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
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
@@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
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
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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
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
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 TH-cam
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
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
Ok, my proof was correct and I freaked out for nothing.