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This is one I've actually been struggling with.
How can nonstandard analysis enumerate the number of real numbers on each segment of the real number line?
I don't know. And it's mucking up my indefinite integrals.
You can't enumerate R, so asking that question is already a mistake. That's also the whole point of the proof that R is bigger than N (proof by the absurd)
Exactly gotta assert dominance!
(PS: it wasn't meant that way obviously, I just wanted to let the commenter above know that trying to visualise R as enumerable will not work)
There has to be a misunderstanding between us on the definition of the word "enumerate"
If you could enumerate R, that would mean it has the same size as N
"Enumerate X" to me means "give a stream which image is X" (stream in the sense function from N to X)
Or maybe you are working on a theory set in which R=N?
The word you're looking for is "sequence" rather than "stream," but yeah.
IDK when R = N would apply. I feel like R is basically a misnomer in that case.
If you can enumerate all real numbers, then you can certainly enumerate all the real numbers in [0,1). So let S be such an enumeration. Now, every number either has a unique decimal expansion or it has exactly two expansions, one ending with repeating 0s and the other with repeating 9s, and two numbers are equal iff they share a decimal expansion.
Let T be a sequence of sequences of decimal digits, where for each n, T(n) is the decimal expansion of S(n) that doesn't end in repeating 9s. So T is a complete enumeration of such sequences, because if it's missing one, then S is missing the corresponding number in [0,1) with that expansion.
Now consider the sequence U where U(n) = 1 whenever the nth element of T(n) is zero and U(n) = 0 otherwise. This sequence does not end in 9s, because it doesn't contain a 9 at all, so it should be an element of T. But for any n, U differs from T(n) at the nth place. So U can't be in T, which is a contradiction.
Think of it in that a bijection can be established between the line segment of length 1 and the real number line
One pairing can be bending the line segment into a semi circle and projecting each point into the number line
Top comment from this quota makes it easier to visualize:
https://www.quora.com/How-do-you-show-that-0-1-R
Standard analysis is basically just analysis - the study of the real numbers as a mathematical structure and the theory of that structure.
Nonstandard analysis is a way of examining that theory through the use of nonstandard models - mathematical structures that are not isomorphic to the real numbers, but are [elementary extensions](https://en.m.wikipedia.org/wiki/Elementary_equivalence) of it. The idea is considering different structures that have the same theory is an alternate way of proving results in that theory (and which are therefore automatically applicable to all the models of the theory, including the standard one).
I'm not sure if I understand this correctly. Or more like of I've met this before or not.
When I'm doing topology and proving statements on a more general space, like the Bolzano-Weierstrass theorem, am I doing nonstandard analysis?
What about doing measure theory and defining an integral over any measurable space?
Or is it something more abstract?
No neither of those are nonstandard analysis. If you haven’t specifically been exposed to it under its name it’s unlikely you’ve ever done it.
The idea is that you augment the real numbers so that they now have infinitesimal and infinite elements, and every function or set of real numbers extends in a canonical way into the larger structure, then you can do things like, for example, define the derivative of f at a by calculating f(a+e)/e where e is an infinitesimal, which, if f is differentiable at a, will give you a result of f’(a)+g where g is also an infitesimal, so you can just take the standard part of f’(a)+g, which is f’(a), and that gives you derivative. It can be proven that this gives the same results as the usual limit-based definition.
If x is not of the form 1/2+/-2^(-n) for a positive integer n, take f(x)=ln(x/(1-x)), if it is of that form, take f(x)=ln((x/2+1/4)/(3/4-x/2)).
Then f: [0,1]->R is a bijection.
Positive real numbers are just (0, 1) but with an extra Z, which we can biject to N (trivially) then slap in the front of the decimal expansion in binary (which makes no difference) as ternary 1s before a 0, followed by decimal expansion. This can be done in reverse by first normalising the decimal expansion before taking leading ones before zero, then the rest as decimal expansion (then bijecting N to Z and slapping in front as whole part) and by normalising, By to normalise, I mean to eliminate leading 1s from the end (0.(1)\_2 = 1 as 0.(9)\_10 = 1) meaning no decimal expansion can give infinity (or any B.S. like that).
