# Thread: closure union identity

1. ## closure union identity

I'm trying to prove that the closure of the union is the union of the closures.

Where for topological space $(X, \mathcal T)$,
$\overline{A} = \cap \{F \subseteq X : A \subseteq F\, and\, F\, is \, closed\}$
denotes the closure of $A$.

I'm trying to prove $\overline{A\cup B} = \overline{A} \cup \overline{B}$

===========

This is what I have so far:

First, proving $\overline{A\cup B} \subseteq \overline{A} \cup \overline{B}$:
We know $\overline{A}, \overline{B}$ are closed,
so $\overline{A} \cup \overline{B}$ is closed.
Furthermore, $A \subseteq \overline{A}, B \subseteq \overline{B}$,
so $A \cup B \subseteq \overline{A} \cup \overline{B}$.
Since $\overline{A} \cup \overline{B}$ is a closed set containing $A \cup B$, we have $\overline{A\cup B} \subseteq \overline{A} \cup \overline{B}$.

Now to prove $\overline{A} \cup \overline{B} \subseteq \overline{A\cup B}$:

... This is where I'm stuck ...

2. Originally Posted by CatNoob
Now to prove $\overline{A} \cup \overline{B} \subseteq \overline{A\cup B}$:
... This is where I'm stuck ...
Recall that $A\subseteq A\cup B$ so $\overline{A}\subseteq \overline{A\cup B}$. Likewise $\overline{B}\subseteq \overline{A\cup B}$.

3. Wow, thanks Plato. Why didn't I see that.

Also, is the first part of the proof correct? Is there an easier way?

4. Originally Posted by CatNoob
Also, is the first part of the proof correct? Is there an easier way?
That's the easiest way to prove it here! That said, it can be proven just by using the axioms of a closure operator, if that in any way interests you.

5. Originally Posted by Drexel28
That's the easiest way to prove it here! That said, it can be proven just by using the axioms of a closure operator, if that in any way interests you.
What are the axioms of the closure operator? I just know the definition of closure given above.

6. Originally Posted by CatNoob
What are the axioms of the closure operator? I just know the definition of closure given above.
It's not super important for this case, but it's an intertesting alternative way to define a topological space. Namely, call a closure space to be an ordered pair $\left(X,\text{cl}_X\right)$ where $X$ is a non-empty set and $\text{cl}_X:2^{X}\to 2^{X}$ (here $2^X$ is the power set) such that for every $E,G\in 2^X$

\begin{aligned}&\mathbf{(1)}\quad \text{cl}_X\left(\varnothing\right)=\varnothing\\ &\mathbf{(2)}\quad E\subseteq \text{cl}_X\left(E\right)\\ &\mathbf{(3)}\quad \text{cl}_X\left(E\cup G\right)=\text{cl}_X\left(E\right)\cup\text{cl}_X\ left(G\right)\\ &\mathbf{(4)}\quad \text{cl}_X\left(\text{cl}_X\left(E\right)\right)= \text{cl}_X\left(E\right)\end{aligned}

Clearly every topological space induces a closure operator, the normal one. What's more interesting is that if one defines on a closure space $\left(X,\text{cl}_X\right)$ the set $\mathcal{T}=\left\{E\in 2^X:\text{cl}_X\left(X-E\right)=E\right\}$ then $\mathcal{T}$ is a topology on $X$. What's really interesting about this whole things is that these two notions of topological spaces (the normal definition and the definition which makes closure space and topological space synonomous) are really equivalent. What I mean by this is if you start with a topological space $\left(X,\mathcal{J}\right)$ and creat from this the usual closure operator $\text{cl}_X$ (the intersection of all the closed sets containing the set in question) and then create from this the topology $\mathcal{T}$ induced by the closure operator (as I previously mentioned) then one can see that $\mathcal{T}=\mathcal{J}$. Conversely, if one starts with a closure space $\left(X,\text{cl}_X\right)$ and create the topology $\mathcal{T}$ induced by this closure operator and then defines the usual closure operator $\text{cl}'_X$ on the topological space $\left(X,\mathcal{T}\right)$ (the intersection of all closed supersets) then $\text{cl}_X=\text{cl}'_X$. Kind of interesting, huh? What's weird is that while the formulation of topological spaces seen today is probably the most natural the concept of a closure space predates it, as can be seen in the works of Kuratowski.

Sorry for the rant, just thought I'd share.