# Thread: conclude that the closure of a union is the union of the closures.

1. ## conclude that the closure of a union is the union of the closures.

For this question, we are allowed to assume the following is true and has been proven:

Let A,B$\displaystyle \subseteq$R and p$\displaystyle \in$R. Prove that p is an accumulation point of A $\displaystyle \cup$B if and only if p is an accumulation point of A or p is an accumulation point of B: (A$\displaystyle \cup$B)'=A' $\displaystyle \cup$ B'.

Now since we assumed the above is true,

Conclude that the closure of a union is the union of the closures:
(A $\displaystyle \cup$ B)^cl=A^cl $\displaystyle \cup$ B^cl. How would I go about proving this?

2. The closure of a set is the ‘smallest’ closed that contains the set.
We know that $\displaystyle A\subseteq\overline{A}~\&~ B\subseteq\overline{B}$.
The union of two closed sets is a closed set.
That forces $\displaystyle \overline{A\cup B}\subseteq\overline{A}\cup \overline{B}$. Why and how?

Can you do the converse?

3. After working on this for a little, is this proof correct?

Assume p$\displaystyle \notin$A^cl and that p$\displaystyle \notin$ B^cl. That is, there exists $\displaystyle \epsilon$$\displaystyle _1>0 such that A has no point in (p-\displaystyle \epsilon$$\displaystyle _1$,p+$\displaystyle \epsilon$_1) and there exists $\displaystyle \epsilon$$\displaystyle _2>0 such that B has no point in (p-\displaystyle \epsilon$$\displaystyle _2$,p+$\displaystyle \epsilon$_2). Let $\displaystyle \epsilon$$\displaystyle _3=min{\displaystyle \epsilon$$\displaystyle _1$,$\displaystyle \epsilon$$\displaystyle _2}, and clearly \displaystyle \epsilon$$\displaystyle _3$>0 and satisfies the requirement.

Then Assume p$\displaystyle \in$A^cl $\displaystyle \cup$ B^cl.
case 1: p$\displaystyle \in$A^cl which means that every neighborhood of p contains a point of A. Hence every neighborhood contains a point of A$\displaystyle \cup$B .

case 2: p$\displaystyle \in$B^cl which means that every neighborhood of p contains a point of B. Hence every neighborhood contains a point of A$\displaystyle \cup$B .
Thus, (A$\displaystyle \cup$B)^cl=A^cl$\displaystyle \cup$B^cl