Valid argument?

[Imum Coeli]

Hi, I was wondering if this is a sufficient argument. It just seems rather too simple...

Q) If vectors x and y in R^n and the intersection of B(x,delta) and B(y,epsilon) is not equal to the empty set then show ||x-y||<= epsilon + delta

A) Consider the extreme cases:
1) If the boundary of B(x,delta) lies on the boundary of B(y,epsilon) then ||x-y||= epsilon + delta
2) If the centre of B(x,delta) lies on the centre of B(y,epsilon) such that x = y then 0 = ||x-y||
Thus ||x-y||<= epsilon + delta

Thanks.

 

[pka]

Hi, I was wondering if this is a sufficient argument. It just seems rather too simple...

Q) If vectors x and y in R^n and the intersection of B(x,delta) and B(y,epsilon) is not equal to the empty set then show ||x-y||<= epsilon + delta

\(\displaystyle \exists t \in {\mathfrak{B}}\left( {x;\delta } \right) \cap {\mathfrak{B}}\left( {y;\varepsilon } \right)\)

Now \(\displaystyle \|x-y\|\le\|x-t\|+\|t-y\|\).

 

[Imum Coeli]

\(\displaystyle \exists t \in {\mathfrak{B}}\left( {x;\delta } \right) \cap {\mathfrak{B}}\left( {y;\varepsilon } \right)\)

Now \(\displaystyle \|x-y\|\le\|x-t\|+\|t-y\|\).

So then \(\displaystyle \|t-x\|\le \delta\) and \(\displaystyle \|t-y\|\le \epsilon\) ?