$B = \{(x, y) \in \Bbb{R}^2 | y = x^2\}$
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
$\begingroup$
$\endgroup$
3
-
2$\begingroup$ That's a function (or "it's the graph of a function", if you prefer). But the corresponding set with $x^2 = y$ is not. $\endgroup$– goblin GONECommented Oct 7, 2016 at 14:46
-
4$\begingroup$ For every $x$, there is exactly one $y$ such that $(x,y)$ is in $B$. It is a function... $\endgroup$– StackTDCommented Oct 7, 2016 at 14:47
-
$\begingroup$ Seems to me that we need to know exactly what your definition of a function is. $\endgroup$– LubinCommented Oct 7, 2016 at 15:58
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
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.
