Recent content by NickAlger

Heh, my counterexample has somehow managed to morph into something basically equivalent to [0,w1). I'd be interested in hearing a hint about the other example you mention.

Ahh yes, interesting. Technically for the proof I posted, you need to check all combinations of sets involving only the numbers 1,2,3, and 4. I just (stupidly) assumed there were no other small solutions. I wonder how many other solutions there are for big odd N, as for N variables the proof...

Thanks, but not right now. Maybe in a day or so if my current strategy doesn't pan out.

I am new to topology and have not read about that yet. I'll look into it, thanks. So far my search for counterexamples has been mostly with variations on discrete topologies, I think I am close to coming up with an example of my own that works, so I'm going to keep thinking about my own ideas...

I'm not sure you mean by [0, \omega_1). Could you elaborate?

Ok, I think I might have a counterexample to it as written, but I'm not totally convinced and need to mull it over a bit. Let me run it by yall. Let Y be a uncountable set and let \Re be the the real line. Construct a basis for the topology of the combined sets as follows: 1) All singleton...

I'm not really sure what a net is, though from a quick googling it seems to be a generalization of a sequence, so I will look into it. However, the http://www.ma.utexas.edu/~arbogast/driver.pdf" I'm working from (which could be wrong...) defines a strict limit as follows (page 9)...

I am trying to show the following proposition, but I can only do so in the special case where the basis for the topology is countable. I posted this in the calc & analysis forum and didn't get any responses - I think that this is a more appropriate forum judging by the other posts. Proposition...

Another idea about this, For n variables, X1 x X2 x X3 x ... x Xn = X1 + X2 + X3 + ... + Xn if n is even then {1, 1, 1, ..., 2, n-2} is the solution. eg: 1,1,2,4 1,1,1,1,2,6 1,1,1,1,1,1,2,8 why? Because 1 + 1 + ... + 2 + n = (n - 2) + 2 + n = 2*n You could also do this by induction on n...

No clue. Perhaps someone else will know? I can say that the above proof strategy will work for any number of variables in the same form (ABCDEFG... = A+B+C+D+E+F+G+...).

lol, this proof can be made much simpler: A' x B' x C' x D' = (A + n)(B + m)(C + o)(D + p) = ABCD + ABCp + ABoD + AmCD + nBCD + other nonnegative terms = A + B + C + D + ABCp + ABoD + AmCD + nBCD + other (remember ABCD=A+B+C+D still) > A + n + B + m + C + o + D + p = A' + B' + C' + D'

Suppose there is another solution set A', B', C', D' with larger numbers. A' = A + n (=1+n) B' = B + m (=1+m) C' = C + o (=2+o) D' = D + p (=4+p) where n,m,o,p are some nonnegative integers that you add to the original numbers to get the new solution numbers. Then A' x...

I have been thinking about this as well, and I think RC4 would be feasible given a few months of training. The key to performing the encryption at speed in your mind will be memorizing huge tables of precomputed operations.

If you want to improve the GPA, I think you should identify precisely what is stopping you from getting higher marks, and then work on fixing that. Are the low scores from test scores? Homework scores? Look back at what you got wrong on homework and test problems - were your mistakes...

Y'all are forgetting Archimedes.