Is this a well-formed set-builder notation?

[Dethrone]

Are these three sets equivalent?

$$A=\left\{(x,y):x,y\in\Bbb{R},y\ge x^2-1\right\}$$
$$B=\left\{x,y\in\Bbb{R}:y\ge x^2-1\right\}$$
$$C=\left\{(x,y)\in\Bbb{R}^2:y\ge x^2-1\right\}$$

I am thinking that $A$ and $C$ are, but not $B$ as it might be ambigious as to which dimension it is in, i.e it could be in $\Bbb{R}^3$, where any value of $z$ will satisfy. Am I right? :D

 

[evinda]

Are these three sets equivalent?

$$A=\left\{(x,y):x,y\in\Bbb{R},y\ge x^2-1\right\}$$
$$B=\left\{x,y\in\Bbb{R}:y\ge x^2-1\right\}$$
$$C=\left\{(x,y)\in\Bbb{R}^2:y\ge x^2-1\right\}$$

I am thinking that $A$ and $C$ are, but not $B$ as it might be ambigious as to which dimension it is in, i.e it could be in $\Bbb{R}^3$, where any value of $z$ will satisfy. Am I right? :D

(Wave) Happy new year! (Party)

Yes, you are right.

At the set $A$ we have the ordered pair $(x,y)$ such that $x,y \in \mathbb{R}$ so it is meant that $(x,y) \in \mathbb{R}^2$.
The set $B$ isn't equivalent to the other two because of the fact that at the sets $A,C$ we consider an ordered pair and and so the order in which the objects appear is significant, but for the set $B$ this doesn't hold.
For example, if we are given $x=5$, $y=1$, we check if $(5,1) \in A$ that does not hold since it doesn't hold that $1 \geq 5^2-1$.
For the set $B$ we check if $y \geq x^2-1$ which holds by taking $y=5$ and $x=1$.

 

[Evgeny.Makarov]

Are these three sets equivalent?

Sets can be equal, and definitions can be equivalent.

$$B=\left\{x,y\in\Bbb{R}:y\ge x^2-1\right\}$$

This is not a well-formed set-builder notation. A well-formed notation has the form $\{x\mid P(x)\}$, $\{x\in A\mid P(x)\}$ or $\{f(x)\mid P(x)\}$, but not $\{x,y\in A\mid P(x,y)\}$.