1
$\begingroup$

I want to show that $\Vert \check x = \check y \Vert = 1$ implies $x = y$, where $\check x$ is the canonical name for $x$ in $V^B$.

I'd like try induction over the ranks of $\check x$ and $\check y$ and so far I have deduced that for every $t \in \operatorname{dom}(\check x): \sum_{s \in \operatorname{dom}(\check y)} \Vert s = t \Vert = 1$.

Here I is where I'm having difficulties. I'm unable show that there actually is an $s \in \operatorname{dom}(\check y)$ satisfying $\Vert s = t \Vert = 1$.

The fact that $V^B$ is full yields an $m \in V^B : \Vert s = t \Vert = 1$, so I guess I could try to go over the proof of the fullness of $V^B$ to determine, whether m is in $\operatorname{dom}(\check y)$ or not.

Should I do that or am I missing something much simpler?

  • 0
    It is easier to prove that $x\neq y$ implies $\|\check{x}=\check{y}\|=0$. This is equivalent to what you want and the assumption $x\neq y$ gives you more to work with.2012-06-12

0 Answers 0