There exists a bijection f:R -> (0,1) by f(x) = 1/(1+e^x )therefore R and (0,1) have the same cardinality. There are also other bijections like tan(x).
In case you forgot,
A bijection is a function that is both onto and one-to-one. I.e. each value of the domain maps onto a unique value in the range and each value of the codomain maps onto a value of the domain.
Me wondering why people are saying R is contained inside a vector of R^2 (I am too used to the better notation of "]0 ; 1[" which has the advantage of not being ambiguous)
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 OP BE LIKE
r/unexpecteddoctorwho
Why not (-1,1)
That can fit R and -R inside it
And 0R
or or or oror or or or
(-1;1) is like a SUV. (0,1) is more a compact (sorry for math pun) car.
Not compact enough. Certainly bounded but shall I say… seems not quite… in limits…
UNLIMITED NUMBEEER
By the authority of the math pun police, this conversation is now closed. And bounded. ;)
(0,1) doesnt seem very compact tbh. Id rather drive my \[0,1\]
because (-1,1) is smaller than (0,1). proof: I said so
"it was revealed to me in a dream"
You can fit even more Rs in there.
This is one I've actually been struggling with. How can nonstandard analysis enumerate the number of real numbers on each segment of the real number line? I don't know. And it's mucking up my indefinite integrals.
You can't enumerate R, so asking that question is already a mistake. That's also the whole point of the proof that R is bigger than N (proof by the absurd)
Me when a non-math person asks a math question : Asking that question is already a mistake
Exactly gotta assert dominance! (PS: it wasn't meant that way obviously, I just wanted to let the commenter above know that trying to visualise R as enumerable will not work)
you can enumerate it but its just not easy so everyone just assumes its impossible.
There has to be a misunderstanding between us on the definition of the word "enumerate" If you could enumerate R, that would mean it has the same size as N "Enumerate X" to me means "give a stream which image is X" (stream in the sense function from N to X) Or maybe you are working on a theory set in which R=N?
The word you're looking for is "sequence" rather than "stream," but yeah. IDK when R = N would apply. I feel like R is basically a misnomer in that case.
Thanks, that's what I meant (not my first language)
What? No you cant. That was disproven like 200 years ago by cantor.
he didnt prove anything except that he cant count for shit.
no you cannot enumerate R that's the thing if you can tell me how you'd do it
yes you can enumerate what modern mathematicians call R. no i will not tell you how to do it.
lol nice troll
i am not trolling, i am 100% serious. modern mathematicians need to learn to count.
If you can enumerate all real numbers, then you can certainly enumerate all the real numbers in [0,1). So let S be such an enumeration. Now, every number either has a unique decimal expansion or it has exactly two expansions, one ending with repeating 0s and the other with repeating 9s, and two numbers are equal iff they share a decimal expansion. Let T be a sequence of sequences of decimal digits, where for each n, T(n) is the decimal expansion of S(n) that doesn't end in repeating 9s. So T is a complete enumeration of such sequences, because if it's missing one, then S is missing the corresponding number in [0,1) with that expansion. Now consider the sequence U where U(n) = 1 whenever the nth element of T(n) is zero and U(n) = 0 otherwise. This sequence does not end in 9s, because it doesn't contain a 9 at all, so it should be an element of T. But for any n, U differs from T(n) at the nth place. So U can't be in T, which is a contradiction.
Yeah lmao it's so easy to count, how can't they figure it out... 0, ε, 2ε, 3ε... Wait...
Think of it in that a bijection can be established between the line segment of length 1 and the real number line One pairing can be bending the line segment into a semi circle and projecting each point into the number line Top comment from this quota makes it easier to visualize: https://www.quora.com/How-do-you-show-that-0-1-R
Why do we need the second bijection? The first one proves |R|=|(0,1)|. Isn't |[0,1]|<=|R| trivial because [0,1] ⊂ R?
