Open and Closed Sets

This scene shows what an open set is. Move the point and the radius, and check whether the small ball fits inside.

Open means no point clings to the boundary. Around every point there is a small interval lying inside the set. At a boundary point, every interval, however small, pokes outside.

For example, kick a ⚽ ball in the middle of a yard. A gentle kick keeps it inside. Kick from right at the fence, and it flies out at once. That is the difference between an interior point and a boundary point.

U open   ⟺  ∀x∈U ∃ε>0:(x−ε,x+ε)⊆UU \text{ open } \iff \forall x \in U\ \exists \varepsilon > 0 : (x - \varepsilon, x + \varepsilon) \subseteq U
F closed   ⟺  Fc openF \text{ closed } \iff F^c \text{ open}

Examples & Exercises

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

ε=min⁡(x,1−x)>0\varepsilon = \min(x, 1-x) > 0
x=0.5x = 0.5, ε=0.2\varepsilon = 0.2 — the ball stays inside (0,1)(0,1).
Show proof

  1. Let x∈(0,1)x \in (0,1). Since x>0x > 0 and 1−x>01 - x > 0, we may put ε=min⁡(x,1−x)>0\varepsilon = \min(x, 1-x) > 0.

  2. Let y∈(x−ε,x+ε)y \in (x-\varepsilon, x+\varepsilon). From ∣y−x∣<ε≤x\lvert y - x\rvert < \varepsilon \leq x we get y>x−ε≥0y > x - \varepsilon \geq 0, and from ∣y−x∣<ε≤1−x\lvert y - x\rvert < \varepsilon \leq 1 - x we get y<x+ε≤1y < x + \varepsilon \leq 1.

  3. Hence (x−ε,x+ε)⊆(0,1)(x-\varepsilon, x+\varepsilon) \subseteq (0,1). Since xx was arbitrary, (0,1)(0,1) is open.

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

⋂n=1∞(−1n,1n)={0}\bigcap_{n=1}^{\infty} \left(-\frac{1}{n}, \frac{1}{n}\right) = \{0\}
{0}\{0\} — no ε\varepsilon-ball fits inside a single point.
Show proof

  1. Let x∈⋂n(−1/n,1/n)x \in \bigcap_n (-1/n, 1/n). Then ∣x∣<1/n\lvert x\rvert < 1/n for every nn. If ∣x∣>0\lvert x\rvert > 0, the Archimedean property gives an nn with 1/n<∣x∣1/n < \lvert x\rvert, contradicting ∣x∣<1/n\lvert x\rvert < 1/n.

    Hence ∣x∣=0\lvert x\rvert = 0, i.e. x=0x = 0.

  2. Conversely 00 lies in every (−1/n,1/n)(-1/n, 1/n), so the intersection is exactly {0}\{0\}.

  3. {0}\{0\} is not open: at 00, for any ε>0\varepsilon > 0 we have ε/2∈(−ε,ε)\varepsilon/2 \in (-\varepsilon, \varepsilon) but ε/2∉{0}\varepsilon/2 \notin \{0\}, so no ε\varepsilon-ball fits inside {0}\{0\}. — infinite intersections of open sets need not be open.

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

−ε2∈(−ε,ε),−ε2∉[0,1]-\frac{\varepsilon}{2} \in (-\varepsilon, \varepsilon), \quad -\frac{\varepsilon}{2} \notin [0,1]
x=0x = 0 — at a boundary point every ε\varepsilon-ball leaks out.
Show proof

  1. 0∈[0,1]0 \in [0,1]. Let ε>0\varepsilon > 0 be arbitrary.

  2. −ε/2∈(−ε,ε)-\varepsilon/2 \in (-\varepsilon, \varepsilon) but −ε/2<0-\varepsilon/2 < 0, so −ε/2∉[0,1]-\varepsilon/2 \notin [0,1].

  3. Hence no ε\varepsilon-ball at 00 stays inside [0,1][0,1]; [0,1][0,1] is not open.

5. Topology