Is a circle a closed subset of $\mathbb{R}^2$?

Question

Intuitively, I think that a circle is a closed subset of $\mathbb{R}^2$. But, how I can prove this?

Answer 1

It is the zero-locus of the continuous function $(x,y) \mapsto x^2+y^2-1 : \Bbb R^2 \to \Bbb R$, and $\{0\}$ is closed in $\Bbb R$, so $\{ (x,y) \mid x^2+y^2-1 \}$ is closed in $\Bbb R^2$.

Answer 2

We need to be careful with terminology. Centre on the origin and set the radius to $1$. The open disk is $r<1$; the closed disk is $r\ge 1$; the circle is $r=1$. As you can probably guess from the name (and can easily prove yourself), the first of these is open not closed, while the latter is the reverse. As for the circle, it's not open in $\mathbb{R}^2$, although it is closed because $r\ne 1$ is also open.