# Thread: Closure Operation Properties? Kinda lost...

1. ## Closure Operation Properties? Kinda lost...

I'm sorta lost and sorta have an idea, so I don't know what to do really. The problem is to show that the closure operation has the following properties:
(a) if E1 is a subset of E2, then closure(E1) is a subset of closure(E2)
(b) closure(E1 U E2) = closure(E1) U closure (E2)
(c) closure(E1 intersect E2) is a subset of closure(E1) intersect closure(E2)

I do not need the full out proofs, but I do need some direction...

on (a) I'm not sure how to show it at all, it's the one I'm totally lost on.

(b) I get part of this one. I know you start with an element that is in the left side which leads to it being in E1 or E2 or (E1 U E2)', or the accumulations points of the union of E1 and E2 but do not know how to rewrite it so I can get the element in E1' and E2', or if (E1UE2)' is just E1' U E2.

(c) I can get this once I can get b.

2. Originally Posted by mathgirl13
I'm sorta lost and sorta have an idea, so I don't know what to do really. The problem is to show that the closure operation has the following properties:
(a) if E1 is a subset of E2, then closure(E1) is a subset of closure(E2)
(b) closure(E1 U E2) = closure(E1) U closure (E2)
(c) closure(E1 intersect E2) is a subset of closure(E1) intersect closure(E2)

I do not need the full out proofs, but I do need some direction...

on (a) I'm not sure how to show it at all, it's the one I'm totally lost on.

(b) I get part of this one. I know you start with an element that is in the left side which leads to it being in E1 or E2 or (E1 U E2)', or the accumulations points of the union of E1 and E2 but do not know how to rewrite it so I can get the element in E1' and E2', or if (E1UE2)' is just E1' U E2.

(c) I can get this once I can get b.
In what kind of space are we working?

Note for the first one that if $E\subseteq E_1$ then every closed neighborhood which contains $E_1$ contains $E$. So the intersection of all neighborhoods which contain $E$ has to contain the intersection of all the closed neighborhoods which contain $E_1$

For the second one it looks like you got the one way, the other way comes from saying let $x\in \overline{E_1}\cup\overline{E_2}$ then it is either an adherent point of $E_1$ or of $E_2$. Assume WLOG that it is one of $E_1$ then every neighborhood intersects $E_1$ and thus every neighborhood intersects $E_1\cup E_2$ and so $x$ is an adherent point of $E_1\cup E_2$ or equivalently $x\in\overline{E_1\cup E_2}$