# Let $A \subset X$. Show that if $C$ is a closed set of $X$ and $C$ contains $A$, then $C$ contains the closure of $A$.

I am using Analysis on Manifolds by Munkres to study for a course and the following question comes from an early section in topology.

Definition of limit point: Let $A \subset R^n$ and let $x_o \in R^n$. $x_o$ is a limit point of $A$ if, for every $r>0,B(x_o,r)$ contains a point of $A \setminus \{x_o\}$

Exercise: Let $A$ be a subset of $X$ where $X$ is a metric space. Show that if $C$ is a closed subset of $X$ and $C$ contains $A$, then $C$ contains the closure of $A$.

This is what I have so far:

$C$ is closed implies that $C$ contains its limit points. $A \subset C$, thus $A$ contains all points of $A$ (by Theorem from book). Since $\bar{A} = A \cup \{\text{limit points of$A$}\}$, we only have left to show that the set of limit points of $A$ is in $C$.

Let $p$ be a limit point of $A$. Suppose $p \notin C$. Then $p \in X \setminus C$. We know that $X \setminus C$ is open...

EDITED (changed proof strategy after doing more research)

let $p$ be a limit point of $A$ and if $p$ is not in $C$, then $X \setminus C$ is an open set containing $p$ but not intersecting $C$, which implies that $X \setminus C$ does not intersect $A$, which contradicts the fact that $p$ is a limit point of $A$. This is since any neighborhood of a limit point $A$ must intersect a point other than $x_o$ in $A$ (by definition of limit point; and since $X \setminus C$ is open, every neighborhood of a point contained in $X \setminus C$ has a radius in $x \setminus C$ (by definition of open).

• the adherence of $A$ is the intersection of the closed spaces which contain $A$. – Tsemo Aristide Jan 9 '16 at 19:56
• How do you define a limit point? – A.P. Jan 9 '16 at 20:00
• A "limit point" is defined as such:Let $A \subset R^n$ and let $x_o \in R^n$. $x_o$ is a limit point of $A$ if, for every $r>0, B(x_o,r)$ contains a point of $A \setminus \{ x_o \}$ – Old J Jr. Jan 9 '16 at 20:05
• @OldDave: If this is the right approach depends on the setting and definitions you are using: Is $X$ a metric space, or more generally a topological space? What is your definition of the closure / what definitions do you already know. – Jendrik Stelzner Jan 9 '16 at 20:07

A closure operator $A\mapsto cl(A)$ satisfies the Kuratowski axioms:

• $A\subseteq cl(A)$
• $A\subseteq B \implies cl(A)\subseteq cl(B)$
• $cl(cl(A)) = cl(A)$.

$A\subseteq X$ is closed iff $A = cl(A)$.

If you know that $A\mapsto A\cup A'$ is a closure operator, then the proof is straightforward: if $A\subseteq C = cl(C)$, then $cl(A)\subseteq cl(C) = C$.

If you don't know this, then in effect the exercise asks you to prove that closure really is a closure operator.

• We don't know of the Kuratowski axioms. So, let $p$ be a limit point of $A$ and if $a$ is not in $C$, then $X\setminus C$ is an open set containing $p$ but not intersecting $C$, which implies that $X\setminus C$ does not intersect $A$, which contradicts the fact that $p$ was a limit point of $A$. Is this correct? – Old J Jr. Jan 9 '16 at 21:12
• Yes, correct. And actually I like the proof by contradiction ($p\notin C$) [$p$ not $a$]. – BrianO Jan 10 '16 at 0:02
• To be precise, what exactly is the underlying reason why this contradicts? Is it because any neighbourhood of a limit point of $A$ must intersect with at least one point other than $x_o$ of $A$? – Old J Jr. Jan 10 '16 at 6:46
• Because you want to show that $p\in C$, and then suppose that $p\notin C$ to reach a contradiction. – BrianO Jan 10 '16 at 16:51

Hint: If a point in is in the closure of $A$, that means it is the limit of some (convergent) sequence of points in $A$. But a sequence of points in $A$ is also a sequence of points in $C$ ...

Alternative hint: The closure of $A$ is the smallest closed set containing $A$. As such, it is the intersection of all closed sets containing $A$.