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$D= \{(x,y)~|~x \ne 0, y \ne 0\}$

I'm thinking I can show that the set is open for $x \gt 0$ and $y \gt 0$ using disks. Maybe I could do the same for $x \lt 0$ and $y \lt 0$. But this process seems too long.

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up vote 7 down vote accepted

Consider the function $f(x,y)=xy$. Note that $f^{-1}[\mathbb R\setminus \{0\}]=D$. Since $\mathbb R\setminus \{0\}$ is open (which is easier to prove) and $f$ is continuous it follows that $D$ is open.

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+1 Short and sweet. – JavaMan Mar 10 '13 at 4:48

I don't see why you consider using open disks to take long; given any $(a,b)\in D$, just show that the open disk centered at $(a,b)$ of radius $\frac{\min\{a,b\}}{2}$ is contained in $D$.

Here are some examples; note how they don't touch the axes, i.e. $\{(x,y)\in\mathbb{R}^2\mid x=0\text{ or }y=0\}$.

enter image description here

MyCircle[point_] := {RegionPlot[(x - point[[1]])^2 + (y - point[[2]])^2 <  
(Min[Abs[point[[1]]], Abs[point[[2]]]]/2)^2, {x, -1, 1}, {y, -1, 1}, 
PlotPoints -> 40, PlotStyle -> {Hue[SeedRandom[Floor[(point[[1]] + point[[2]])*
1000]^2]; RandomReal[]]}], ListPlot[{point}, PlotStyle -> Directive[Black,

listofpoints = Select[RandomReal[{-1,1},{60,2}], Min[Abs[#[[1]]], Abs[#[[2]]]] 
> 0.04 &]; listofplots = {}; For[n = 1, n <= 30, n++, 
AppendTo[listofplots, Show[Plot[{}, {x, -1, 1}, Axes -> True, Ticks -> None, 
AspectRatio -> Automatic], Flatten[Map[MyCircle, Take[listofpoints, n]]]]]]; 

Export["animation.gif", listofplots, "DisplayDurations" -> {1}]
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I'd say min{|a|, |b|}/2 – RiaD Mar 10 '13 at 13:29

Note that $$ \mathbb{R}^2\setminus D=\{(x,y)\;;\;x=0\;\mbox{or}\;y=0\}=f^{-1}(\{0\})\cup g^{-1}(\{0\}) $$ where $$ f:(x,y)\longmapsto x\qquad g:(x,y)\longmapsto y $$ are both continuous whatever norm you put on $\mathbb{R}^2$, since they are linear and $\mathbb{R}^2$ is finite-dimensional.

Now $\{0\}$ is easily seen to be closed in $\mathbb{R}$. So $f^{-1}(\{0\})$ and $g^{-1}(\{0\})$ are both closed, hence $\mathbb{R}^2\setminus D$ is closed.

Thus $D$ is open.

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