# Description of sets

#### mathmari

##### Well-known member
MHB Site Helper
Hey!! • I want to describe in words the following sets:

1. $A:=\{(x,y)\in \mathbb{R}^2\mid x>0, y\leq 1\}$

$A$ is the set of all pointgs where the first coordinate is positiv and the second one is less or equal to $1$.

It is the subarea of the plane that is under the point $(0/1)$ to the right, without the y-axis.

Is this description enough or can we say also something else? The graphical representation is: 2. $C:=\mathbb{N}\times \{x\in \mathbb{R}\mid 0\leq x\leq 2\}$

$C$ is the cartesian product of the natural number and the interval $[0,2]$. It s the set of points with two coordinates $(n,x)$, where the first coordinate is a natural number and the second coordinate is a real number in the interval $[0,2]$.

What else can we say here?

How does the graphical representation look like? • I want to give also the corresponding set fo the following: It is $\{(x,y)\mid 0\leq x \leq 2, 1\leq y\leq 3\}$, right? Is this enough, or could we also justify that it is like that? • $F$ is the set of all points in the plane that are at least as far from the origin as $P = (3 \mid 0)$. Does this mean that we have the set $$F=\{(x,y)\mid x^2+y^2\leq (x-3)^2+y^2\}$$ or have I understood wrong the definition of $F$ ? #### Klaas van Aarsen

##### MHB Seeker
Staff member
Hey!! • I want to describe in words the following sets:

1. $A:=\{(x,y)\in \mathbb{R}^2\mid x>0, y\leq 1\}$

$A$ is the set of all pointgs where the first coordinate is positiv and the second one is less or equal to $1$.

It is the subarea of the plane that is under the point $(0/1)$ to the right, without the y-axis.

Is this description enough or can we say also something else? The graphical representation is:
Hey mathmari !!

Seems fine to me. Btw, shouldn't it be $(0,1)$ instead of $(0/1)$?

• 2. $C:=\mathbb{N}\times \{x\in \mathbb{R}\mid 0\leq x\leq 2\}$

$C$ is the cartesian product of the natural number and the interval $[0,2]$. It s the set of points with two coordinates $(n,x)$, where the first coordinate is a natural number and the second coordinate is a real number in the interval $[0,2]$.

What else can we say here?

How does the graphical representation look like?
It's a set of vertical line segments isn't it? • 
• I want to give also the corresponding set fo the following:

It is $\{(x,y)\mid 0\leq x \leq 2, 1\leq y\leq 3\}$, right? Is this enough, or could we also justify that it is like that?
It seems as if it should be $1 < y < 3$, shouldn't it?

• $F$ is the set of all points in the plane that are at least as far from the origin as $P = (3 \mid 0)$. Does this mean that we have the set $$F=\{(x,y)\mid x^2+y^2\leq (x-3)^2+y^2\}$$ or have I understood wrong the definition of $F$ ?
What is $P = (3 \mid 0)$? #### mathmari

##### Well-known member
MHB Site Helper
Seems fine to me. Btw, shouldn't it be $(0,1)$ instead of $(0/1)$?
Oh yes It's a set of vertical line segments isn't it? So, we get the following, or not?  What is $P = (3 \mid 0)$? Oh I meant $P(3, 0)$. #### Klaas van Aarsen

##### MHB Seeker
Staff member
So, we get the following, or not?
Yes, although I'd be inclined to draw them vertically.
That is because we usually draw the first coordinate along the x-axis, and the second along the y-axis. Oh I meant $P(3, 0)$.
Okay, so we're talking about the points that have at least the same distance to the origin as $(3,0)$ yes?
Its distance is $3$.
So we're looking at all points with $\sqrt{x^2+y^2} \ge 3$, don't we? .

#### mathmari

##### Well-known member
MHB Site Helper
Yes, although I'd be inclined to draw them vertically.
That is because we usually draw the first coordinate along the x-axis, and the second along the y-axis. Oh yes, you're right! So, we get the following, don't we?  Okay, so we're talking about the points that have at least the same distance to the origin as $(3,0)$ yes?
Its distance is $3$.
So we're looking at all points with $\sqrt{x^2+y^2} \ge 3$, don't we? .
Ahh ok!! So we have the set $$F=\{(x,y)\mid \sqrt{x^2+y^2} \ge 3\}$$ right? #### Klaas van Aarsen

##### MHB Seeker
Staff member
Oh yes, you're right! So, we get the following, don't we?

Ahh ok!! So we have the set $$F=\{(x,y)\mid \sqrt{x^2+y^2} \ge 3\}$$ right?
Yep. 