Open sets and closed sets in Rn

Open sets

In $\mathbb{R}$, we have the notion of an open interval such as $(a,b)=\{x\in \mathbb{R}|a<x<b\}$. We want to extend this idea to apply to ${\mathbb{R}}^n$. We also introduce the notions of bounded sets and closed sets in ${\mathbb{R}}^n$.

Recall that the $\epsilon$-neighbor of a point $x\in {\mathbb{R}}^n$ is defined as $N_{\epsilon }(x)=\{y\in {\mathbb{R}}^n|d(x,y)<\epsilon \}$ where $d(x,y)$ is the distance between $x$ and $y$.

Definition (Open set)

A subset $U$ of ${\mathbb{R}}^n$ is said to be an open set if the following holds:

$$\begin{array}{}\text{(Eq:OpenSet)}& \mathrm{\forall }x\in U\mathrm{\exists }\delta >0(N_{\delta }(x)\subset U).\end{array}$$

That is, for every point in an open set $U$, we can always find an open ball centered at that point, that is included in $U$. See the following figure.

Perhaps, it is instructive to see what is not an open set. Negating (Eq:OpenSet), we have $$\mathrm{\exists }x\in U\mathrm{\forall }\delta >0(N_{\delta }(x)\not\subset U).$$ This means that there is some point in $U$ such that its neighbor cannot be included in $U$ (i.e., some part of the neighbor crosses the "boundary" to go out of $U$.) 

Example. The open interval $(0,1)$ in $\mathbb{R}$ is an open set. To see this, take any point $x\in (0,1)$ and let $\delta =min\{x,1-x\}$ (note: $x>0$ and $1-x>0$). Then, we have the $\delta$-neighbor $N_{\delta }(x)=(x-\delta ,x+\delta )\subset (0,1)$. □

Example. The closed interval $[0,1]$ is not an open set. To see this, let $x=1\in [0,1]$. For any $\delta >0$, we have the neighbor $N_{\delta }(1)=(1-\delta ,1+\delta )\not\subset [0,1]$ because $1+\delta >1$.  □

Definition (Path-connected)

1. For each $i=1,2,\cdots ,n$, let ${\sigma }_i(t)$ be a continuous function on $[0,1]$. Then the $n$-tuple of these functions, $\sigma (t)=({\sigma }_1(t),{\sigma }_2(t),\cdots ,{\sigma }_n(t))$ is a point in ${\mathbb{R}}^n$, parameterized by $t\in [0,1]$. The set defined by $$\sigma =\{\sigma (t)\mid t\in [0,1]\}$$ is called a path in ${\mathbb{R}}^n$. The points $\sigma (0)$ and $\sigma (1)$ are called the initial and terminal points, respectively, and both are collectively called the endpoints.
2. If the path $\sigma$ in ${\mathbb{R}}^n$ is included in the subset $S$ of ${\mathbb{R}}^n$, that is, $\sigma \subset S$, then $\sigma$ is said to be a path in $S$.
3.  For the subset $S$ of ${\mathbb{R}}^n$, for any $x,y\in S$, if there exists at least one path $\sigma$ with $x$ and $y$ being its endpoints, $S$ is said to be path-connected.

Example. Let $o=(0,0)$, $a=(3,0)$ in ${\mathbb{R}}^2$ and consider the union $S$ of  1-neighbors (i.e., $\epsilon =1$): $S=N_1(o)\cup N_1(a)$ (Draw pictures!). No matter how we connect the two points $o$ and $a$ by a path (continuous line or curve), there are always some points not in $S$. Therefore, $S$ is not path-connected. □

Definition (Open region in ${\mathbb{R}}^n$)

A path-connected open set in ${\mathbb{R}}^n$ is called an open region in ${\mathbb{R}}^n$.

Example. ${\mathbb{R}}^2$ is an open region. So is $\{(x,y)|y>0\}$. However, $\{(x,y)|y\ge 0\}$ is not. □

Theorem (Any open ball is an open region). 

The $r$-neighbor of $x=(x_1,x_2,\cdots ,x_n)$ in ${\mathbb{R}}^n$, $N_r(x)$, is an open region.

Proof.   First, we show that $N_r(x)$ is an open set. Take an arbitrary $y\in N_r(x)$ and let $d=d(x,y)$. Noting $d<r$, let $\delta =r-d(>0)$. It suffices to show that $N_{\delta }(y)\subset N_r(x)$. By the triangle inequality, for any $z\in N_{\delta }(y)$, $$d(x,z)\le d(x,y)+d(y,z)<d+\delta =r$$ so that $z\in N_r(x)$, and hence $N_{\delta }(y)\subset N_r(x)$.

