Exercise 7.A.17

Proof. Suppose $w\in \mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->T$ and $w\ne 0$. By 7.7 (d), we see that $w\in {(\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T^{\text{∗}})}^{\text{⊥}}$. In the previous exercise, we saw that $\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T=\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T^{\text{∗}}$, thus $w\in {(\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T)}^{\text{⊥}}$. Since $\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T\cap {(\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T)}^{\text{⊥}}=\{0\}$, it follows that $w\notin \mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T$.

With this reasoning in mind, one can easily see (or show by induction on $k$) that if $v\notin \mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T$ then $v\notin \mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T^k$ for any positive integer $k$. Therefore, taking the contrapositive of this statement, we see that $\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T^k\subset \mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T$. The inclusion in the other direction is obvious, hence $\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T^k=\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T$.

Therefore

where the first line follows from 7.7 (d), the second and the last were proved in the previous exercise and the fourth follows from 7.7 (b). □