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

Question

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.

Answer 1

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).