$U = [{(x_1 , x_2) \in \Bbb R^2 : 0 < x_1^2 + x_2^2 < 1}]$

I have to prove U is an open set in $(\Bbb R^2 , d)$, the Euclidean metric.

What I have so far:

Let $x = (x_1,x_2) \in U$. Let $r < min[\sqrt{x_1^2+x_2^2}, 1-\sqrt{x_1^2+x_2^2}]$

Claim: $B_{(x)}(r) \subseteq U$.

Let $y = (y_1,y_2) \in B_{(x)}(r).$ Then $\sqrt{|x_1-y_1|^2 + |x_2-y_2|^2} < r \Rightarrow |x_1-y_1|^2 + |x_2-y_2|^2 < r^2$

$|x_1-y_1|^2 < r^2$ and $|x_2-y_2|^2 < r^2$

$|x_1-y_1| < r$ and $|x_2-y_2| < r$

$x_1 - r < y_1 < x_1 + r$ and $x_2 - r < y_2 < x_2 + r$

From here, I know I have to show $y \in U$, but I'm not sure where to go with it. Am I way off base in choosing my r or am I just not seeing my next step?


Once you have

$$ \sqrt{|x_1-y_1|^2 + |x_2-y_2|^2} < r,$$

you can get

$$\sqrt{y_1^2+y_2^2}\le \sqrt{|x_1-y_1|^2 + |x_2-y_2|^2}+\sqrt{x_1^2+x_2^2}<r+\sqrt{x_1^2+x_2^2}\le 1.$$

This shows that $y_1^2+y_2^2<1.$ Can you show that $y_1^2+y_2^2>0?$

  • $\begingroup$ I'm not sure exactly how you reached that conclusion. Could you explain this step a bit further? Specifically how we can say $$\sqrt{y_1^2+y_2^2}\le \sqrt{|x_1-y_1|^2 + |x_2-y_2|^2}+\sqrt{x_1^2+x_2^2}$$ $\endgroup$ – clutchcityreturns Oct 20 '14 at 17:29
  • 1
    $\begingroup$ It is the triangle inequality for the distance $d((a,b),(c,d))=\sqrt{(c-a)^2+(d-b)^2}.$ In our case, it is $d((y_1,y_2),(0,0))\le d((y_1,y_2),(x_1,x_2))+d((x_1,x_2),(0,0)).$ $\endgroup$ – mfl Oct 20 '14 at 17:35

Metric space approach - have a look at this.

Fix a point in your set $\vec{z}\in U$.

By definition its norm is bounded: $$0<\|\vec{z}\|<1$$

Now, the appropriate choice is: $$\varepsilon:=\min\{\|\vec{z}\|,1-\|\vec{z}\|\}>0$$

Consider a point in that ball: $$\|\vec{x}-\vec{z}\|<\varepsilon$$ Then it follows: $$\|\vec{x}\|\leq\|\vec{x}-\vec{z}\|+\|\vec{x}\|<1-\|\vec{z}\|+\|\vec{z}\|=1$$ $$\|\vec{x}\|\geq\|\vec{z}\|-\|\vec{x}-\vec{z}\|>\|\vec{z}\|-\|\vec{z}\|=0$$

See how this generalizes to annuli centered around arbitrary points in any metric space: $$B_\varepsilon(z)\subseteq A_{r,R}(a)\quad(\varepsilon:=\min\{d(z,a)-r,R-d(z,a)\}>0)$$


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