Why is $\{(x, y) \in \Bbb{R}^2 | y = x^2\}$ not a function

Question

$B = \{(x, y) \in \Bbb{R}^2 | y = x^2\}$
Why is this not a function?
I understand that to be a function it must pass the Vertical Line Test, my only thought is that for every number in $\Bbb{R}^2$ does not satisfy the condition of $y=x^2$, not sure if that is right though

Answer 1

Based on the notation and the (limited) context in your question, I'm assuming that you have seen a definition of a function as a (sub)set with certain properties. It would however be useful if you could clarify the definition of function that you are using.

A function from and to the reals, so $\mathbb{R} \to \mathbb{R}$, can be seen a subset of $\mathbb{R}^2 = \mathbb{R} \times \mathbb{R}$ that satisfies a specific condition: for every $x \in \mathbb{R}$, there is exactly one $y \in \mathbb{R}$ such that the ordered pair $(x,y)$ is in the set that defines the function.

If you would make a graphical representation of a subset of $\mathbb{R}^2$ by plotting all the points in an $xy$-plane ($x$ horizontal, $y$ vertical), then this condition comes down to what you call the Vertical Line Test: every vertical line will intersect the graph of a function in one point and not in more.

As for the set $B$ in your question: you can plot its graph and see that it does in fact pass the Vertical Line Test or, equivalently, for $y=x^2$ all the elements $(x,x^2)$ satisfy the condition I mentioned before: for every first component $x$ in an ordered pair, the second component $x^2$ is unique.

Changing the relation between $x$ and $y$ from $y=x^2$ to $x=y^2$ would give you a set $B$ that doesn't define a function since, for example, the ordered pairs $(4,2)$ and $(4,-2)$ would be in $B$: this does not satisfy the earlier condition and its graph would also fail the Vertical Line Test.

Long story short: unless you are using a strange definition of a function, the example $B$ in your question is a function.