In my general topology textbook there is the following proposition:
Let $A$ be a subset of a topological space $(X, τ)$. Then $A$ is closed in $(X, τ )$ if and only if $A$ contains all of its limit points.
And then they give the following proof:
Assume that $A$ is closed in $(X, \tau)$. Suppose that $p$ is a limit point of $A$ which belongs to $X \setminus A$. Then $X \setminus A$ is an open set containing the limit point $p$ of $A$. Therefore $X \setminus A$ contains elements of $A$ (1).This is clearly false and so we have a contradiction to our supposition. Therefore every limit point of $A$ must belong to $A$
Conversely, assume that $A$ contains all of its limit points. For each $z \in X \setminus A$, our assumption implies that there exists an open set $U_z \ni z$ such that $U_z \cap A = \emptyset$; that is, $U_z \subseteq X \setminus A$. Therefore $X \setminus A = \bigcup_{z \in X \setminus A} U_z$. So $X \setminus A$ is the union of open sets and hence is open. Consequently $A$ is closed.
My question is in the marker (1). How do they conclude that $X \setminus A$ contains an element of $A$?
Best Answer
The definition of a limit point that your text uses (I'd call that notion an adherence point BTW) going by the second part of the proof:
But in your first part $X\setminus A$ is an open set (because we assumed $A$ is closed) that contains $p$ but and we also assumed that $p$ is a limit point so the definition can be applied to the open set $X\setminus A$ to reach that conclusion.