๐งต Real number arithmetic
Anonymous at Mon, 15 Apr 2024 13:47:30 UTC No. 16130087
Can someone explain proposition A1.6 here? Apparently it's used to define arithmetic between real numbers as infinite decimals.
Anonymous at Mon, 15 Apr 2024 19:59:06 UTC No. 16130576
Is this from that nonstandard real-analysis book?
what is the point of identifying real numbers or real scalar functions with decimal expansions of arbitrary precision? like what is advantage of doing this over usual point-set topology?
Anonymous at Mon, 15 Apr 2024 20:03:01 UTC No. 16130580
>>16130576
It's from the appendix to "Vector Calculus, Linear Algebra, and Differential Forms" by Hubbard & Hubbard.
Anonymous at Mon, 15 Apr 2024 20:04:11 UTC No. 16130583
D is the set of finite decimals by the way.
Anonymous at Mon, 15 Apr 2024 20:28:15 UTC No. 16130600
>>16130087
A1.5 for k=10000 give inf of many function value, each change by <10^(-10000) for all coordinates
The inf get bigger for k larger because its smaller set. So sup k give the "limit" with more and more decimal places of x
It work with continuous function to get closer, and A1.5 is the only extended continuous for R, now prove it
Anonymous at Mon, 15 Apr 2024 20:32:31 UTC No. 16130607
>>16130583
how is |x| defined? number of digits?
Anonymous at Mon, 15 Apr 2024 20:55:47 UTC No. 16130644
>>16130600
>for k=10000 give inf of many function value, each change by <10^(-10000) for all coordinates
I don't get what you mean. Are you saying that the value of the function will be less than 10^-10000? How do you prove that? I don't get how we can bound the function when we don't know what it does.
Anonymous at Mon, 15 Apr 2024 20:59:11 UTC No. 16130648
homework thread
Anonymous at Mon, 15 Apr 2024 21:00:45 UTC No. 16130652
>>16130644
the coordinate change by 10^(-10000), not output
but, this is related to definition of continuity from A1.6
read about epsilon delta first if it's confused
Anonymous at Mon, 15 Apr 2024 21:22:39 UTC No. 16130683
>>16130652
How do you know that the set of f([x_1]_l,...[x_n]_l) is bounded for l<=-k ?
Anonymous at Mon, 15 Apr 2024 21:38:01 UTC No. 16130713
>>16130683
proposition A1.6 is lack the hypothesis f satisfies Definition A1.4, so if k large it's true now
Anonymous at Mon, 15 Apr 2024 21:45:18 UTC No. 16130725
>>16130713
I don't get why it being finite decimal continuous means that it must be bounded for a choice of l.
Anonymous at Mon, 15 Apr 2024 22:12:22 UTC No. 16130752
>>16130725
use equation A1.4
Anonymous at Tue, 16 Apr 2024 08:02:06 UTC No. 16131192
>>16130752
How do we know it has a lower bound?
Op Messi fan at Tue, 16 Apr 2024 09:16:20 UTC No. 16131236
>>16130648
>Hw thread
The material is interesting enough to deserve it's own thread, and /sqt/ can't handle anything past undergrad anyway. And why should we care? It's not like we volunteered to clean up /sci/ for free.
Anonymous at Tue, 16 Apr 2024 11:39:57 UTC No. 16131395
>>16131192
use equation A1.4, has absolute value
Anonymous at Tue, 16 Apr 2024 11:46:05 UTC No. 16131405
>>16131395
But that says that for any k there exists l, here we know an l, but we don't necessarily know that there exists a k. I mean, for given k, all values of [x_i]_l are at least k-close, but that doesn't mean that all values of f([x_1]_l, ..., [x_n]_l) are close. Does it?
Anonymous at Tue, 16 Apr 2024 15:00:59 UTC No. 16131692
>>16130576
Point is to give a definition of the reals and their arithmetic. Here the approach is to define them as decimals (which is painful as can be seen, hence why sane people define them as equivalent classes of Cauchy sequences or Dedekind cuts).
Anonymous at Tue, 16 Apr 2024 21:34:58 UTC No. 16132155
>>16131405
Bump
Anonymous at Wed, 17 Apr 2024 07:44:49 UTC No. 16132725
I still don't get how we know the function is bounded.
Anonymous at Wed, 17 Apr 2024 08:08:42 UTC No. 16132737
>>16132725
I think it is a similar argument to f is uniformly continuous on a bounded subset
Anonymous at Wed, 17 Apr 2024 16:00:21 UTC No. 16133105
>>16130600
>>16130683
How about:
The set of ([x_1]_l, ..., [x_n]_l) for l<=-k are k-close, i.e. |[x_i]_l - [x_i]_-k| <= 10^-k so they are a compact subset C (closed and bounded) of D^n.
Assume f is unbounded. Then for any integer N, no matter how large, there exists a point X_N (being an element of C) such that |f(X_N)| > N. By the theorem of the existence of a convergent subsequence in a compact subset, the sequence N->X_N must contain a convergent subsequence j->X_N(j) which converges to some point b (being an element of C). Since f is D-continuous at b, for any k, there exists l such that when |x_i - b_i| <= 10^-l, then |f(X) - f(b)| <= 10^-k; that is, |f(X)| <= |f(b)| + 10^-k.
Since j->X_N(j) converges to b, we have |X_N(j) - b| <= 10^-l for j sufficiently large. But as soon as N(j) > |f(b)| + 10^-k, we have |f(X_N(j))| > N(j) => |f(b)| + 10^-k. A contradiction. Therefore the set of values of f is bounded, which means f has a supremum.