Next, we show that $N_r(x)$ is path-connected. Let ${\sigma }_y(t)$ denote a path (if exists) that starts at $y\in N_r(x)$ (i.e., ${\sigma }_y(0)=y$) and ends at $x$ (i.e., ${\sigma }_y(1)=x$). For any $y,z\in N_r(x)$, define the following path:

$${\sigma }_{yz}(t)=\{\begin{array}{lc}{\sigma }_y(2t)& (0\le t\le \frac{1}{2}),\\ {\sigma }_z(2-2t)& (\frac{1}{2}<t\le 1),\end{array}$$ then this is a path from $y$ to $z$ in $N_r(x)$. We still need to show such path as ${\sigma }_y(t)$ for each $y=(y_1,y_2,\cdots ,y_n)\in N_r(x)$ exists in $N_r(x)$. Let us define, for example,

$$\begin{array}{rcl}{\sigma }_y(t)& =& (t{x}_1+(1-t)y_1,\cdots ,t{x}_n+(1-t)y_n)\\ & =& tx+(1-t)y\\ & =& y+t(x-y)\end{array}$$ for $t\in [0,1]$. This is a path (line segment) from $y={\sigma }_y(0)$ to $x={\sigma }_y(1)$. Furthermore, $$d(x,{\sigma }_y(t))=\Vert {\sigma }_y(t)-x\Vert =(1-t)\Vert y-x\Vert =(1-t)d(x,y)<r.$$ Thus, ${\sigma }_y(t)\in N_r(x)$ for all $t\in [0,1]$. 

This proves that $N_r(x)$ is path-connected. ■

Closed sets

The Intermediate Value Theorem can be extended to multivariate functions on a path-connected set. To extend the Extreme Value Theorem to multivariate functions, we need the notion of bounded closed sets.

Definition (Bounded set)

The subset $S$ of ${\mathbb{R}}^n$ is said to be bounded if there exist $x\in {\mathbb{R}}^n$ and $r>0$ such that $S\subset N_r(x)$.

That is, a set is bounded if it can be "covered" by an open ball.

Remark. Let $d$ be the distance between $x$ and the origin $O(0,0,\cdots ,0)$. Then (draw a figure!) $$N_r(x)\subset N_{d+r}(O).$$ 

Therefore, in the above definition, we can replace the point $x$ with the origin.

Example. In ${\mathbb{R}}^2$, the rectangular region $[a,a^{\prime }]\times [b,b^{\prime }]$ where  $-\mathrm{\infty }<a<a^{\prime }<\mathrm{\infty },-\mathrm{\infty }<b<b^{\prime }<\mathrm{\infty }$ is bounded. To see this, let

$$r=max\{\frac{a^{\prime }-a}{2},\frac{b^{\prime }-b}{2}\}$$ and $$P(\frac{a+a^{\prime }}{2},\frac{b+b^{\prime }}{2}).$$

Then, for any real number $R$ such that $R>\sqrt{2}r$, we have $$[a,a^{\prime }]\times [b,b^{\prime }]\subset N_R(P).$$ □

Definition (Closed set)

The subset $F$ of ${\mathbb{R}}^n$ is said to be a closed set if ${\mathbb{R}}^n\setminus F$ (i.e., the complement of $F$ in ${\mathbb{R}}^n$), is an open set. A bounded closed set is a closed set that is bounded.

Remark. Be careful: "The set $A$ is not an open set" does not necessarily imply that $A$ is a closed set! See the following proposition.

Proposition

The semi-closed ("clopen") interval $[0,1)\subset \mathbb{R}$ is neither open nor closed.

 Proof. Let $x=0\in [0,1)$. For any $\delta >0$ and any point $y\in (-\delta ,0)\subset N_{\delta }(0)$, we have $y\notin [0,1)$. Hence $N_{\delta }(0)\not\subset [0,1)$. Accordingly, $[0,1)$ is not open. 

Next, consider the complement $[0,1{)}^c=(-\mathrm{\infty },0)\cup [1,\mathrm{\infty })$. Take $x=1\in [0,1{)}^c$. For any $\delta >0$ and any point $y\in (1-\delta ,1)\subset N_{\delta }(1)$, we have $y\notin [0,1{)}^c$. Hence $N_{\delta }(1)\not\subset [0,1{)}^c$. That is, $[0,1{)}^c$ is not open, and therefore, $[0,1)$ is not closed. ■

Example. Let us show that the rectangular region $[a,a^{\prime }]\times [b,b^{\prime }]$ in ${\mathbb{R}}^2$ is a closed set. Let $F=[a,a^{\prime }]\times [b,b^{\prime }]$ and $U={\mathbb{R}}^2\setminus F$. We need to show that $U$ is open. Take an arbitrary point $p=(x,y)\in U$. Since $p\notin F$, it is either $x\notin [a,a^{\prime }]$ or $y\notin [b,b^{\prime }]$. 

Consider the case that $x\notin [a,a^{\prime }]$ (the other case is similar). Then, either $x<a$ or $x>a^{\prime }$. Consider the case that $x>a^{\prime }$ (the other case is similar). Let $\delta =x-a^{\prime }$. Then $\delta >0$. For any point $q\in N_{\delta }(p)$, its $x$-coordinate is greater than $a^{\prime }$. Thus, $q\notin F$, that is, $q\in U$. This implies that $N_{\delta }(p)\subset U$. Hence $U$ is open. Accordingly, $F$ is closed. □