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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 $\delta$?

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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.

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

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