How a Mathematician Would Prove These 4 Results
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- Опубликовано: 8 фев 2025
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I love what you are doing! Could you do some on homology and cohomology? :0 cheers!
@@looped89yesssss, that’s a great idea actually. There not many high quality videos on these subjects in RUclips
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.
@@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
Ok, my proof was correct and I freaked out for nothing.
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
Can't this be proved without using contradiction? I worry about using proof by contradiction when working with infinite sets.
@@JGLambourne it can definitely be proved without using contradiction. But what is the problem with using it?
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.
@@jammasound How do you prove that ? The real numbers are equivalence classes of Cauchy sequences of rational numbers.
@@JGLambourne Not sure, perhaps its taken for granted like the existence of parallel lines.
@@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
It would be better if you started with deep examples and then did the general explanation.
Christ, give me patience to write a proof here!
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