So if the point (x - ε) on the lower end of the domain of the set or (x + ε) on the upper isn't possible, since if any x falls on the boundary, being less or more would become outside the set, this closes the set & vice versa? If so, then for the set (2,10), what if x=2, how does (x - ε) remain in the set since (x - ε) < 2?
In the set (2, 10), 2 is not included in the set, so x can not be 2. Neither can it be 10. However, in the set [2, 10), 2 is included in the set, so x can equal 2. This is why (2, 10) is open, but [2, 10) is not.
@@copypasta1585 following the idea of my first sentence, then fine, let x be the lowest number that isn't 2 and let epsilon = 5, then how is x-epsilon in the set when it goes past the open boundary? x = 2.00000000000000000000000000000001 - 5 = -2.9999999999999999 or whatever, but -2 isn't in the set. How?
@@SequinBrain Well, there simply isn’t a “lowest number that isn’t 2”, is there? There are an infinite number of real numbers between any two given real numbers, so you can always tack on another decimal place. So, if we say x is 2.000001, then simply set epsilon to something like 0.0000009, and the epsilon neighborhood still remains within the set. Edit: I should clarify that the criterion is not that every value of x must satisfy this property for every value of epsilon, but rather for *some* value epsilon, so you are free to make epsilon arbitrarily small
@@copypasta1585 I get lost in the non-specifics is the point, doesn't matter what low value that isn't 2 is, but that setting 5 as epsilon goes outside of the boundary. Where is the proof this is possible while still remaining in the set? Where are the rules for setting the value for epsilon? Point is there's too much guessing and no apparent rules, which makes all tests on the subject as arbitrary as is the ruleset, amounting to nonsense.
@@SequinBrain I’m sorry if I misunderstood what you were asking. I don’t actually think there’s much guesswork, though. At least, for the set of all reals, there isn’t. And, as I said, the “rules” for choosing a value epsilon are simply that it has to be a positive number such that x+epsilon and x-epsilon are elements in the subset. And for the general case of a subset (a, b) of the set of reals, there’s a pretty simple way to show that this is always possible. First, we consider the element of (a, b) x. From a to x, and from x to b, there are two intervals. The length of said intervals can be calculated via the differences x-a and b-x. Then, we take the minimum of the two, and in order for the set to be considered open, we must be able to find some positive number epsilon such that x +- epsilon remains within (a, b). Based on the characteristics of the real numbers, we know that this must be possible, because we know that in the intervals (a, x) and (x, b), there are infinitely many reals, so selecting epsilon is as simple as arbitrarily selecting one of the reals found in the smaller interval. Of course, this is specifically possible because we are applying it to a subset of the reals. It doesn’t necessarily hold for any set of numbers, which is why some sets (along with the set of all open subsets of said set) constitute topological spaces and some don’t. I hope that makes sense. I worry that in text it’s easy to end up talking past each other
Why at 9:25 you said that U is contained in the set containing zero, instead of directly saying that U is equal to the set containing zero?? is there something i am missing?
![How to Prove that the Interval (a, b] is Not an Open Set](http://i.ytimg.com/vi/aAYMxtLjNc0/mqdefault.jpg)
![How to Prove that the Interval (a, b] is Not an Open Set](/img/tr.png)
Thanx man i was kinda lost every time i see this example but now its clear
awesome!
Great video man! These videos have helped me a lot
Awesome!
So if the point (x - ε) on the lower end of the domain of the set or (x + ε) on the upper isn't possible, since if any x falls on the boundary, being less or more would become outside the set, this closes the set & vice versa? If so, then for the set (2,10), what if x=2, how does (x - ε) remain in the set since (x - ε) < 2?
In the set (2, 10), 2 is not included in the set, so x can not be 2. Neither can it be 10. However, in the set [2, 10), 2 is included in the set, so x can equal 2. This is why (2, 10) is open, but [2, 10) is not.
@@copypasta1585 following the idea of my first sentence, then fine, let x be the lowest number that isn't 2 and let epsilon = 5, then how is x-epsilon in the set when it goes past the open boundary? x = 2.00000000000000000000000000000001 - 5 = -2.9999999999999999 or whatever, but -2 isn't in the set. How?
@@SequinBrain Well, there simply isn’t a “lowest number that isn’t 2”, is there? There are an infinite number of real numbers between any two given real numbers, so you can always tack on another decimal place. So, if we say x is 2.000001, then simply set epsilon to something like 0.0000009, and the epsilon neighborhood still remains within the set.
Edit: I should clarify that the criterion is not that every value of x must satisfy this property for every value of epsilon, but rather for *some* value epsilon, so you are free to make epsilon arbitrarily small
@@copypasta1585 I get lost in the non-specifics is the point, doesn't matter what low value that isn't 2 is, but that setting 5 as epsilon goes outside of the boundary. Where is the proof this is possible while still remaining in the set? Where are the rules for setting the value for epsilon? Point is there's too much guessing and no apparent rules, which makes all tests on the subject as arbitrary as is the ruleset, amounting to nonsense.
@@SequinBrain I’m sorry if I misunderstood what you were asking. I don’t actually think there’s much guesswork, though. At least, for the set of all reals, there isn’t. And, as I said, the “rules” for choosing a value epsilon are simply that it has to be a positive number such that x+epsilon and x-epsilon are elements in the subset. And for the general case of a subset (a, b) of the set of reals, there’s a pretty simple way to show that this is always possible.
First, we consider the element of (a, b) x. From a to x, and from x to b, there are two intervals. The length of said intervals can be calculated via the differences x-a and b-x. Then, we take the minimum of the two, and in order for the set to be considered open, we must be able to find some positive number epsilon such that x +- epsilon remains within (a, b). Based on the characteristics of the real numbers, we know that this must be possible, because we know that in the intervals (a, x) and (x, b), there are infinitely many reals, so selecting epsilon is as simple as arbitrarily selecting one of the reals found in the smaller interval.
Of course, this is specifically possible because we are applying it to a subset of the reals. It doesn’t necessarily hold for any set of numbers, which is why some sets (along with the set of all open subsets of said set) constitute topological spaces and some don’t.
I hope that makes sense. I worry that in text it’s easy to end up talking past each other
Why at 9:25 you said that U is contained in the set containing zero, instead of directly saying that U is equal to the set containing zero?? is there something i am missing?
thank you very much.
You are welcome !
thank u for this video
😄
Life saver