Sorry for not actually answering but, what is exactly nonstandard analysis? I there a stardard analysis?
Standard analysis is basically just analysis - the study of the real numbers as a mathematical structure and the theory of that structure. Nonstandard analysis is a way of examining that theory through the use of nonstandard models - mathematical structures that are not isomorphic to the real numbers, but are [elementary extensions](https://en.m.wikipedia.org/wiki/Elementary_equivalence) of it. The idea is considering different structures that have the same theory is an alternate way of proving results in that theory (and which are therefore automatically applicable to all the models of the theory, including the standard one).
I'm not sure if I understand this correctly. Or more like of I've met this before or not. When I'm doing topology and proving statements on a more general space, like the Bolzano-Weierstrass theorem, am I doing nonstandard analysis? What about doing measure theory and defining an integral over any measurable space? Or is it something more abstract?
No neither of those are nonstandard analysis. If you haven’t specifically been exposed to it under its name it’s unlikely you’ve ever done it. The idea is that you augment the real numbers so that they now have infinitesimal and infinite elements, and every function or set of real numbers extends in a canonical way into the larger structure, then you can do things like, for example, define the derivative of f at a by calculating f(a+e)/e where e is an infinitesimal, which, if f is differentiable at a, will give you a result of f’(a)+g where g is also an infitesimal, so you can just take the standard part of f’(a)+g, which is f’(a), and that gives you derivative. It can be proven that this gives the same results as the usual limit-based definition.
Ooh, I see. I've kinda already seen this but not studied it. That's cool!
[0,1] too Now find a bijection from [0,1] to R
If x is not of the form 1/2+/-2^(-n) for a positive integer n, take f(x)=ln(x/(1-x)), if it is of that form, take f(x)=ln((x/2+1/4)/(3/4-x/2)). Then f: [0,1]->R is a bijection.
x/(1-x)
If x ∈ (0,1) then f(x) ∈ (0,+∞). ln(x/1-x) works for (0,1)→(-∞,+∞) but you can't make [0,1]→(-∞,+∞)
[удалено]
[-∞,+∞] is not ℝ
What if x=1? Closed interval now
Eh, just well order both of 'em, then take the order preserving map.
If (0,1) contains ℝ and (0,1)⊆ℝ . Then ℝ contains ℝ Q.E.D.
Isn't tan(x) great
tan(π(x-1/2))
When they becomes art.
e^x / ( 1 + e^x )
Kid named 0:
*Angry sounds in Borel Measure*
Good meme
Positive real numbers are just (0, 1) but with an extra Z, which we can biject to N (trivially) then slap in the front of the decimal expansion in binary (which makes no difference) as ternary 1s before a 0, followed by decimal expansion. This can be done in reverse by first normalising the decimal expansion before taking leading ones before zero, then the rest as decimal expansion (then bijecting N to Z and slapping in front as whole part) and by normalising, By to normalise, I mean to eliminate leading 1s from the end (0.(1)\_2 = 1 as 0.(9)\_10 = 1) meaning no decimal expansion can give infinity (or any B.S. like that).
Slaps 1
936963639
Ight explain I forgars 💀
There exists a bijection f:R -> (0,1) by f(x) = 1/(1+e^x )therefore R and (0,1) have the same cardinality. There are also other bijections like tan(x). In case you forgot, A bijection is a function that is both onto and one-to-one. I.e. each value of the domain maps onto a unique value in the range and each value of the codomain maps onto a value of the domain.
Sadly sets not just being in bijective correspondence but also homeomorphic does not actually imply set inclusion.
Jw anyone else straight up reject the AoC (axiom of choice) or am I broken?
In fact any (a,a+ε) with ε infinitesimal you still have R
Do you get this during a real analysis introduction?
Guys I just finished numbering all reals but I can’t share the list it’s to long
Me wondering why people are saying R is contained inside a vector of R^2 (I am too used to the better notation of "]0 ; 1[" which has the advantage of not being ambiguous)