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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 $\mathbb{R}^2$

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

2 Answers 2

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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.

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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 $\mathbb R^3$.

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