Let $A= \{(x,y,z) \in \Bbb R^3 \vert x+y
Let $A= \{(x,y,z) \in \Bbb R^3 \vert x+y<z < x^2+y^2 \}$. Show that $A$ is an open set in $\Bbb R^3$ defined by the Euclidean metric.
So $A$ can be written as $A = \{(x,y,z) \in \Bbb R^3 \vert x+y-z<0, x^2+y^2-z>0 \} = \{(x,y,z) \in \Bbb R^3 \vert x+y-z<0\} \cap \{(x,y,z) \in \Bbb R^3 \vert x^2+y^2-z>0\}$.
Now the book I'm reading had solved this defining  $f,g : \Bbb R^3 \to \Bbb R$, $f(x) = x+y-z$ and $g(x) = x^2+y^2-z$. Showing that these two functions are continuous seemed to imply that $A$ is open? I'm not yet on the chapter that introduces continuity in metric spaces so I was thinking if there's any other way to show that $A$ would be open? I know the definition of continuity in metric spaces, but here they used some projections, etc. which I'm not familiar yet.
 A: The method using continuous functions is by far the least messy way to solve such problems.
Define $f(x,y,z) = x+y-z$ and $g(x,y,z) = x^2+y^2-z$. Then $x+y < z$ can be described as $f(x,y,z) < 0$ and $z < x^2+y^2$ as $0 < g(x,y,z)$. So
$$A = f^{-1}[(\leftarrow, 0)] \cap g^{-1}[(0,\rightarrow)]$$ which is the intersection of two open sets (as $f,g$ are continuous) and thus open (axiom of topology).
We do use that projections are continuous, which is pretty easy.
You could also show that in any ordered space (like $\Bbb R$ is) the set $\{(x,y,z) \mid f(x,y,z) < g(x,y,z)\}$ is open for any $f,g: \Bbb R^3 \to \Bbb R$ that are continuous, and apply it twice for functions $x+y$, $z$ and $x^2+y^2$ on $\Bbb R^3$. Now we use that $\Bbb R$ is a group instead. But using that addition and squaring is continuous is the most economical: if you'd give a full $\epsilon$-proof from first principles you'd be taking much longer.
A: Take $(a,b,c)\in A$. Then $a^2+b^2>c$. If $\delta>0$ and if $\|(x,y,z)-(a,b,c)\|<\delta$, then\begin{align}x^2+y^2-z&=\bigl((x-a)+a\bigr)^2+\bigl((y-b)+b\bigr)^2-(z-c)+c\\&=(x-a)^2+(y-b)^2-(z-c)+2a(x-a)+2b(y-b)+a^2+b^2-c.\end{align}And, since $|x-a|,|y-b||z-c|<\delta$,$$\bigl|(x-a)^2+(y-b)^2-(z-c)+2a(x-a)+2b(y-b)\bigr|<2\delta^2+(2|a|+2|b|+1)\delta.$$So, if $\delta$ is so small that $2\delta^2+(2|a|+2|b|+1)\delta<a^2+b^2-c$, the computations above show that$$(x,y,z)\in B_\delta\bigl((a,b,c)\bigr)\implies x^2+y^2-z>0.$$Suppose now that $\delta$ is also so small that $3\delta<c-(a+b)$. Then a similar (indeed, simpler) computation show that you also have$$(x,y,z)\in B_\delta\bigl((a,b,c)\bigr)\implies z-(x+y)>0$$and therefore$$(x,y,z)\in B_\delta\bigl((a,b,c)\bigr)\implies(x,y,z)\in A.$$This proves that $A$ is an open set.
