you don't actually need the axiom of choice, do you? if the Fs are closed and bounded, then you can define a choice function, e.g. pick the minimum. in fact i'd think that makes the proof easier to grasp since then the sequence is nondecreasing
This is sorta like the generalization to the nested interval property. Also, why do you use closed and bounded instead of compact ? Aren't they the same thing
@@drpeyam oh I guess you didn't prove that yet in the video series lol. I just thought it would be more elegant if you phrased things in terms of compact sets
My favorite theorem, in which it has a property of incompleteness with several hypotheses as the proof of this is possible and at the same time not, the indeterminate independence is always there in all the problems defined with simple axioms in which they break and not. I think cantor do a good theorem that people underestimated. Also This theorem can give us a logic definition of the undefined everything is nothing. That is the same that I said in a “simple ways”.
Cool video.. One of my favorite theorems in analysis. Just a quick clarification- In single variable analysis, the point x that's contained in all the closed and bounded intervals F_n is *unique* if the lengths of these intervals goes to 0. I guess the same would be true for R^k { k >= 2} with an appropriate notion of "length". If the area/volume of closed and bounded sets (like generalized rectangles or hyper-spheres) goes to 0, then this point x is unique. Am I right?
I don't think this is right in one variable (oops- just realized you noted it was for intervals only, so this counterexample doesn't work). Consider F_n = [0, 1/n] union [1, 1 + 1/n]. The intersection of it all is just the points 0 and 1. I think, if you wanted to prove your claim, a connectedness argument might work. Maybe show that the intersection of all the sets is connected, and that the only connected set of measure 0 is just a singleton. (Although measure, defined in the usual way, wouldn't work because something like a line segment would have measure 0 in R^2)
"Press F to pay respects" good night everyone.
After watching one by one I addicted to your vedioes
Ingliszh
yea that makes things clear
calling the subsequence the "express train" was pretty fun!
You have really helped me in every aspect of math
you don't actually need the axiom of choice, do you? if the Fs are closed and bounded, then you can define a choice function, e.g. pick the minimum. in fact i'd think that makes the proof easier to grasp since then the sequence is nondecreasing
That’s true, good point
Excellent ! Beautiful presentation of the topic . Thanks .DrRahul Rohtak Haryana India
He is a great Mathematician
Excellent ! Thank you very much.
This is sorta like the generalization to the nested interval property. Also, why do you use closed and bounded instead of compact ? Aren't they the same thing
Yeah but let’s keep it simple
@@drpeyam oh I guess you didn't prove that yet in the video series lol. I just thought it would be more elegant if you phrased things in terms of compact sets
Elegant yes, but quite traumatizing if you introduce it this way hahaha
My favorite theorem, in which it has a property of incompleteness with several hypotheses as the proof of this is possible and at the same time not, the indeterminate independence is always there in all the problems defined with simple axioms in which they break and not. I think cantor do a good theorem that people underestimated. Also This theorem can give us a logic definition of the undefined everything is nothing. That is the same that I said in a “simple ways”.
Thanks so much Dr
Great teaching Payamji. Need a bigger whiteboard
Buy me one then, haha
Cool video.. One of my favorite theorems in analysis. Just a quick clarification- In single variable analysis, the point x that's contained in all the closed and bounded intervals F_n is *unique* if the lengths of these intervals goes to 0. I guess the same would be true for R^k { k >= 2} with an appropriate notion of "length". If the area/volume of closed and bounded sets (like generalized rectangles or hyper-spheres) goes to 0, then this point x is unique. Am I right?
I don't think this is right in one variable (oops- just realized you noted it was for intervals only, so this counterexample doesn't work). Consider F_n = [0, 1/n] union [1, 1 + 1/n]. The intersection of it all is just the points 0 and 1.
I think, if you wanted to prove your claim, a connectedness argument might work. Maybe show that the intersection of all the sets is connected, and that the only connected set of measure 0 is just a singleton. (Although measure, defined in the usual way, wouldn't work because something like a line segment would have measure 0 in R^2)
For the first non-ex, wouldn't [0,1/n) also work even though it's not closed?
No since 0 is in all of them hence in their intersection
@@drpeyam Thank you!
Funky