How to Prove that the Interval (a, b] is Not an Open Set
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- Опубликовано: 18 дек 2024
- In this video I will show you how to prove that the interval (a, b] is not an open set. I do a proof by contradiction and I go over all of the steps very carefully. I hope this video helps someone.
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Thank you:)
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Real analysis 1 brought me here and man am I so grateful for your videos. This stuff CAN be simple so long as its explained well! Thanks for doin what you do I am almost certain you have saved many a mathematics student's grade!
Was just literally studying about this exact definition for multivariable analysis. Amazing how you always upload stuff I'm working on
LOL so cool 😄
Perfect timing, my upcoming test has this topic
awesome!
I like the way you make calculus seem easy
Thank you!
Thank you, I have a midterm tm and this has helped so much. Thank YOU!
This is cool, thanks for sharing
:)
thanks for this vid. i always have hard time proving these kind of obv. stuff. would love more analysis content.
Beautiful explanation!
Thank you!
Very good explanation 👍👍
Thank you!
this guy is literally amazing
Very clear, thank you.
You are welcome!
I am trying to figure out why it is necessary to go to the trouble of bringing ε/2 into the argument rather than just using ε. The only difference appears to be the assertion that b+ε/2 is an element of (b-ε,b+ε)⊆ (a,b] and therefore b+ε/2 is an element of (a,b] and therefore b+ε/2 is
yeah that's correct
implies b + e/2 is in (a, b], so b + e/2
I removed my comments but my name is still showing in some comments above. So for the record, this is what I posted:
For all sets A and B of (extended) real numbers, A ⊆ B implies sup A ≤ sup B. Since sup(c,d) = d and sup(a,b] = b, we have (c,d) ⊆ (a,b] implies d ≤ b.
So in the case of this video: (b-ε, b+ε) ⊆ (a,b] implies b+ε ≤ b. Which is the more direct contradiction you are talking about in your first comment.
Of course in some classes at this level you may have to additionally prove these facts, for instance that sup(c,d)=d. In which case the argument of The Math Sorcerer is much shorter and direct. This is probably the reason he chose the approach of bringing ε/2 into the argument. So the argument in the video is not necessary, but may be preferred in some settings.
Now I have answered all of the questions of your first comment.
If you are satisfied with these answers, please remove the comments in which you mention my name. Good luck in your studies and have a good day!
Yeh, that wiggle room thing can be a slippery little sucker 👍
and in (a,b], b doesn't wiggle. So if it were the case that b < d, then there would be more of (c,d) wiggling past b. This would mean (c,d) would exceed the constraint of being contained in (a,b], thus contradicting the statement (c,d) ⊆ (a,b] .Such is the image my mind has conjured up.... feel free to help me out here 😂. Good times 👍
does not open imply closed or do you have to prove closedness separately?
Beautiful proof
Who said mathematicians can't be cool😁 Thanks for the video
Is there a video like this, showing that the set is neither closed?
Learned a new symbol ( c with line below it ) thx
ahh yes the all important subset symbol! let's call it
I sense a little bit of snark in the proof of contradiction LOL XD
Lol
thanks
YES! topology and ... we get to stay home because of corona. What better to do than study math.
Max Percer Get some good sleep lol!
Nice
thx:)
How to prove that set of rational numbers is not closed??
its dense in R and it doesn't contain any x € R so it doesnt contain all of its boundary points therefore its not closed.
😊clear
Gzb
♥️🇧🇷
:)