APPLIED ANALYSIS Fall 97 Comments

Please, use aknyazev@math.ucdenver.edu, if you want to send a comment to Andrew Knyazev personally.

Name: Andrew Knyazev
Email: aknyazev@math.ucdenver.edu
WWW: http://math.ucdenver.edu/~aknyazev
Comments: Remember, that comments are ABSOLUTELY anonymous and do not forget to submit your e-mail if you want a email reply.

Wed Sep 3 17:01:09 1997
Name: Andrew Knyazev
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Comments: Our next class on Monday, Sept. 8, and all classes then will be in our main (Dravo) building.

Tue Sep 9 14:27:12 1997
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Comments: For Problem 1 ,Chp.2: It's not the set of polygons. It's the set of lines that form polygons. So the smallest set would be 3 lines? Is this right? In class, you drew a disconnected series of lines. It didn't form a polygon but you called them polygonal lines? I was confused.

Wed Oct 8 21:34:42 1997
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Comments: Concerning problem #1 for homework #2. I assumed that the smallest polygonal line would have 2 segments, not 3 since the final line is not drawn.

Thu Oct 9 10:28:33 1997
Name: Andrew Knyazev
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Comments: Concerning problem #1 for homework #2.
I do not see why not to consider the smallest polygonal line to be even with one segment, but it should not matter anyway

Sat Oct 25 17:04:49 1997
Name: Andrew Knyazev
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Comments: Question:
Prof. Andrew Knyazev,
I have a question about the solutions to homework #2. There are two methods you have written down for number 7. The second method proves that the set of all sequences consisting only of zeros and ones is uncountable. The question asks to prove that this set has the power of the continuum. To do that I thought that we needed to prove a one to one correspondence with the interval [0,1], or a one to one correspondence with a set that we know has the power of the continuum. In my class notes I wrote down that it is possible to find a set that is not power continuum and not countable. I must be missing something.
Answer:
You are missing a point, I mean the bonus point that I promised to give everybody who finds a flaw in my solutions. You are absolutely right, and the point is yours. You are not the first student who noticed that, but I asked the first fellow to keep it a secret for a while. It is now the right time to annonce it, and I am going to put a corresponding message on the Web. Thanks! Try to find more flaws! Andrew

Wed Nov 19 09:26:52 1997
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Comments: A nondecreasing function can be discontinous at all rational numbers,why can't I have another nondecreasing function that is discontinuous at all irrational numbers?

Wed Nov 19 11:08:58 1997
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Comments: I'm having problems with chapter 5 question 5. I can prove discontinuity at rational points but don't see why it's continuous at irrational points. Anyone have insight/hints?