Open and Closed Sets

A set $U$ is open when every point is an interior point — each point can be wrapped in an $\varepsilon$-ball. Move the point $x$ and the radius $\varepsilon$ on the line and watch whether the ball stays inside.

Examples & Exercises

Prove directly from the definition that $(0,1)$ is open.

$$\varepsilon = \min(x, 1-x) > 0$$

Show proof

1. Let $x \in (0,1)$. Since $x > 0$ and $1 - x > 0$, we may put $\varepsilon = \min(x, 1-x) > 0$.
2. Let $y \in (x-\varepsilon, x+\varepsilon)$. From $\lvert y - x\rvert < \varepsilon \leq x$ we get $y > x - \varepsilon \geq 0$, and from $\lvert y - x\rvert < \varepsilon \leq 1 - x$ we get $y < x + \varepsilon \leq 1$.
3. Hence $(x-\varepsilon, x+\varepsilon) \subseteq (0,1)$. Since $x$ was arbitrary, $(0,1)$ is open.

Show $\bigcap_{n=1}^{\infty} (-1/n, 1/n) = \{0\}$, and prove it is not open.

$$\bigcap_{n=1}^{\infty} \left(-\frac{1}{n}, \frac{1}{n}\right) = \{0\}$$

Show proof

1. Let $x \in \bigcap_n (-1/n, 1/n)$. Then $\lvert x\rvert < 1/n$ for every $n$, so letting $n \to \infty$ gives $\lvert x\rvert = 0$, i.e. $x = 0$.
2. Conversely $0$ lies in every $(-1/n, 1/n)$, so the intersection is exactly $\{0\}$.
3. $\{0\}$ is not open: at $0$, for any $\varepsilon > 0$ we have $\varepsilon/2 \in (-\varepsilon, \varepsilon)$ but $\varepsilon/2 \notin \{0\}$, so no $\varepsilon$-ball fits inside $\{0\}$. — infinite intersections of open sets need not be open.

Prove $[0,1]$ is not open.

$$-\frac{\varepsilon}{2} \in (-\varepsilon, \varepsilon), \quad -\frac{\varepsilon}{2} \notin [0,1]$$

Show proof

1. $0 \in [0,1]$. Let $\varepsilon > 0$ be arbitrary.
2. $-\varepsilon/2 \in (-\varepsilon, \varepsilon)$ but $-\varepsilon/2 < 0$, so $-\varepsilon/2 \notin [0,1]$.
3. Hence no $\varepsilon$-ball at $0$ stays inside $[0,1]$; $[0,1]$ is not open.

4. Topology