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

Question

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 $\mathbb{R}^2$

Answer 1

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.

Answer 2

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 $\mathbb R^3$.