Elementary Classical Analysis, Marsden, 2nd ed, Chapter 2, Problem 1

Problem

Discuss whether the following sets are open or closed:

(a) (1,2) in $\mathbb{R}$

(b) [2,3] in $\mathbb{R}$

(c) $\bigcap_{n=1}^\infty [-1,1/n)$ in $\mathbb{R}$

(d) $\mathbb{R}^n$ in $\mathbb{R}^n$

(e) A hyperplane in $\mathbb{R}^n$

(f) $\{r \in (0,1) \;|\; \text{$r$ is rational} \}$ in $\mathbb{R}$

(g) $\{(x,y) \in \mathbb{R}^2 \;|\; 0<x \leq 1\}$ in $\mathbb{R}^2$

(h) $\{x \in \mathbb{R}^n \;|\; \|x\|=1\}$ in $\mathbb{R}^n$.

Answer

(a) Open: For $x \in (1,2)$, we can choose $\epsilon = \min(x-1,2-x)$ so that $D(x,\epsilon) \subset (1,2)$.

(b) Closed: For $x \in [2,3]^c$, we can choose $\epsilon = \min(2-x,x-3)$ so that $D(x,\epsilon) \subset [2,3]^c$.

(c) Closed: We first prove that $[-1,0] = A:=\bigcap_{n=1}^\infty [-1,1/n)$. Since $[-1,0] \subset [-1,1/n)$ for all $n \in \mathbb{N}$, $[-1,0] \subset A$.

To prove $A \subset [-1,0]$, we assume that $x \in A$. Then obviously, $x \geq -1$. If $x > 0$, then there exists $n \in \mathbb{N}$ such that $x > 1/n$, which means $x \notin [-1,1/n)$. Hence, $x \leq 0$ and this implies $x \in [-1,0]$.

We can prove that $[-1,0]$ is closed set as in (b).

(d) Open and closed: For all $x \in \mathbb{R}^n$, $D(x,1) \subset \mathbb{R}^n$, obviously, which means $\mathbb{R}^n$ is an open set. Furthermore, $(\mathbb{R}^n)^c$ is a empty set, which is a trivial open set.

(e) Closed: Let $H \subset \mathbb{R}^n$ be a hyperplane. For $x \in H^c$, there is a distance $d$ between $x$ and $H$. Then $D(x,d/2) \subset H^c$.

(f) Neither open nor closed: For $1/2 \in A:=\{r \in (0,1) \;|\; \text{$r$ is rational} \}$ and $\epsilon >0$, $D(1/2,\epsilon)$ contains some irrational numbers in $(0,1)$ since $D(1/2,\epsilon)$ is uncountable.

Conversely, for $\sqrt{2}/2 \in A^c$ and $\epsilon > 0$, $D(\sqrt{2}/2,\epsilon)$ contains some rational number in $A$ since rational number is dense in $\mathbb{R}$

(g) Neither open nor closed: For $(1,0) \in A:=\{(x,y) \in \mathbb{R}^2 \;|\; 0<x \leq 1\}$ and $\epsilon>0$, $D((1,0) ,\epsilon)$ contains $(1+\epsilon/2,0) \in A^c$.

Conversely, For $(0,0) \in A^c$ and $\epsilon >0$, $D((0,0),\epsilon)$ contains $(\epsilon/2,0) \in A$.

(h) Closed: Define $A:=\{x \in \mathbb{R}^n \;|\; \|x\|=1\}$. For $y \in A^c$, we can choose $\epsilon=|1-\|y\||/2$ so that $D(y,\epsilon) \subset A^c$.