The closure of a set is closed

Question

Definition: The closure of a set $A$ is $\bar A=A\cup A'$, where $A'$ is the set of all limit points of $A$.

Claim: $\bar A$ is a closed set.

Proof: (my attempt) If $\bar A$ is a closed set then that implies that it contains all its limit points. So suppose to the contrary that $\bar A$ is not a closed set. Then $\exists$ a limit point $p$ of $\bar A$ such that $p\not \in \bar A.$ Clearly, $p$ is not a limit point of $A$ because if it were then $p\in \bar A$. This means that $\exists$ a neighborhood $N_r(p)$ which does not contain any point of $A$. But $p$ is a limit point of $\bar A$ so it must contain an element $y\in \bar A-A$ in its neighborhood $N_r(p).$ Of course, $y$ is a limit point. Now, $0<d(p,y)=h<r.$ If we choose $0<\epsilon<r-h$, then $N_{\epsilon}(y)$ will contain no point $x\in A$, which is a contradiction since $y$ is a limit point.

I want to know if this proof is correct.

Answer 1

Your proof is correct, maybe we can make it slightly faster.

Let $z$ be a limit point of $\overline A$. Every open set $U$ containing $z$ must contain a point $x$ in $\overline A$. If this point $x$ is in $A^\circ$ then $U$ must intersect $A$ because it contains a limit point of $A$. If $x$ is in $A$ then $U$ obviously intersects $A$.

So every limit point of $\overline A$ is a limit point of $A$, and $\overline A$ contains all of its limit points.

Answer 2

Let $X$ be a metric space with the distance-metric $d()$. Let $E\subset X$ and let $E'$ be the set of limit-points of $E$. Then the closure $\bar{E}$ of $E$ is defined as $\bar{E} = E\,\cup\,E'$.

Prove: $\bar{E}$ is closed.

Proof by contradiction: Let $\bar{E}$ is not closed, i.e. $\exists x\in X$ such that it's a limit-point of $\bar{E}$ but $x\notin\bar{E}$ Or more formally

$$ N_r(x)\,\cap\,\bar{E} \neq \phi\qquad\forall r > 0 $$

where, $N_r(x) = \{y\in X\,|\,d(x,y) < r\}$ is the neighborhood of $x$. Then,

\begin{align*} (N_r(x)\,\cap\,\bar{E})\,\backslash\{x\} & = (N_r(x)\,\cap\,(E\,\cup\,E'))\backslash\{x\} \\ & = ((N_r(x)\,\cap\,E)\,\cup\,(N_r(x)\,\cap\,E'))\backslash\{x\} \\ & = \underbrace{((N_r(x)\,\cap\,E)\backslash\{x\})}_{\text{term-1}}\,\cup\,\underbrace{((N_r(x)\,\cap\,E')\backslash\{x\})}_{\text{term-2}} \neq \phi\qquad\forall r > 0 \end{align*}

The above equation implies that either term-1 or term-2 or both are non-empty.

Term-1: if $(N_r(x)\,\cap\,E)\backslash\{x\}\neq\phi\quad\forall r>0$, then it would mean that $x$ is a limit-point of $E$, i.e. $x\in E'$ and therefore, $x\in\bar{E}$.

Term-2: if $(N_r(x)\,\cap\,E')\backslash\{x\}\neq\phi\quad\forall r > 0$, then it would mean that $x$ is a limit-point of $E'$. Now, we need to show that any limit-point of $E'$ is also a limit-point of $E$.

• Let $x\in X$ be a point that is a limit-point of $E'$ but not a limit-point of $E$. Then, $$ \exists q>0:(N_q(x)\,\cap\,E'\neq\phi)\wedge(N_q(x)\,\cap\,E=\phi) $$ Let $y\in (N_q(x)\,\cap\,E')$, and $0 < d(x,y) = h < q$. Then, for any $0 < \epsilon < q - h$ we must have $N_{\epsilon}(y)\,\cap\,E = \phi$, since $N_{\epsilon}(y)\subset N_q(x)$. This, contradicts the fact that $y$ is a limit-point of $E$.

Therefore, any limit-point of $E'$ is also a limit-point of $E$, i.e. it also belongs to $E'$ and therefore belongs to $\bar{E}$.

QED: We have shown that every limit-point of $\bar{E}$ also belongs to $\bar{E}$, therefore $\bar{E}$ is closed!