# Thread: Proof of Intersection of an open set

1. ## Proof of Intersection of an open set

Prove that the intersection of an open set of $R^2$ with the x-axis is an open set of R $\times$ {0} (where the latter is endowed with the distance d given by d(x, 0) := |x|, for x $\in$ R).

I don't understand what exactly this means.... any help is much appreciated

2. Hi Panda,

let $A\subseteq \mathbb{R}^2$ be the open set. We have to show that $A \cap (\mathbb{R} \times \{0\})$ is an open set of the metric space $\mathbb{R} \times \{0\}$ (endowed with the metric $p((x,0),(y,0)) = |x-y|$ which is the metric induced by the norm $d(x,0) = |x|$. Note that the statement of the problem says $d$ is a distance but it is wrong, $d$ such defined is a norm. We used this norm to induce the distance function (or metric) $p$ ).

This means that for every $(x,0) \in A \cap (\mathbb{R} \times \{0\})$ we must find an open ball in the space $\mathbb{R} \times \{0\}$ with center at $(x,0)$ such that the whole ball is a subset of $A \cap (\mathbb{R} \times \{0\})$.

How do open balls of $\mathbb{R} \times \{0\}$ look like? Given some $\varepsilon > 0$ an open ball of $\mathbb{R} \times \{0\}$ with center at some point $(y,0) \in \mathbb{R} \times \{0\}$ and radius $\varepsilon$ is the set $\{(z,0) \in \mathbb{R} \times \{0\};\, p((z,0),(y,0))<\varepsilon\} = \{(z,0) \in \mathbb{R} \times \{0\};\, |z-y|<\varepsilon\}$. So they're open intervals.

Since $(x,0) \in A \cap (\mathbb{R} \times \{0\})$ implies $(x,0) \in A$ and $A$ is an open set in $\mathbb{R}^2$, there must exist an open ball of $\mathbb{R}^2$ with center at $(x,0)$ and some radius $\varepsilon >0$ such that this whole ball is a subset of $A$. This open ball is the set

$\{(z_1,z_2)\in \mathbb{R}^2;\, ||(z_1,z_2)-(x,0)||<\varepsilon\} = \{(z_1,z_2)\in \mathbb{R}^2;\, ||(z_1-x,z_2)||<\varepsilon\} =$

$=\{(z_1,z_2)\in \mathbb{R}^2;\, \sqrt{(z_1-x)^2+z_2^2}<\varepsilon\}$

Now if we intersect this open ball with $\mathbb{R} \times \{0\}$, we get exactly an open ball of the space $\mathbb{R} \times \{0\}$ centered at $(x,0)$:

$\{(z_1,z_2)\in \mathbb{R}^2;\, \sqrt{(z_1-x)^2+z_2^2}<\varepsilon\} \cap (\mathbb{R} \times \{0\}) = \{(z_1,0)\in \mathbb{R}^2;\, \sqrt{(z_1-x)^2+0}<\varepsilon\}$

$=\{(z_1,0)\in \mathbb{R}^2;\, |z_1-x|<\varepsilon\}$

This ball is obviously a subset of $A \cap (\mathbb{R} \times \{0\})$ (because the original ball is a subset of $A$).
We conclude that the set $A \cap (\mathbb{R} \times \{0\})$ is an open set of the space $\mathbb{R} \times \{0\}$.