How a Mathematician Would Prove These 4 Results

[@looped89]

I love what you are doing! Could you do some on homology and cohomology? :0 cheers!

[@dibeos]

@@looped89yesssss, that’s a great idea actually. There not many high quality videos on these subjects in KZbin

[@jammasound]

Pretty much knew you were going to do contradiction for the other 3, after seeing the 1st one. Nice little intro to analysis for me. I'm still in Calculus but I want to jump into analysis at some point.

[@dibeos]

@@jammasound yeah, even though it is kind of “repetitive”, because the idea is the same for all 4, I believe that it’s a very good exercise to reinforce the argument 4 times in slightly different proofs. Honestly , that’s the way to learn how to prove stuff in math, by practicing over and over again

[@JGLambourne]

Can't this be proved without using contradiction? I worry about using proof by contradiction when working with infinite sets.

[@dibeos]

@@JGLambourne it can definitely be proved without using contradiction. But what is the problem with using it?

[@jammasound]

I think its OK here, because for any two real numbers you have exactly one of the following is true: a < b, a = b, a > b.

[@JGLambourne]

@@jammasound How do you prove that ? The real numbers are equivalence classes of Cauchy sequences of rational numbers.

[@jammasound]

@@JGLambourne Not sure, perhaps its taken for granted like the existence of parallel lines.

[@willnewman9783]

​@@JGLambourneI don't know why you would worry about proof by contradiction with infinite sets. Anyway, to show that for all real numbers either a=b, ab, using the Cauchy sequence definition, one needs to define

[@mairc9228]

by using fairly routine tricks like multiplying my -1 squared you can actually show a and b just from c and d. also, using similar methods, along with a special case of c where for all -inf(-A)=sup(A) and similarly for d, you can show that a and c imply d and that b and d imply c, and by what we saw above, only ac, bd, or cd need to be proven in order to make sure all 4 are correct i will leave the proofs below: suppose c and d and x>0, then inf(xA)=inf(-(-x)A). using c, inf(-(-x)A)=-sup(-xA), and then using d: -sup(-xa)=-(-x)inf(A)=xinfa, thus we obtain b similarly, sup(xA)=sup(-(-x)A). using d, sup(-(-x)A)=-inf(-xA), and then using c: -inf(-xa)=-(-x)sup(A)=xsup(A), and so a is true suppose x

[@omargaber3122]

It would be better if you started with deep examples and then did the general explanation.

[@samueldeandrade8535]

Ok, my proof was correct and I freaked out for nothing.

[@samueldeandrade8535]

Christ, give me patience to write a proof here!

[@samueldeandrade8535]

Hummm. Very ... questionable. I could get mad criticising this video, but I guess I am too old. I will just leave this here: I will use "ext" for sup or inf. Then ext A satisfies an inequality x * ext A * b (★), for all x in A, for all b * bound, meaning IF ext = sup, then * = ≤ and b is upper bound IF ext = inf, then * = ≥ and b is lower bound Also, txe A will denote the other guy and ° will denote the other inequality symbol. Ok. Take μ≠0 (μ will be my lambda, I can't write lambda) and some * bound b for μA This means μx * b, for all x in A Let s=sign of μ, then x (s*) b/μ, for all x in A So, b/μ is an s* bound for A, with s* = *, if s=1 (μ>0), s* = ° , if s=-1 (μ0), sext = txe , if s=-1 (μ0, μ ext A = ext μA if μ