Proposition 1.15

If $$x \neq y \in X$$ for some metric space $$X$$, then there is some $$\epsilon$$ so that $$B(x,\epsilon) \cap B(y,\epsilon) = \emptyset$$.

baruch_shahi
Take $$x\neq y$$ in the metric space $$X$$. Set $$\varepsilon=\frac{d(x,y)}{4}$$. We claim that $$B(x,\varepsilon)\cap B(y,\varepsilon)=\varnothing$$. To this end, suppose not. Then there exists $$z\in B(x,\epsilon)\cap B(y,\epsilon)$$. This implies both that $$d(x,z) < \epsilon=\frac{d(x,y)}{4}$$ and $$d(z,y) < \epsilon=\frac{d(x,y)}{4}$$. By the triangle inequality then, we have:

$$d(x,y)\le d(x,z)+d(z,y) < \varepsilon+\varepsilon =\frac{d(x,y)}{4}+\frac{d(x,y)}{4} = \frac{d(x,y)}{2}$$, which is a contradiction.

Therefore, $$B(x,\varepsilon)\cap B(y,\varepsilon)=\varnothing$$.

baruch_shahi 23:18, February 27, 2011 (UTC -5)

'''Very good! --Steven.clontz 16:07, March 1, 2011 (UTC)'''