Exercise 1.9

Show that with the standard metric on $$\mathbb{R}^2$$, $$\{(x,y): x^2+y^2 < 1\}$$ is open.

AddemF
Let $$(a,b) \in B \big( (0,0), 1 \big) = B$$ then consider $$B \big( (a,b), 1-\sqrt{a^{2}+b^{2} - \varepsilon} \big) = B_{1}$$ where $$\varepsilon$$ is some arbitrarily small quantity which is still greater than $$1-\sqrt{a^{2}+b^{2}}$$. For any $$(c,d) \in B_{1}$$, distance $$d \big( (0,0), (c,d) \big) \leq \sqrt{a^{2} + b^{2}} + 1 - \sqrt{a^{2} + b^{2}} - \varepsilon < 1$$ by the triangle inequality proved earlier, where the first summand of the middle of the inequality is distance from origin to $$(a,b)$$ and the second summand is distance from there to $$(c,d)$$.

Since $$d\big( (0,0), (c,d) \big) < 1$$, we have $$B_{1} \subseteq B$$.

littlespoon1
Let $$U = \{(x,y) : x^2 + y^2 < 1\}$$. Let $$p=(a,b) \in U$$ and let $$d = \sqrt{a^2 + b^2}$$. Consider the line segment from (0,0) through p and on to the circle's edge at length 1.

d     1-d 0    p         1

Then for all $$p \in U, B(p,1-d) \in U$$ therefore U is open.

'''This is correct though it appeals to our geometric intuitions, which may be a little less rigorous than we desire. --AddemF 184.32.143.29 02:15, March 23, 2011 (UTC)'''

Jamie
Goal: Using the standard distance metric in $$\mathbb{R}^{2}$$, $$d\left(\left(x_{1},y_{1}\right),\left(x_{2},y_{2}\right)\right)=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$$, prove that the ball $$B=B\left(\left(0,0\right),p\right)$$ is open.

Proof: A set $$S$$ is open if for all points $$p$$ in $$S$$, there exists a ball $$B_{p}=B\left(p,\epsilon\right)$$, where $$\epsilon>0$$, such that $$B_{p}\subseteq S$$. Let $$P$$ be an arbitrary point in $$B$$, and $$D=d\left(\left(0,0\right),P\right)$$. Because the radius of $$B$$ is $$p$$, $$D < p$$. Let $$s=p-D$$. Because $$D < p$$, $$s>0$$. Consider the ball $$B_{s}=B\left(P,s\right)$$. I claim that $$B_{s}$$ is a subset of $$B$$.

This can be proved using the triangle inequality. Let $$Z$$ be an arbitrary point in $$B_{s}$$. $$d\left(P,Z\right) < s$$ because the radius of $$B_{s}$$ is $$s$$. Now, the triangle inequality states that $$D+d\left(P,Z\right)\geq d\left(\left(0,0\right),Z\right)$$. Because $$d\left(P,Z\right) < s$$ and $$D+s=p$$, we can safely say that $$D+d\left(P,Z\right) < p$$, and therefore $$d\left(\left(0,0\right),Z\right) < p$$. It follows that $$Z\in B$$. Since this holds true for any point in $$B_{s}$$, $$B_{s}\subseteq B$$ and therefore $$B$$ is open.

I think that this should work for any metric space as long as the triangle inequality holds.

--203.118.163.84 09:39, March 2, 2011 (UTC)

'''Great. Notationally it's a little confusing because you mixed upper and lowercase letters for elements of a set, while we're going to try and stick to calling sets uppercase letters and set elements lowercase letters, but it's a solid proof. I updated $$\subset$$ to $$/subseteq$$ (as you intended it to mean) to stick with our convention. As for the discrete/square metric, I'm not convinced your argument is enough (although it is open in those metrics), but I made that a separate discussion. --Steven.clontz 18:20, March 2, 2011 (UTC)'''

Yay! Thank you :) -Jamie