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Closure of an Set

Julio

Member
Feb 14, 2014
71
Let $(X,\tau)$ an topological space. Show that

$\overline{A}=\{x\in X: U\cap A\ne \varnothing, \text{ for all } U\in\tau \text{ and } x\in U\}.$


Hello, if $x\in\overline{A}$, then for all $V$ neighborhood of $x$, we have $V\cap A\ne \varnothing.$ My question is if its neccesary the inclusion of sets for showed.
 

Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,890
Hi Julio ,

What definition is given to you for $\overline{A}$?
 

Julio

Member
Feb 14, 2014
71
Hi Euge :).

The definition that I have is:

Def: Let $(X,\tau)$ a topological space. We will call closure of $A$ the intersection of all closed sets containing $A$. That is to say,
$\overline{A}:=\displaystyle\bigcap_{F \text{ closed in } X}F.$​

Characterizations...

$x\in \overline{A}\iff \forall U \text{ open of } X \text{ such that } x\in U \text{ we have } U\cap A\ne \varnothing.$

Or

$x\in \overline{A}\iff (\forall F)(F\subseteq X)(F \text{ closed})(A\subseteq F \text{ and } x\in F).$

Or

$x\in \overline{A}\iff (\forall V \text{ neighborhood of } x)(V\cap A\ne \varnothing)$.


But what or how can we use it?
 

Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,890
You can prove your proposition in two parts:

1. Prove that given $x\in \overline{A}$, and $U$ is an open neighborhood of $x$, then $U \cap A = \emptyset$ implies $\overline{A} \subset X\setminus U$. This leads to a contradiction.

2. Prove that if $x\notin \overline{A}$, then there is some open neighborhood $N$ of $x$ that is disjoint from $A$.