Proof: Maximum of a Set is the Supremum | Real Analysis

[@WrathofMath]

Many apologies for the error in this video I am about to explain. I usually take videos down when there are errors, but people haven't seemed to be confused by this one, and I don't say anything incorrect, just write something a bit sloppy. At about 3:52, I am explaining that since m is in A, and we assumed m' < m, we have contradicted the fact that m' is greater than or equal to a for all a in A. So what I am saying is that it is NOT true that m' is greater than or equal to all a in A, since we see it isn't true for m. However I write that m' is not greater than a for all a in A. In plain English, that may sound like the same statement, but in my written symbols it is clearly incorrect. Please ask if this is not clear, apologies for that sloppiness! And thanks to Froglet for making me notice this. Check out this proof that the minimum of a set is the infimum for an analogous proof without that awkwardness, my explanation is better here: kzbin.info/www/bejne/sJDGi2Z9i9OXd5I

[@shoopinc]

Thanks for the correction, happens to the best of us. Analysis takes some deep precision.

[@AFCOE]

Okay sir. Understood.❤❤🎉🎉

[@JRTO_X]

So in essence you are saying because we have assumed m' < m and m' is greater than every element of A. Given m is also in A (by def of max) thus m < m' which is not true by the assumption => contradiction. Is this correct? Cheers.

[@rmw6151]

All is forgiven. The proof is succinct and understandable. Thanks!

[@maxmillian8251]

The problem I have with most KZbin teachers just like you is the rush when explaining. If someone is watching this video is a proof he/she has real difficulty in understanding this topic, so why the rush??????

[@mahmoudalbahar1641]

Many thanks for this good video.

[@jacksonmadison9994]

but it's possible to have a supremum/infimum that is not a max/min, correct?

[@WrathofMath]

Yes indeed! For example every open interval has, as its supremum and infimum, a number which is not in the set and thus not a max or min.

[@MrHamtroll]

Thank you, king.

[@WrathofMath]

My pleasure, also King. Thanks for watching! Let me know if you have any questions.

[@anonymousvevo8697]

bro , do you have a video of the following demo A intersects B

[@WrathofMath]

Thanks for watching and I am not sure about your question! What do you mean by "demo"? Assuming you just want to prove the given inequality, if A and B are sets, writing A intersect B

[@Bedoroski]

​@@WrathofMathit must have been that way. Do you have any hints or video on that?

[@tagerauen5378]

Wonderful!

[@WrathofMath]

Thank you!

[@shivangitomar5557]

Hey, can you please prove this in your next video as soon as possible: Let G be a simple graph without an even cycle. Prove that the graph G has an independent set of size at least n/3. PLEASE!! Thank you so much!!

[@anonymousvevo8697]

Sorry i want to say sup( A intersects B )

[@WrathofMath]

I see, thanks for clarifying! I'll try to make a lesson on that soon. In the meantime, you may find this related proof helpful: kzbin.info/www/bejne/l5TVYZeojdqBjKc

[@WrathofMath]

kzbin.info/www/bejne/iGO6pKyeesR5prM

[@anonymousvevo8697]

@@WrathofMath you are the best

[@froglet827]

why is m' not greater or equal than a?

[@WrathofMath]

Thanks for watching and for the question! I'm sorry I think I presented that in a sloppy manner. m' is greater than or equal to a for every element a of A. The problem is that m is also in A, and so m' is greater than or equal to m, but we had assumed that m' was less than m. That is the contradiction. Near the end of the proof, I was trying to say that since m' is less than m, m' cannot be greater than or equal to every element of A (again, that is the contradiction). However what I wrote is that m' is not greater than or equal to a, for ever a in A, which is not necessarily true. I was just trying to say that it is not the case, that m' is greater than or equal to a for all a in A, since we saw it isn't true for m. It still may be true for other elements of A, but since it isn't true for ALL of them, that is a contradiction. Does that help? Sorry about that confusion!

[@WrathofMath]

Also, here is a link to a lesson which releases later this week: kzbin.info/www/bejne/sJDGi2Z9i9OXd5I This goes over the analogous result that the minimum of a set is the infimum, and I explain this one more cleanly. The ideas are basically the same just with flipped inequalities, so it may help.

[@froglet827]

Thanks for the quick reply!

[@davidgao3005]

I think the set should be finite

[@adnansohail8160]

We have to show sup(S) =max(s) So for contradiction sup(s) /=max(s)

[@WrathofMath]

Thanks for watching and no it is not. To be precise, we did not suppose that max S is not equal to sup S, that would only mean that max S is not an upper bound or max S is not the least upper bound. Instead, we pointed out that max S is certainly an upper bound, and supposed for contradiction that max S is not the least such bound. Showing this is impossible demonstrates that max S is in fact the least upper bound, and thus, by definition, the supremum. Do you reject the law of non-contradiction or is there an error in the video I missed?