[Math] How to determine whether domain is an open or a closed region (or both open and closed)

functionsmultivariable-calculus

Right now I am studying Partial Derivative in my university, but I am confused in its basic topic; Multi-variable Functions

More specifically, I am having difficulty in understanding open or closed domain regions in XY Plane and whether they are bounded or not?

For Example:

Question

Given that $f(x, y)=x^2 – y^2$

(a) Find the function's domain
(b) Find the function's range
(c) Describe function's level curves
(d) Find the boundary of the function’s domain
(e) Determine if the domain is an open region, a closed region, or neither
(f) Decide if the domain is bounded or unbounded

Solution

(a) Domain: Entire XY Plane
(b) Range: $(-\infty, \infty)$
(c) Level Curves: $x^2-y^2=c$
(d) Since the domain is entire plane, there is no boundary
(e) Following is my understanding of open and closed domain in XY Plane:

  • A domain (denoted by region R) is said to be closed if the region R contains all boundary points

  • If the region R does not contain any boundary points, then the Domain is said to be open

  • If the region R contains some but not all of the boundary points, then the Domain is said to be both open and closed

The answer in book is: both open and closed, but how can we even say that if there is no boundary as answered in part (d)?

(f) Following is my understanding of bounded and unbounded domains:

  • If the boundary points form a closed loop, then the domain is said to be bounded, otherwise they are considered unbounded.

OR

  • Books Definition: A region in the plane is bounded if it lies inside a disk of finite radius. A region is unbounded if it is not bounded.

The answer of this part in book is: Unbounded, and this is because I think (please correct me if I am wrong), since there is no boundary, the region can not be bounded, so by default, the region is unbounded.

Best Answer

The expression "there is no boundary" must be read as "the boundary is empty".

For region $R$ there is always a set $\partial R$ of boundary points. We have $x\in\partial R$ if every neighborhood of $x$ has a non-empty intersection with $R$ and has a not empty intersection with $R^c=\mathbb R^2\setminus R$.

$R$ is open if it contains none of these boundary points, i.e. if $R\cap\partial R=\varnothing$.

$R$ is closed if it contains all of these boundary points, i.e. if $\partial R\subseteq R$.

In your example $R=\mathbb R^2$ so that $\partial R=\varnothing$. Consequently $R\cap\partial R=\varnothing$ as well as $\partial R\subseteq R$.

So $R$ is open and closed.

There is no $r\in\mathbb R$ such that $R=\mathbb R\subseteq\{\langle x,y\rangle\in\mathbb R^2\mid x^2+y^2\leq r^2\}$. This justifies the statement that $R$ is unbounded.

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