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

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• February 9th 2011, 11:10 AM
alice8675309
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 $\subseteq$R and p $\in$R. Prove that p is an accumulation point of A $\cup$B if and only if p is an accumulation point of A or p is an accumulation point of B: (A $\cup$B)'=A' $\cup$ B'.

Now since we assumed the above is true,

Conclude that the closure of a union is the union of the closures:
(A $\cup$ B)^cl=A^cl $\cup$ B^cl. How would I go about proving this?
• February 9th 2011, 11:26 AM
Plato
The closure of a set is the ‘smallest’ closed that contains the set.
We know that $A\subseteq\overline{A}~\&~ B\subseteq\overline{B}$.
The union of two closed sets is a closed set.
That forces $\overline{A\cup B}\subseteq\overline{A}\cup \overline{B}$. Why and how?

Can you do the converse?
• February 13th 2011, 06:50 PM
alice8675309
After working on this for a little, is this proof correct?

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

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

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