How can the y-axis in $\mathbb{R^2}$ be open?

Question

I have read that $\{(x, \frac{1}{x}): x \neq 0\}$ is closed in $\mathbb{R^2}$.

So hence the complement of this set, $\{x = 0\}$, i.e. the y-axis must be open? But we cannot put an open ball with radius $r > 0$ on any point in the $y$ axis without that ball no longer being fully contained in the y-axis. So how can it be open?

Answer 1

The $y$-axis can be characterized as the set $$ \{(x,y):x=0\} $$ and so it's closed, because the projection to the first coordinate is continuous.

This set is definitely not the complement of $$ H=\left\{\left(x,\frac{1}{x}\right):x\ne0\right\} $$ because, for instance, $(2,2)$ belongs to the complement of $H$ but not to the $y$-axis.

Answer 2

If you draw the set $\{(x,1/x) \, | \, x \neq 0\}$ you get a very familiar graph, namely a hyperbola. (It is also the set $\{(x,y) \, |, \, xy = 1$.)

If you google "graph of xy = 1" (as I just did) you should get lots of images. Once you see them, you won't have trouble seeing that this graph is closed, that it's complement is open, and that it's complement is much bigger than just the $y$-axis.