Prove that $E$ closure is closed

Question

Always thank you for helping me out I was following my professor's proof and this is how it goes

Prove that $E$ closure is closed let $x$ be a limit point of $\operatorname{cl}(E)$, claim: $x$ is in $\operatorname{cl}(E)$ By the definition of a limit point, $(N_e(x)\setminus\{x\})\cap \operatorname{cl}(E)$ is nonempty, let $y$ be the element in $\operatorname{cl}(E) = E \cup E'$

1. $y$ is in $E$ $\implies$ $x$ is a limit point of $E$, then $x$ is in $E'$ thereby in $\operatorname{cl}(E)$ since $E'$ is a subset of $\operatorname{cl}(E)$

2. $y$ is in $E'$ $\implies$ $x$ is a limit point of $E'$ and $y$ is a limit point of $E$ since $y$ is an interior point of $N_e(x)\setminus\{x\}$, there exists $z$ in $E_e'(y)\setminus\{y\}$ and $z$ is in $N_e(x)\setminus\{x\}$ $\implies$ $x$ is a limit point of $E$

I fully understood up to the first line of 2. where $y$ is a limit point of $E$, but how did he come up with the idea of $y$ is an interior point of $N_e(x)\setminus\{x\}$? Is it because a neighborhood is an open set, so any element in it should be an interior point?

Thank you guys in advance!

Answer 1

The problem with this proof is that it's a bit messy. I don't know what is $e$, what is $E_e'(y)$.

If I understand correctly, the definitions that are used are the following:

• $\bar E=E\cup E'$ where $E'=\{x:\forall (\varepsilon>0) N_\varepsilon(x)\setminus\{x\}\cap E\neq\emptyset\}$.
• $E$ is closed iff $E'\subset E$ (which is equivalent to: $\bar E=E$.

Fact to prove: $\bar E$ is closed.

Proof. We need to prove that $\bar E'\subset \bar E$. Take any $x\in\bar E'$. We are going to prove that $x\in E'\subset \bar E$. Take any $\varepsilon>0$. Our goal is to show that $N_\varepsilon(x)\setminus\{x\}\cap E\neq\emptyset$. From the fact that $x\in \bar E'$ we have some $y\in N_\varepsilon(x)\setminus\{x\}\cap \bar E$. We see that $y\in E\cup E'$.

• If $y\in E$ then we are done.
• If $y\in E'$ then then there exists $z$ in a set $N_\delta(y)\setminus\{y\}\cap E$ for any $\delta>0$. Since $N_\varepsilon(x)\setminus\{x\}$ is open and $y$ belongs to it, we have $z\in N_\delta(y)\subset N_\varepsilon(x)\setminus\{x\}$ for some small $\delta$. Therefore $z\in N_\varepsilon(x)\setminus\{x\}\cap E\neq\emptyset$.

Answer 2

This is another approach, different than one from the lecture. In my opinion it's much simpler, but it needs some preparation, which is profitable and will pay off later.

Fact (equivalent definitions of the closure) The following statements are equivalent:

1. $x\in \bar E=E\cup E'$,
2. $N_\varepsilon(x)\cap E\neq\emptyset$ for all $\varepsilon>0$
3. $U\cap E\neq\emptyset$ for every $U$ - open neighbourhood of $x$.

Proof

$1.\implies 2.$: If $x\in E$ then $x\in N_\varepsilon(x)\cap E\neq\emptyset$. If $x\in E'$ then $$N_\varepsilon(x)\cap E \supset N_\varepsilon(x)\setminus\{x\}\cap E\neq\emptyset.$$

$2.\implies 1.$: If $x\in E$ then we are done. If not then $$N_\varepsilon(x)\setminus\{x\}\cap E = N_\varepsilon(x)\cap E \neq\emptyset.$$ Therefore $x\in E'$.

$2.\implies 3.$: Take any $U$ as above. Therefore, since $U$ is open and $x\in U$, we have $N_\varepsilon(x)\subset U$ for some $\varepsilon>0$. Therefore $$ U\cap E\supset N_\varepsilon(x)\cap E\neq\emptyset.$$

$3.\implies 2.$: Obvious, since $N_\varepsilon(x)$ is an open nbhood of $x$. $\square$

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Now we are ready to show that $\bar E$ is closed, that is $\bar{\bar E}\subset \bar E$. We'll use the third condition.

Proof Take any $x\in \bar{\bar E}$. Take any open nbhood $U$ of $x$. Since $x\in \bar{\bar E}$, we have $y\in U\cap \bar E$. Since $y\in \bar E$ and $U$ is an open nbhood of $y$, we have $U\cap E\neq \emptyset$. This shows that $x\in\bar E$.