to conclude that accumulation point = adherent points.

Supposed they are not equal, then since having all accumulation points inside the set already makes it a closed set, then a closed set doesn't necessarily contains all its adherent points which contradicts the definition.

Sorry if this is an easy question, but then if my reasoning is wrong, please help me to correct it. Thanks.

 

[HallsofIvy]

A point, p, is an "adherent" point of a set, A, if and only if every neighborhood of p contains a member of A.

A point, p, is an "accumulation" point of a set, A, if and only if every neighborhood of p contains a member of A other than p itself.

The difference is that if p is in A, then it can be the "member of A" in every neighborhood in the definition of "adherent" point. There need not be any other member of A anywhere close to p.

That is, a point is an "adherent point" of a set, A, if it is either an accumulation point of A or a member of A. The two "if and only ifs": "if and only if it contains all of its adherent points"= "if and only if it contains all of its accumulation points" works because any adherent point that is not an accumulation point is already in the set. However, you certainly can have "adherent points" that are NOT accumulation points. That is, points in the set that are not accumulation points- they are called "isolated" points.

For any point, p, the singleton set, {p}, is closed. Its only "adherent" point is p itself. It has no accumulation points.

For another example, consider the subset of the real line $A= (0, 1)\cup \{2\}$- that is, the open interval from 0 to 1 and the point 2. The set of all accumulation points is the closed interval [0, 1]. The set of all adherent points is $[0, 1]\cup \{2\}$. 2 is an isolated point- it is in the set, so an adherent point, but not an accumulation point.

 

[kadas]

ah, awesome. Now I understand how the "if and only if" parts works. How stupid of me not to think of an example of singleton. Thanks!