# How to show that the set of points satisfying an arbitrary circle in a 2D Euclidean plane $(\mathbb R \times \mathbb R)$ is infinite?

For a problem that I am currently working on, I want to demonstrate that the following set, with $$d \gt 0$$, is of infinite size (whether or not the set is countable or not is unimportant):

$$S=\left\{(x',y') \in \mathbb R \times \mathbb R : \sqrt{(x'-x_0)^2+(y'-y_0)^2}=d \gt 0\right\}$$

The above equation should be recognizable as the formula for a circle in a 2D Euclidean plane that has its origin at $$(x_0,y_0)$$.

To show that this $$S$$ has infinite members, I created the following function: $$f: \mathbb N \to [0,\frac{\pi}{2})$$, where $$f(n)=\frac{\pi}{2n}\cdot(n-1)$$. Note that $$f$$ is a strictly increasing function, which means that $$f$$ is injective. For notation convenience, let $$\theta_n:=f(n)$$.

Next, to each unique $$\theta_n$$, assign the ordered pair $$\left(\cos(\theta_n))\cdot d+x_0,\sin(\theta_n)\cdot d+ y_0 \right)$$. Because $$\cos$$ and $$\sin$$ are functions, clearly the assignment of $$\theta_n$$ to $$\left(\cos(\theta_n))\cdot d+x_0,\sin(\theta_n)\cdot d+ y_0 \right)$$ is unambiguous. This means that we can consider the function $$g: [0,\frac{\pi}{2}) \to \mathbb R \times \mathbb R$$, where $$g(\theta_n)=\left(\cos(\theta_n))\cdot d+x_0,\sin(\theta_n)\cdot d+y_0 \right)$$. It should be straightforward to see that setting $$x'=\cos(\theta_n)\cdot d+x_0$$ and $$y'=\sin(\theta_n)\cdot d+y_0$$ will solve the desired equation of $$S$$.

Note that over the interval $$[0,\frac{\pi}{2})$$, $$\cos$$ is injective (because it is strictly decreasing). This means that for $$n_1 \neq n_2 \in [0,\frac{\pi}{2})$$: $$\cos(\theta_{n_1}) \neq \cos(\theta_{n_2})$$ ...which means that $$\left(\cos(\theta_{n_1}))\cdot d+x_0,\sin(\theta_{n_1})\cdot d+y_0 \right) \neq \left(\cos(\theta_{n_2}))\cdot d+x_0,\sin(\theta_{n_2})\cdot d+y_0 \right)$$. We can then say that $$g$$ is injective.

Because $$f$$ is injective and $$g$$ is injective, we must have that their composition is injective. That is to say that the function $$h = g \circ f : \mathbb N \to \mathbb R \times \mathbb R$$ is injective. Given that $$\mathbb N$$ is an infinite set, this necessarily means that the $$\text{image}(h)$$ must also be a set of infinite size. Further, $$\text{image}(h) \subseteq S$$, by construction. This means that $$S$$ must be an infinite set, as well.

I think the logic for this argument is fine, but I was wondering if there were more standard approaches to show that a set has an infinite size.

• That is not how to specify that $d>0$. "I want to demonstrate that if $d>0$, then the following set is of infinite size". Commented Mar 19, 2022 at 19:13

For any $$x\in[x_0-d,x_0+d]$$, let $$y=y_0+\sqrt{d^2-(x-x_0)^2}$$. Then $$(x,y)\in S$$. Assuming $$d>0$$ (which you forgot to specify), there are an infinite (indeed, uncountable) number of such $$x$$. Hence $$S$$ is infinite (indeed, uncountable).
• Could you please explain why $(x,y) \in S$? Plugging in your values for $x'$ and $y'$, I get: $$\sqrt{(x-x_0)^2+\left(\sqrt{(x-x_0)^2}\right)^2}$$, which equals $$\sqrt{(x-x_0)^2+(x-x_0)^2}$$, equaling $$(x-x_0)\cdot \sqrt{2}$$ Why should this equal $d$? Thanks~ Commented Mar 19, 2022 at 19:25
• Took a few to derive, but, yes, I agree with your claim. Note that we assumed $d \gt 0$, $|x-x_0| \lt d$, and $y \gt y_0$ in order to arrive at your result...starting from the equation: $$d^2=(x-x_0)^2+(y-y_0)^2$$ Commented Mar 19, 2022 at 20:14