1. ## Open Set

Show that $U=\{ (x,y,z) \in \mathbb{R}^3 \ : \ \frac{e^{x+y^2-z}-1}{x^2+y^2-z^3}>7 \}$ is open in $\mathbb{R}^3$.
$U$ can be rearranged to get:

$U=\{(x,y,z) \in \mathbb{R}^3 \ : \ e^{x^2+y^2-z}-1-7x^2-7y^2+7z^3>0 \}$

The definition of an open ball is $(\forall u \in U)(\exists \varepsilon_w>0)[B_{\varepsilon_w}(u,d) \subset U]$ where $w \in U$.

Define $\varepsilon_w:=\frac{1}{2} \left(e^{x+y^2-z}-1-7x^2-7y^2+7z^3 \right)$

The fact that it is halved means that any point in any ball around a point $u$ will be greater than 0 and will therefore be in $U$. This ball is open.

Since this is valid $\forall u \in U$, $U$ is expressed as a union of open balls ie. $\cup_{w \in U} B_{\varepsilon_w}(u,d)=U$

Hence $U$ is open.

Is this right? I'm wondering if I have made any syntax errors!

2. Originally Posted by Showcase_22
$U$ can be rearranged to get:

$U=\{(x,y,z) \in \mathbb{R}^3 \ : \ e^{x^2+y^2-z}-1-7x^2-7y^2+7z^3>0 \}$

The definition of an open ball is $(\forall u \in U)(\exists \varepsilon_w>0)[B_{\varepsilon_w}(u,d) \subset U]$ where $w \in U$.

Define $\varepsilon_w:=\frac{1}{2} \left(e^{x+y^2-z}-1-7x^2-7y^2+7z^3 \right)$

The fact that it is halved means that any point in any ball around a point $u$ will be greater than 0 and will therefore be in $U$. This ball is open.

Since this is valid $\forall u \in U$, $U$ is expressed as a union of open balls ie. $\cup_{w \in U} B_{\varepsilon_w}(u,d)=U$

Hence $U$ is open.

Is this right? I'm wondering if I have made any syntax errors!
In a metric space $\left(X,d\right)$ as et $O$ is open if and only if for every $x\in O$ there exists a $\delta>0$ such that $B_{\delta}(x)\subseteq O$. Now, have you done that?

I am honestly not even sure what you are doing here. Are we working with the usual metric?

For your definition of an open ball, what is $\varepsilon_w$, I don't understand the $w$ part?

3. I'm trying to show that my set is a union of open balls so it's open. I'm not sure if it's the easiest way of doing it, but I think it might work.

I'm afraid that my question does not say. Is it possible to prove it for every metric? I not, it probably means Euclidean.

I was thinking that for every point (x,y,z) $e^{x+y^2-z}-1-7x^2-7y^2+7z^3
$
will produce a single number. If I create a ball of radius $

\delta=\frac{1}{2} \left(e^{x+y^2-z}-1-7x^2-7y^2+7z^3 \right)
$
and centre (x,y,z), (the $B_{\delta}(x,y,z)$ that you wrote) then this will be an open ball that is always a subset of U.

Is there something wrong with my idea?

4. Originally Posted by Showcase_22
I'm trying to show that my set is a union of open balls so it's open. I'm not sure if it's the easiest way of doing it, but I think it might work.

I'm afraid that my question does not say. Is it possible to prove it for every metric? I not, it probably means Euclidean.

I was thinking that for every point (x,y,z) $e^{x+y^2-z}-1-7x^2-7y^2+7z^3
$
will produce a single number. If I create a ball of radius $

\delta=\frac{1}{2} \left(e^{x+y^2-z}-1-7x^2-7y^2+7z^3 \right)
$
and centre (x,y,z), (the $B_{\delta}(x,y,z)$ that you wrote) then this will be an open ball that is always a subset of U.

Is there something wrong with my idea?
What you are saying is analogous to this. Let $O=\left\{x\in\mathbb{R}:x^3>0\right\}$. To see that this is open we merely note that choosing $\frac{x^3}{2}=\delta$ guarantees that $B_{\delta}(x)\subseteq O$. Take $x=2\implies \delta=4$. Then, $B_\delta\left(2\right)\nsubseteq O$

5. ah, so I was wrong!

In your example of $

O=\left\{x\in\mathbb{R}:x^3>0\right\}
$
, does $\delta=\frac{x}{2}$ work in the ball $B_{\delta}(x)$?

It works in the case $x=2$. The set is bounded below by 0 so if $a \in O$, then $\frac{a}{2}$ is always in the set since it's the same distance from a as it is 0.

So extending this idea to the set $U$ gives $\delta= \frac{\frac{e^{x+y^2-z}-1}{x^2+y^2-z^3}-7}{2}$.

I think i'm starting to get it!