Rudin's Proof -- The Closure of a Set is Closed

[OhMyMarkov]

Hello everyone!

I was reading Rudin's proof for the theorem that states that the closure of a set is closed. I'll write the proof and the parts I'm having trouble connecting:

if $p\in X$ and $p\notin E$ then $p$ is neither a point of $E$ nor a limit point of $E$. (So far so good).

Hence, $p$ has a neighborhood which does not intersect $E$. (Great)

...

The compliment of $\overline{E}$ is therefore open. Hence $\overline{E}$ is closed.


Now, this is what I believe should go into the dots, but still not enough to conclude the proof:

The neighborhood around $p$ does not intersect $E$ so it lies completely in $E^c$.


Now, I can't complete the proof.

Any help is appreciated!

 

[Amer]

want to prove that the complement of the closure is open

let x in $\bar A^c $ x is not in the closure of A , so there exist an open set U and $x \in U $ such that
$ U \cap A = \phi $
claim $ \bar A \cap U = \phi $ suppose not
there exist y in $ \bar A \cap U$
y in U and $ U \cap A = \phi $ so y is not in $\bar A$
contradiction so
$ \bar A \cap U = \phi $
which mean $ U \subseteq \bar A^c $ ends
we take an arbitrary point in A closure complement and found open set containing it contained in A closure complement so A closure complement is open which mean A closure is closed

 

[Plato]

I was reading Rudin's proof for the theorem that states that the closure of a set is closed. I'll write the proof and the parts I'm having trouble connecting:

Suppose that $x\notin\overline{A}$ then $\exists O_x$ an open set such $x\in O_x~\&~O_x\cap A=\emptyset$

Clearly $\overline{A}^c = \bigcup\limits_{x \in \overline{A}^c } {O_x } $. The union of open sets is an open set.

 

[Amer]

Suppose that $x\notin\overline{A}$ then $\exists O_x$ an open set such $x\in O_x~\&~O_x\cap A=\emptyset$

Clearly $A^c = \bigcup\limits_{x \in A^c } {O_x } $. The union of open sets is an open set.

but the question want $\bar A^c$ open not A ?$A^c$ need not to be open
I think you have also to prove that $O_x \cap \bar A = \phi $

 

[OhMyMarkov]

Hello Amer, thanks for replying!

if $y\notin A$, $y$ could still be in $A'$, the set of limit points of $A$. I don't see how none of the replies talked about anything related to the limited points of $A$ or even of $\overline{A}$, because the definition of closed is as follows:

A set is closed every every limit point is a point of this set.

 

[Amer]

Hello Amer, thanks for replying!

if $y\notin A$, $y$ could still be in $A'$, the set of limit points of $A$. I don't see how none of the replies talked about anything related to the limited points of $A$ or even of $\overline{A}$, because the definition of closed is as follows:

A set is closed every every limit point is a point of this set.

I prove it in other way i proved that the complement is open which means the closure is closed if you want to use that definition
Solution
Let $x \in \bar A' $ , let U be any open set containing x ( what i will do i will show that $U \cap A \neq \phi$ which implies $x \in \bar A$)
Remember that
$ U- \{x\} \cap \bar A \neq \phi $ ( this because $x \in \bar A' $), that means $\exists y \in U$ and $ y \in \bar A $ and $ y \neq x $
suppose $U \cap A = \phi $ this will give that y dose not belong to $\bar A $ contradiction ( since if y is not in $\bar A$ there exist an open set$ U\ni y $such that $U \cap A = \phi$ )
so $ U \cap A \neq \phi $ ,
U is arbitrary open set so that holds for every U containing x
so $x \in \bar A $

 

[sweatingbear]

Here is a thorough proof for future inquirers:

We wish to prove $\overline{E}$ is closed. This is true exactly when $\overline{E}^c$ is open, so let us attempt to show $\overline{E}^c$ is open. Choose $x \in \overline{E}^c$ i.e. $x \not{\in} \overline{E}$. In order to show $\overline{E}^c$ is open, we need to find a neighbourhood $N$ about $x$ such that $N \subset \overline{E}^c$.

Now $x \not{\in} \overline{E}$ means that $x$ is neither a member of $E$ nor a limit point of $E$. The latter is of particular interest, because it means that there exists some neighbourhood $N$ about $x$ which contains no members of $E$. So all members of $N$ do not lie in $E$.

We would like all members of $N$ to lie in $\overline{E}^c$ in order to complete our proof. Thus we must ascertain every member of $N$ neither lies in $E$ nor is a limit point of $E$. We just concluded every member of $N$ is no member of $E$; what remains then is to show that there are no limit points of $E$ in $N$.

Suppose that was the case i.e. $q \in N$ is a limit point of $E$. Since neighbourhoods are open, $N$ is open and there is some neighbourhood $M\subset N$ about $q$. This neighbourhood $M$ has point $e \neq q$ such that $e \in E$, because $q$ is a limit point of $E$. But then $e \in M \subset N$ i.e. we have found a member $e$ of $N$ which lies in $E$, which contradicts the property $N \cap E = \emptyset $ of $N$.

Hence every member of $N$ is no member of $E$ nor a limit of point $E$ or, in other words, every member $p$ of $N$ satisfies $p \in \overline{E}^c$. So there indeed exists a neighbourhood $N$ about $x \in \overline{E}^c$ such that $N \subset \overline{E}^c$. $\overline{E}^c$ is open and $\overline{E}$ is closed, as desired. $\blacksquare$