In general Topology by Kelley (1955), Theorem 5 states:
A subset of a topological space is closed if and only if it contains the set of its accumulation points.
a point $x$ is an accumulation point (aka limit point) of a subset $A$ of a topological space $(X, \tau)$ if and only if every neighborhood of $x$ contains points of $A$ other than $x$.
I'm trying to understand the theorem 5.
Suppose $a,b,c \in A$ and $A$ is closed in topological space $(X, \tau)$. Does it then imply that $\{b,c\},\{a,c\},\{a,b\}\in A$?
Edit
Okay, I got it. I did not understand definition of accumulation point correctly – I was under impression that it contains all points other than $x$ and does not contain $x$.
Best Answer
It means that "you can't get infinitely close" to a point that's not in the set.
Example: the set $(0, 1]$ is not closed, because you can get as close to $0$ as you want while staying in the set (ie: in every open neighbourhood of $0$, you can find a point of $(0,1]$), but you couldn't reach it without leaving the set! However, if you now include $0$, the set $[0,1]$ is closed.
Other example: the unit disc in dimension $2$. If you include the boundary, then your set is closed, because you "contain" everything. But if you remove the boundary (ie, if you only consider the open disc $D = \{(x,y) : x^2 + y^2 < 1\}$), then you'll be able to get points to converge to the boundary, but could not reach it.
This is what we commonly call closure! In a general topological space, a set $A$ is closed iff it contains all its points of closure: ie the set of points $x$ such that for every neighbourhood $\mathcal{U}$ of $x$: $\mathcal{U}\cap A \ne \emptyset$.
note that this notion is slightly different from limit point, since we do not exclude the case where $\mathcal{U}\cap A$ is simply $\{a\}$. The equivalence "$A$ is closed $\iff A$ contains all its limits points is only true in certain types of topological spaces: they need to be first-countable (but all metric spaces do satisfy this property, so altogether it's not a bad way to represent it to yourself).
And $a, b, c \in A$ in your example are a finite example, so it doesn't really tell us much. However, say you take a converging sequence $(a_n$) in $A$, then this property tells you that its limit will also be in $A$