Closed sets and closures

A closed set is a set that contains all of its accumulation points. Equivalently, a closed set is the complement of an open set.

The closure of any set $E$ is the smallest closed set that contains $E$—it contains $E$ as well as all its limit points. By smallest, we mean precisely the intersection of all closed sets that contain $E$. It follows that the closure of a closed set is simply the set itself.

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In metric spaces §

Definition: Closure

The closure of a set $E\subset X$, denoted $\bar{E}$, is the set containing $E$ and all of its limit/accumulation points.

Theorem: Rudin 2.27

1. $\bar{E}$ is a closed set.
2. $E$ is closed if and only if $E=\bar{E}$.
3. If $E\subset F$ is a subset and $F$ is closed, then $\bar{E}\subset F$.

Proposition

Let $A\subseteq X$ be bounded. Then $\text{diam}(A)=\text{diam}(\overline{A})$, where $\overline{A}$ is the closure of $A$.

Proposition: Sequence characterization of closed sets

Let $(X,d)$ be a metric space and $A\subseteq X$. Then $A$ is closed if and only if every convergent sequence in $A$ converges to some $a\in A$.

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In topological spaces §

For the following definitions and theorems, recall the terminology:

• Given $x\in X$ and an open set $U\subseteq X$, we say $U$ is a neighborhood of $x$ if $x\in U$.
• Given $A,B\subseteq X$, we say $A$ intersects $B$ if $A\cap B\ne \varnothing$.

Closed sets §

Definition: Closed set in topological space

Given a topological space $X$, we say a subset $A\subseteq X$ is closed if its complement $X\backslash A$ is open.

In principle, one can “do topology” using closed sets instead of open sets; most mathematicians just prefer the latter.

Theorem (Munkres 17.1): Doing topology with closed sets

Let $X$ be a topological space. Then:

1. $\varnothing$ and $X$ are closed;
2. Arbitrary intersections of closed sets are closed;
3. Finite unions of closed sets are closed.

Closure and interior §

Definition: Closure and interior of a set

Given a subset $A$ of a topological space $X$:

• The interior of $A$, denoted $\text{Int}(A)$, is the union of all open sets contained in $A$;
• The closure of $A$, denoted $\bar{A}$, is the intersection of all closed sets containing $A$.

By definition, $\text{Int}(A)\subset A\subset \bar{A}$. Further, $\text{Int}(A)$ is the biggest open set contained in $A$, and $\bar{A}$ is the smallest closed set containing $A$.

Theorem (Munkres 17.4): Closure of a subset in a subspace

Let $Y\subseteq X$ be a subspace, and let $A\subset Y$ be any subset. Then the closure of $A$ in $Y$ is equal to $\bar{A}\cap Y$, where $\bar{A}$ is the closure of $A$ in $X$.

Transclude of (Theorem)-Describing-the-closure-of-a-set-using-a-topological-basis#^111747

Transclude of (Theorem)-The-closure-of-a-set-is-the-union-of-the-set-with-its-limit-points#^f956e9

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Examples §

Graphs of continuous functions are closed §

If $f:D\subseteq \mathbb{R}\to \mathbb{R}$ is continuous, then the set

$$\text{Graph}(f)=\{(x,f(x))|x\in \mathbb{D}\}\subseteq \mathbb{R}\times \mathbb{R}$$

is closed. If we have a convergent sequence $x_n\to x$ in $\mathbb{R}$, then the image has $f(x_n)\to f(x)$ as well, so any convergent sequence in the graph $(x_n,f(x_n))$ should have a limit point $(x,f(x))$ in the graph.

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Review §

Topology

• Prove that given a subspace $Y\subseteq X$ and a subset $A\subseteq Y$, the closure of $A$ in $Y$ is equal to $Y$ intersected with the closure of $A$ in $X$. ⭐
• Prove (Theorem) Describing the closure of a set using a topological basis. (Hint: for (i), prove the contrapositive.) ⭐
• Prove (Theorem) The closure of a set is the union of the set with its limit points. (Hint: use (Theorem) Describing the closure of a set using a topological basis for one of the containments). ⭐

Honors Mathematics B

• Is $\mathbb{Q}$ dense in $\mathbb{R}$? Why or why not?

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Proof appendix §

Proposition

Let $A\subseteq X$ be bounded. Then $\text{diam}(A)=\text{diam}(\overline{A})$, where $\overline{A}$ is the closure of $A$.

Link to original

Proof.

Since $A\subseteq \overline{A}$, we know automatically $\text{diam}(A)\le \text{diam}(\overline{A})$. Conversely, let $ϵ>0$. Then there exist $x,y\in \overline{A}$ such that

$$d(x,y)\ge \text{diam}(\overline{A})-ϵ.$$

Further, by definition of $\overline{A}$, there exist $x’,y’\in A$ such that $d(x,x’)<ϵ$ and $d(y’,y)<ϵ$. Then

$$\text{diam}(A)\ge d(x^{\prime },y^{\prime })\ge d(x,y)-d(x^{\prime },x)-d(y^{\prime },y)\ge \text{diam}(\overline{A})-3ϵ.$$

Since this holds for any $ϵ>0$, we must have $\text{diam}(A)\ge \text{diam}(\overline{A})$ as well.

Proposition: Sequence characterization of closed sets

Let $(X,d)$ be a metric space and $A\subseteq X$. Then $A$ is closed if and only if every convergent sequence in $A$ converges to some $a\in A$.

Link to original

Sketch of proof.

• ($⟹$) Use definition of a sequence converging to some point $x\in X$, then show that $x$ is a limit point of $A$ using the sequence characterization of limit points. Since $A$ is closed, the limit point $x$ is also in $A$.

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Notes §

Definition: Sequence definition of closed sets

A set $C\subset {\mathbb{R}}^n$ is closed if, whenever a sequence ${\vec{x}}_m\in C$ converges to some limit ${\vec{x}}_m\to \vec{x}\in {\mathbb{R}}^n$, its limit $\vec{x}$ is also in $C$.

• The sequence definition was used in Honors Mathematics B, and a later exercise required proving that this definition holds if and only if the complement of a closed set is open. In MATH-42X, the latter definition was used to prove the former.