Open and Closed Sets

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We generalize the notion of open and closed intervals to open and closed sets in $\R ^2$.

When we make definitions and discuss several important theorems for functions of a single variable, we need the notion of an open interval or a closed interval. A typical example of an open interval is $(a,b)$, which represents the set of all $x$ such that $a<x<b$, and an example of a closed interval is $[a,b]$, which represents the set of all $x$ such that $a\leq x\leq b$. We need analogous definitions for open and closed sets in $\R ^n$.

A set is a collection of distinct objects.

Given a set $A$, we say that $a$ is an element of $A$ if $a$ is one of the distinct objects in $A$, and we write $a \in A$ to denote this.

Given two sets $A$ and $B$, we say that $A$ is a subset of $B$ if every element of $A$ is also an element of $B$ write $A \subseteq B$ to denote this.

With these ideas in mind, we now discuss special types of subsets.

(Open Balls)

We give these definitions in general, for when one is working in $\R ^n$ since they are really not all that different to define in $\R ^n$ than in $\R ^2$.

An open ball $B_r(\mathbf {a})$ in $\R ^n$ centered at $\mathbf {a} =\left (a_1,\dots a_n\right ) \in \R ^n$ with radius $r$ is the set of all points $\mathbf {x} = \point {x_1,x_2,\dots ,x_n} \in \R ^n$ such that the distance between $\mathbf {x}$ and $\mathbf {a}$ is less than $r$.

In $\R ^2$ an open ball is often called an open disk.

(Interior and Boundary Points) Suppose that $S \subseteq \R ^n$.

A point $\mathbf {p} \in S$ is an interior point of $S$ if there exists an open ball $B_r(\pt {a}) \subseteq S$.

Intuitively, $\pt {p}$ is an interior point of $S$ if we can squeeze an entire open ball centered at $\pt {p}$ within $S$.
A point $\pt {p} \in \R ^n$ is a boundary point of $S$ if all open balls centered at $\mathbf {p}$ contain both points in $S$ and points not in $S$.
The boundary of $S$ is the set $\partial S$ that consists of all of the boundary points of $S$.

Suppose that $S = \left \{ (x,y) \in \R ^2 ~ \big | ~ x \geq 0, y > 0\right \}$.

Which of the following are elements of $S$?

$\point {0,0}$ $\point {0,2}$ $\point {2,0}$ $\point {2,2}$

Which of the following are interior points of $S$?

$\point {0,0}$ $\point {0,2}$ $\point {2,0}$ $\point {2,2}$

Which of the following are boundary points of $S$?

$\point {0,0}$ $\point {0,2}$ $\point {2,0}$ $\point {2,2}$

The boundary $\partial S $ is $\left \{ (x,y) \in \R ^2 ~ \big | ~ x = 0, y > 0\right \}$$ \left \{ (x,y) \in \R ^2 ~ \big | ~ x = 0, y = 0\right \}$$\left \{ (x,y) \in \R ^2 ~ \big | ~ x > 0, y = 0\right \}$ $\left \{ (x,y) \in \R ^2 ~ \big | ~ x > 0, y > 0\right \}$

We can now generalize the notion of open and closed intervals from $\R $ to open and closed sets in $\R ^n$.

(Open and Closed Sets)

A set $O \subseteq \R ^n$ is open if every point in $O$ is an interior point.
A set $C$ is closed if it contains all of its boundary points.

Determine if the following sets are open, closed, or neither.

The set $\left \{ (x,y) \in \R ^2 \big | |x+y| < 1 \right \}$ is openclosedneither open nor closed.
The set $\left \{ (x,y) \in \R ^2 \big | x \leq 0 , y \leq 0 \right \}$ is openclosedneither open nor closed.
The set $\left \{ (x,y) \in \R ^2 \big | x \leq 0 , y<0\right \}$ is openclosedneither open nor closed.

(Bounded and Unbounded)

A set $S$ is bounded if there is an open ball $B_M(\pt {0})$ such that
\[ S\subseteq B. \]

Intuitively, this means that we can enclose all of the set $S$ within a large enough ball centered at the origin.
A set that is not bounded is called unbounded.

Which of the following sets are bounded?

$\left \{ (x,y) \in \R ^2 \big | |x+y| < 1 \right \}$ $\left \{ (x,y) \in \R ^2 \big | x \leq 0 , y < 0 \right \}$ $\left \{ (x,y) \in \R ^2 \big | x+y^2 < 1 \right \}$ $\left \{ (x,y) \in \R ^2 \big | x^2+y^2 \leq 1 \right \}$

Let’s now look at a few examples.

Consider the function $F(x,y)=\sqrt {1-\frac {x^2}9-\frac {y^2}4}$.

The domain $D$ of the function is the set of all $(x,y)$ for which $1-\frac {x^2}9-\frac {y^2}4 \geq 0$, which we can write in set

\[ D = \{(x,y): \answer [given]{x^2/9+y^2/4}\leq 1\}. \]
The point $(1,1)$ is an interior pointa boundary pointnot an element of $D$.
The point $(1,2)$ is an interior pointa boundary pointnot an element of $D$.
The domain is openclosedneither open nor closed and boundednot bounded .

We’ve already found the domain of this function to be

This is the region bounded by the ellipse $\frac {x^2}9+\frac {y^2}4=1$. Since the region includes the boundary (indicated by the use of “$\leq $”), the set containsdoes not contain all of its boundary points and hence is closed. The region is boundedunbounded as a disk of radius $4$, centered at the origin, contains $D$.

Determine if the domain of $F(x,y) = \frac {1}{x-y}$ is open, closed, or neither, and if it is bounded.

As we cannot divide by $0$, we find the domain to be

\[ D = \{(x,y):x-y\neq \answer [given]{0}\}. \]

In other words, the domain is the set of all points $(x,y)$ not on the line $y=x$. For your viewing pleasure, we have included a graph:

Note how we can draw an open disk around any point in the domain that lies entirely inside the domain, and also note how the only boundary points of the domain are the points on the line $y=x$. We conclude the domain is an open seta closed setneither open nor closed set. Moreover, the set is boundedunbounded.