Closure and Boundedness

Drawing a bar over $A$ gives the closure $\overline{A}$ — the set with its escaped limit points pulled back in. Follow the round-song: closed ⟺ contains its limit points ⟺ equals its closure, and closed + bounded ⟺ compact (Heine-Borel).

Examples & Exercises

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

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

Show proof

1. Points of $(0,1)$ are trivially in the closure. Let $x = 0$: for every $\varepsilon > 0$, $\varepsilon/2 \in (0,\varepsilon) \cap (0,1)$, so every neighborhood of $0$ meets $(0,1)$; i.e. $0 \in \overline{(0,1)}$.
2. Likewise every $\varepsilon$-ball at $1$ contains $1 - \varepsilon/2 \in (0,1)$, so $1 \in \overline{(0,1)}$.
3. 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.

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

Show proof

1. A closed set equals its closure: $A$ is closed ⟺ $\overline{A} = A$.
2. By problem 1, $\overline{(0,1)} = [0,1]$. But $0 \in [0,1]$ while $0 \notin (0,1)$, so the closure is strictly larger.
3. 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 $A \cup \{0\}$.

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

Show proof

1. 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.
2. Every $\varepsilon$-ball at $0$ contains $1/N \in A$ for $1/N < \varepsilon$, so $0$ is a limit point of $A$: $0 \in \overline{A}$.
3. 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 $A \cup \{0\}$.

4. Topology