# Theorem 6.10 Rudin PMA, Partition

The third paragraph of the proof begins,

'Now form a partition $$P = \{x_0,x_1,...,x_n\}$$ of $$[a,b]$$, as follows: Each $$u_j$$ occurs in $$P$$. Each $$v_j$$ occurs in $$P$$. No point of any segment $$(u_j,v_j)$$ occurs in $$P$$. If $$x_{i-1}$$ is not one of the $$u_j$$, then $$\Delta x_i < \delta$$.'

I don't understand the final sentence, 'If $$x_{i-1}$$ is not one of the $$u_j$$, then $$\Delta x_i < \delta$$.' I interpret this as meaning that $$|v_j - u_{j+1}| < \delta$$ for each $$j$$. Is my interpretation correct? If so, I don't know what allows him to say this--we don't know apriori where the points of discontinuity are within $$[a,b]$$, so how can rudin assert that the distance between the intervals $$[u_j,v_j]$$ and $$[u_{j+1},v_{j+1}]$$ is less than $$\delta$$?

## 1 Answer

I assume this is what it means: Take all the points $$u_j$$ and $$v_j$$ and throw in to $$P$$. Then in $$P$$, there are end points of intervals $$(u_j,v_j)$$. However outside of all the intervals $$(u_j,v_j)$$ are the complement intervals $$(v_j,u_{j+1})$$. In these complement intervals, chop each interval into many shorter intervals, each of length $$<\delta$$. Now take all the points where you chopped up these intervals $$(v_j,u_{j+1})$$, and add them into $$P$$. Hopefully this makes sense.

• This makes sense. Thanks for the reply – Zachary Candelaria Jan 17 at 5:00
• No problem at all – Zachary Candelaria Jan 17 at 5:28