Is a circle a function?

Question

I have seen that circle is not function in high school. Why? Because, consider the simply circle equation $x^2+y^2=1$ centered at $(0,0)$ and radius is $1$. And notice that whenever $x=1/2$ then $y$ can be $\sqrt 3 /2$ and $-\sqrt 3/2$.So clearly it is not a function.

Now, I have recently seen the following as follows: function $f: [0,2\pi] \to S^1$ define by $f(t) =(\cos t, \sin t)$ is a continuous bijection. (OR it can be defined $e^{2\pi it} $)

$S^1$ states a circle. $S^1=\{(x,y) \in \mathbb R^2 : x^2+y^2=1 \}$

My question is this circle is not a function but how can $f$ be function? What is the thing I can not see?

Answer 1

The following things hold and are not in conflict with each other:

• There is no function $s\colon D\subset \mathbb{R}\rightarrow \mathbb{R}$, such that $S^1=\{(x,s(x)):x\in D\}$ = graph of $s$
• There is a function $f\colon D\subset\mathbb{R}\rightarrow\mathbb{R}^2$, such that $S^1=f(D)$ = image of $f$; if you choose $D=[0,2\pi)$ and $f(x)=(\sin(x),\cos(x))$, then $f$ will be a bijection onto $S^1$

Answer 2

I suppose that what you were told was that a circle is not the graph of a function. That's so because there are vertical lines which intersect a circle more than once. However, that does not prevent the circle $\{(x,y)\in\Bbb R^2\mid x^2+y^2=1\}$ from being the image of the function $f\colon[0,2\pi]\longrightarrow\Bbb R^2$ defined by $f(\theta)=(\cos\theta,\sin\theta)$.

Answer 3

Circle is a set of points. It is not a function.

The question is: can the circle be a graph of a function of one variable, i.e. mapping real $x$ from some domain into a real $y$? Answer: there is no such function, because (as you noted) a single value (say $x=1/2$) would need to map into multiple variables (say $y=\pm\sqrt{3}/2$).

A different question is: can the circle be an image of a function of one variable? In that case, the real $t$ maps to a point on the plane, i.e. into a pair $(x,y)$. Answer: yes, take $x=\cos t, y=\sin t$.

Thus you have two different questions, and no wonder you have two different answers.

Note that, in the second question, the graph of the function is not a circle; the graph is $3$-dimensional as it needs to accommodate for both $t$ and the pair $(x,y)$, and if you look a bit deeper - it looks like a helix (i.e. a coil), whose projection on the $x,y$ plane is our circle.

Answer 4

I do not know which definition of function you have in mind. Formally you can define a function $f : X \to Y$ as a subset of $X \times Y$ with suitable properties. See the section "Relational approach" in https://en.wikipedia.org/wiki/Function_(mathematics). In this interpretation a function is identified with its Graph $G(f) = \{ (x,f(x)) \mid x \in X \} \subset X \times Y$.

The circle $S^1$ is a subset of the plane. Is it possible that it is the graph of some function $f : X \to \mathbb R$ defined on some subset $X \subset \mathbb R$? Obviously we must have $X = [-1,1]$ because this set is precisely the set of all $x$-coordinates of $S^1$. But if we take any function $f : [-1,1] \to \mathbb R$, then $G(f)$ cannot contain both points $(0,1)$ and $(0,-1)$ on $S^1$. Thus $S^1$ is not the graph of a function.

The function $f: [0,2\pi] \to S^1, f(t) =(\cos t, \sin t)$ is a continuous surjection. Note that $f(0) = f(2\pi)$. You can rewrite it as $f(t) = e^{it}$ if you use complex numbers. However, the function $g: [0,2\pi) \to S^1, g(t) =(\cos t, \sin t)$ is a continuous bijection. The circle is the image of $f$ and of $g$. But the graphs of $f$ and $g$ are no subsets of $\mathbb R^2$, they are contained in $J \times S^1 \subset \mathbb R^3$, where $J = [0,2\pi], [0,2\pi)$. So you see again that $S^1$ is not their graph.