Basic topology - explain teacher's solution I'm trying to understand how the teacher solved the following question:
Prove that the closure of a union of two sets is equal to the union of the closures of the sets, but that this does not hold for infinite union: $\cup_{a}\bar{E_a} \subset \bar{\cup{E_a}}$ 
Firstly I must apologize and ask for help with the terminology. We don't study in english, we study in hebrew, so I don't know the correct term for this so we will just call it "nice".
$x \in A$ is a "nice" point in $A$ if in every neighborhood of $x$ there is at least one other point $y \neq x$, such that $y \in A$.
Teacher's solution:
Suppose $x \in \cup_{a}\bar{E_a}$. This means that there is some $b$ such that $x \in \bar{E_b}$, so $x$ is a "nice" point of $E_b$ (WHY????).
From this we can infer that $x$ is a "nice" point of $\cup_{a}E_a$ (WHY?) and as such it is a "nice" point of the closure of $\cup_{a}E_a$.
With the other details in the proof I think I can manage. I just don't understand why what he wrote there is true and would appreciate an explanation.
 A: I don't think the definition you gave for "nice" is what the teacher intended here. It really isn't the property needed to discuss closure. For a general $B \subseteq A$, a point $x\in A$ is called a limit point of $B$ if every neighborhood of $x$ intersects $B$ in a point other than $x$. This is similar to what you give, but differs in two respects: $x \in A$, not necessarily in $B$, and it is about the topology of $A$, not the topology of $B$. The closure of $B$ is the set $B$ together with all of its limit points.
To follow your teacher's logic: if $x \in \bigcup_a \overline E_a$, then there is a $b$ such that $x \in \overline E_b$, so by the definition of closure I gave above, either $x \in E_b$ or $x$ is a limit point of $E_b$.
Now if $x \in E_b$, then obviously $x \in \bigcup_a E_a \subseteq \overline{\bigcup_a E_a}$. If instead $x$ is a limit point of $E_b$, then every neighborhood of $x$ intersects $E_b$, and therefore it intersects $\bigcup_a E_a$, since $E_b$ is a subset. Therefore $x$ is a limit point of $\bigcup_a E_a$, and thus $x \in \overline{\bigcup_a E_a}$.
This proves that $\bigcup_a \overline E_a \subseteq \overline{\bigcup_a E_a}$, regardless of whether the collection is finite or infinite. To complete the proof, you also have to show the opposite inclusion when the collection is finite, and that there are exceptions when the collection is infinite. I assume you had no problems with that part of the proof?
