# Is a circle a function?

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?

• $\Bbb{S^1}=Graph(f)$ Commented Jan 13, 2022 at 11:06
• Your $f$ is a function of $t$ rather than of $x$, with a different codomain Commented Jan 13, 2022 at 11:14
• $S^1$ is not a function but $f$ is a function. Why is that a problem? They're two different things... Commented Jan 13, 2022 at 12:00

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 $$S^1$$

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 $$f(\theta)=(\cos\theta,\sin\theta)$$.

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.

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.