Enjoyable video. 11:44 "a/2 is definitely greater than a"? that doesn't seem right. I'm sure you assumed the details could be filled out by the viewer. Here is my attempt at proving (a,b) is an open set, based on your video. Let x be arbitrarily given in (a,b) , and choose E = 1/2 min { x - a, b - x } . We will show that (x - E , x + E ) ⊆ (a,b). 1. x - E > x - ( x - a ) / 2 , since E < (x-a )/2 and therefore -E > - (x-a )/2 , and then add x to both sides (these are valid by field axioms/theorems). 2. x - ( x - a ) / 2 = x/2 + a/2 , by algebra simplification. 3. x/2 + a/2 > a/2 + a/2 , since we assumed x > a therefore 1/2 * x > 1/2 * a , or x/2 > a/2 . 4. a/2 + a/2 = a , by algebra simplification. By transitivity of inequalities, or putting the inequalities together, we get x - E > a. Similarly we can show x + E < b. So far we have shown that a < x - E < x + E < b . Let y be given in ( x - E , x + E ) . Then a < x - E < y < x + E < b . So a < y < b and therefore y is an element of (a,b) . Since y was given arbitrarily, we have for all y if y∈ (x - E , x + E ) then y ∈ (a,b). Therefore (x - E , x + E ) ⊆ (a,b) Since x in (a,b) was given arbitrarily, we get for all x in (a,b) there exists E such that (x - E , x + E ) ⊆ (a,b).
10:45 This is the picture of their vacation they post on Instagram despite the rest of their life is falling apart Oof, I did not need to be called out like that
best instructor i have ever seen.i hope my exam will be okay. thanks a lot.
I swear you are the best instructor out there. I really appreciate your videos thank you a lot!
I really wish if there are more videos on topology :(
Helpful
Enjoyable video.
11:44 "a/2 is definitely greater than a"? that doesn't seem right.
I'm sure you assumed the details could be filled out by the viewer.
Here is my attempt at proving (a,b) is an open set, based on your video.
Let x be arbitrarily given in (a,b) , and choose E = 1/2 min { x - a, b - x } .
We will show that (x - E , x + E ) ⊆ (a,b).
1. x - E > x - ( x - a ) / 2 , since E < (x-a )/2 and therefore -E > - (x-a )/2 , and then add x to both sides (these are valid by field axioms/theorems).
2. x - ( x - a ) / 2 = x/2 + a/2 , by algebra simplification.
3. x/2 + a/2 > a/2 + a/2 , since we assumed x > a therefore 1/2 * x > 1/2 * a , or x/2 > a/2 .
4. a/2 + a/2 = a , by algebra simplification.
By transitivity of inequalities, or putting the inequalities together, we get x - E > a.
Similarly we can show x + E < b.
So far we have shown that a < x - E < x + E < b .
Let y be given in ( x - E , x + E ) . Then a < x - E < y < x + E < b . So a < y < b and therefore y is an element of (a,b) .
Since y was given arbitrarily, we have for all y if y∈ (x - E , x + E ) then y ∈ (a,b).
Therefore (x - E , x + E ) ⊆ (a,b)
Since x in (a,b) was given arbitrarily, we get for all x in (a,b) there exists E such that (x - E , x + E ) ⊆ (a,b).
Thank you Dr ! Pls can you explain theorem of connectedness ?real numbers =AUB.? A and B nonempty subsets .
You might enjoy the end of this video where I “connect” that idea with clopen sets. ua-cam.com/video/_a7dKWxflvM/v-deo.html
10:45 This is the picture of their vacation they post on Instagram despite the rest of their life is falling apart
Oof, I did not need to be called out like that
Analysis proofs do be like that tho
Your representation is easy to understand, but your proof is incorrect when you prove x-epsilon>a and x+ epsilon>b.
correction: x+ epsilon < b. See my solution above.
also i think the proof is easier if you use ε = min { x-a, b -x }
Sir ! Have you made videos on Free Groups ?