Real Analysis #3 - The Completeness Property

[@BriTheMathGuy]

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[@hazelndlovu3996]

Finally useful Real analysis videos. Suggestion -Do more examples at the end.

[@LukeSequ]

These are the best real analysis videos I can find!

[@BriTheMathGuy]

Great to hear!

[@sinjini3189]

I was confused about why Q does not have the completeness property. And this video cleared my confusion. Thank you so much!

[@cgfam5256]

Thank you so much. your 4 min video is way better than +30 min of others!!!

[@BriTheMathGuy]

Glad it helped!

[@izzywhippet1855]

You said that root 2 is an upper bound of the set defined on Q, but your earlier definition of bounds said that an upper bound on Set S has to be an element of S. But as you go on to say, root 2 is not in Q, so it is not in the set you defined either. Have I misunderstood the definiton of bounds?

[@christopherallen3353]

That's the point. The subset must contain root two, but root two is irrational and therefore not in the set. Thus the rational number not complete.

[@sibusisopgambu9699]

Thanks. Please keep making more videos ...

[@BriTheMathGuy]

I'll do my best. Have a great day!

[@supernova3767]

Does the supremum have to be in the set we're working with? Do the bounds have to be in the sets we're working with?

[@nonexistent6426]

no for both questions. supremum is the least upper bound, and you can have infinitely many upper bounds for a set (for e.g., in the video sqrt(3) can also be an upper bound for the given set, but the *least* value is sqrt(2)) and the set is complete only if the supremum belongs to that set

[@mariamary1116]

What's the completeness property of set IR?

[@jaygourley1818]

Is the word "has" equivocal in this lecture? It could also mean "is contained in."

[@AnhNguyen-zd9bd]

So it means the set { x^2

[@whodafookisdatguy7659]

I think the idea is for completeness supA must belong to Q, and no matter which element q1 in Q we take such that q1 = supA there will always be a q2 in Q such that q1

[@mahmoudalbahar1641]

Many thanks for this good video.

[@SpiritVector]

sup.

[@BriTheMathGuy]

Hey :)