# $\{x^2 + y^2<1\}$ is an open set in $\mathbb{R}^3$?

I am asked to show that $$x^2 + y^2<1$$ is an open set in $$\mathbb{R}^3$$.

I'm not sure about this. I guess what was meant is $$\{(x, y) \in \mathbb{R}^2 \mid x^2 + y^2 < 1\}$$ is an open subset of $$\mathbb{R}^2$$

• I would read it as $\{(x,y,z): x^2+y^2<1\}$, which would be an infinite cylinder extending along the $z$-axis. Apr 10, 2021 at 2:50
• I thought so, but I was not sure how to show that it is an open set. Apr 10, 2021 at 2:56
• If $p$ is in your set, show that $B_{d/2}(p)$ is always a subset. Here $d$ should be something like $1-\sqrt{x^2+y^2}$ Apr 10, 2021 at 2:58

One way would be to view $$\mathbb R^3$$ as the product $$\mathbb R^2\times\mathbb R$$. Then we have that the given set is the direct product of $$B^1$$, the unit ball in $$\mathbb R^2$$, with $$\mathbb R$$. Since these are both open sets, their cross-product is open, in the product topology.
the key is how do you describe the topology of $$\mathbb R^3,$$ using the metric or just the product topology. For the latter, it's easy to check that $$A=\{ (x,y,z)\in\mathbb R^3\mid x^2+y^2<1,z\in\mathbb R\}=D^2\times\mathbb R^1 \subset \mathbb R^2 \times\mathbb R^1$$ since $$D^2$$ and $$\mathbb R^1$$ is naturally open in $$\mathbb R^2$$ and $$\mathbb R^1$$, $$A$$ is open in $$\mathbb R^3$$.