Table of Contents

## MTH631 MIDTERM SOLVED PAPERS

**MTH631 MIDTERM SOLVED PAPERS GET PDF PAPERS FILES FROM THE BELOW LINK: **

**Real Analysis II:**

To start the proof, we consider the possibility that in the covering constructed just prior to the statement of the lemma, there is some square or partial square in which no point zj exists such that inequality (1) holds for all other points z in it. If that subregion is a square, we construct four smaller squares by drawing line segments joining the midpoints of its opposite sides (Fig. 55).

JOIN MY TELEGRAM GROUP FOR ALL ASSIGNMENTS, GDB, MIDTERM PAST PAPERS, AND FINAL TERM PAST PAPERS FROM THE BELOW LINK:

## TELEGRAM GROUP LINK

**MTH631 MIDTERM PAST PAPERS BY MOAAZ:**

If the subregion is a partial square, we treat the whole square in the same manner and then let the portions that lie outside of R be discarded. If in any one of these smaller subregions, no point Zj exists such that inequality (1) holds for all other points z in it, we construct still smaller squares and partial squares, etc.

**MTH631 MIDTERM SOLVED PAPERS:**

To verify this, we suppose that the needed points zj do not exist after subdividing one of the original subregions a finite number of times and reaching a contradiction. We let σ0 denote that subregion if it is a square; if it is a partial square, we let σ0 denote the entire square of which it is a part.

**MTH631 MIDTERM SOLVED PAPERS BY MOAAZ:**

After we subdivide σ0, at least one of the four smaller squares, denoted by σ1, must contain points of R but no appropriate point zj . We then subdivide σ1 and continue in this manner. It may be that after a square σk−1 (k = 1, 2, . . .) has been subdivided, more than one of the four smaller squares constructed from it can be chosen. To make a specific choice, we take σk to be the lowest and then the furthest to the left