A vertical line crossing the x-axis at a point $a$ will meet the set in exactly one point $(a, b)$ if $f(a)$ is defined, and $f(a) = b$.
If the vertical line meets the set of points in two points then $f(a)$ is undefined?
A vertical line crossing the x-axis at a point $a$ will meet the set in exactly one point $(a, b)$ if $f(a)$ is defined, and $f(a) = b$.
If the vertical line meets the set of points in two points then $f(a)$ is undefined?
The highlighted proposition is one way of describing the vertical line test, which determines whether $f$ is a function.
If there is one and only point of intersection between $x = a$ and $f(x)$, then $f$ is a function.
If there are two or more points of intersection between $x = a$ and $f(x)$, then $f$ maps a given $x = a$ to two (or more) distinct values, $f(a) = b, f(a) = c, \; b \neq c$, and hence, fails to be a function, by definition of a function. $f$ may, however, a relation.
(Short answer)
No. Rather, we conclude $f$ is a relation, not a function.
Response to comment:
A real function of one variable is really saying three things:
Does this help?
One definition of a function is Formal Dirichlet-Bourbaki definition. This defines a function $f: X\to Y$ to be any subset $S$ of the set $X\times Y$ satisfying that if $(x,y)\in S$ and $(x,y')\in S$, then $y = y'$.
In more vague terms: a function is a rule $f$ that assigns to each element of the domain $X$ exactly one element $y$ in the range. We write $y = f(x)$. So that means that if $x_1 = x_2$, then you will necessarily have $f(x_1) = f(x_2)$.
Now if we are talking about functions from (subsets of) the real numbers to the real numbers, then we have a graph. If in independent variable ($x$) is our horizontal axis and the dependent variable ($y$) our vertical axis, then we can talk about the vertical line test. The graph is then $\{(x,f(x))\mid x\in X\}$. If a vertical line intersects the graph in two points, then we have one value of $x$ corresponding to two distinct values of $y$. That is we have $f(x) = y_1$ and $f(x) = y_2$ say.