Closure and Boundedness | Real Analysis

This scene builds the closure of a set. Watch what happens when every escaped limit point is pulled back in.

The closure contains every point the set was reaching for. If some point is approached but never touched, closing means adding it. A set already holding all its limit points equals its closure. Add boundedness, and compactness is near.

For example, fill a 📦 box and put the lid on. Only with the lid is the box complete. Closure is like that lid: it adds the boundary points to the set.

Examples & Exercises

Prove the closure of $(0,1)$ is $[0,1]$ .

\overline{(0,1)} = [0,1]

The red dots

0, 1

— limit points added by the closure.

Show proof

Points of $(0,1)$ are trivially in the closure. Let $x = 0$ : for every $\varepsilon > 0$ , the witness $\min(\varepsilon/2, 1/2) \in (0,\varepsilon) \cap (0,1)$ (the $\min$ is needed since $\varepsilon/2$ can leave $(0,1)$ when $\varepsilon \ge 2$ ), so every neighborhood of $0$ meets $(0,1)$ ; i.e. $0 \in \overline{(0,1)}$ .
Likewise every $\varepsilon$ -ball at $1$ contains $1 - \min(\varepsilon/2, 1/2) \in (0,1)$ , so $1 \in \overline{(0,1)}$ .
If $x < 0$ , the ball of radius $-x/2$ around $x$ lies in $(-\infty, 0)$ , disjoint from $(0,1)$ ; similarly for $x > 1$ . Hence the closure is exactly $[0,1]$ .

Prove $(0,1)$ is not closed, using the closure.

0 \in \overline{(0,1)} \setminus (0,1)

M = 2

— widening the bound does not make it closed.

Show proof

A closed set equals its closure: $A$ is closed ⟺ $\overline{A} = A$ .
By problem 1, $\overline{(0,1)} = [0,1]$ . But $0 \in [0,1]$ while $0 \notin (0,1)$ , so the closure is strictly larger.
Hence $(0,1) \neq \overline{(0,1)}$ , so $(0,1)$ is not closed.

Show $A = \{1/n : n \geq 1\}$ is bounded and prove its closure is $A \cup \{0\}$ .

\overline{\{1/n : n \geq 1\}} = \{1/n : n \geq 1\} \cup \{0\}

The red dot

0

— the limit point the sequence approaches but never touches.

Show proof

For all $n$ , $0 < 1/n \leq 1$ , so with $M = 1$ we have $\lvert x\rvert \leq M$ for $x \in A$ — $A$ is bounded.
Every $\varepsilon$ -ball at $0$ contains $1/N \in A$ for $1/N < \varepsilon$ , so $0$ is a limit point of $A$ : $0 \in \overline{A}$ .
Points outside $A \cup \{0\}$ miss $A$ : for $x < 0$ the ball of radius $\lvert x\rvert/2$ avoids $A$ , similarly for $x > 1$ ; a point of $(0,1] \setminus A$ lies strictly between consecutive terms $1/(n+1) < x < 1/n$ , so a ball of the nearer gap radius avoids $A$ . Hence the closure is exactly $A \cup \{0\}$ .