Exercise 2.C.14

Answers

Proof. We will prove by induction on $m$ . For the base case we have $\dim U_{1} \leq \dim U_{1}$ . Suppose $m > 1$ and

\dim (U_{1} + \dots + U_{k}) \leq \dim U_{1} + \dots + \dim U_{k}

for all positive integers $k < m$ . Then

\begin{aligned} \dim (U_{1} + \dots + U_{m}) & = \dim (U_{1} + \dots + U_{m - 1}) + \dim (U_{m}) - \dim ((U_{1} + \dots + U_{m - 1}) \cap U_{m}) \\ \leq \dim (U_{1} + \dots + U_{m - 1}) + \dim U_{m} \\ \leq \dim U_{1} + \dots + \dim U_{m - 1} + \dim U_{m}, \end{aligned}

where first equality follows from 2.43 and third from the induction hypothesis. □

celiopassos

2017-10-06 00:00