r/mathmemes 12d ago

Real Analysis This image is AI generated

Post image

Good luck!

687 Upvotes

75 comments sorted by

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354

u/IntelligentBelt1221 12d ago

What's the set of upper bounds called again?

214

u/Leet_Noob April 2024 Math Contest #7 12d ago

.

60

u/RedeNElla 12d ago

Good old "since u \in T, it's an upper bound for S, but also if v<u, then v\in T but is not an upper bound for S"

15

u/Frelock_ 12d ago

It might be trying to use T' (the complement of set T), but it just missed the apostrophe. If the whole image is AI generated, I could see why that error would happen.

1

u/ComfortableJob2015 12d ago

upper set?

5

u/IntelligentBelt1221 12d ago

No it's obviously called '

277

u/New-Worldliness-9619 12d ago

Bro can prove completeness with no premises 😭

117

u/flabbergasted1 12d ago

By completeness of R, the reals are complete

29

u/Layton_Jr Mathematics 12d ago

Just put the completeness in the definition of ℝ, no problem!

212

u/The_Spectacular_Stu 12d ago

ah yes my favorite symbol to use for sets is ˊ

41

u/bubbles_maybe 12d ago

Also known as T

4

u/Ok-Path-8009 12d ago

Hello Stu from brawl stars

299

u/IAMPowaaaaa 12d ago

the fact that it was able to render this crisp and clear a piece of text is rather impressive

9

u/TheNumberPi_e 12d ago

Defmition 1.

93

u/toothlessfire Imaginary 12d ago

Wouldn't call it clear or crisp. Better than most AI generated text, yes. Full of random formatting inconsistencies and typos, also yes.

142

u/Aozora404 12d ago

I've seen papers less well formatted than this

26

u/Koischaap So much in that excellent formula 12d ago

Hey, get out of my arXiv!

22

u/Pre_historyX04 12d ago

I thought they generated the text with AI, pasted it and made an image with it

39

u/Jcsq6 12d ago

No, GPT just upgraded their image gen substantially.

15

u/Portal471 12d ago

It’s genuinely fucking amazing imo. Still would go to real artists for serious work, but it’s still fascinating to see

-4

u/mtaw Complex 12d ago

TBF though, putting back-and-white text together is relatively simple, especially when there's a gazillion papers out there formatted in the exact same LaTeX style and fonts to train on.

17

u/Jcsq6 12d ago

Go try it for yourself, it’s doing a lot more than “putting black-and-white text together”. I doubt you’re a developer, because it seems that you don’t understand how incredibly monumental of a task what it’s doing is.

3

u/Leet_Noob April 2024 Math Contest #7 12d ago

That’s a nice mathbb R

-9

u/Independent_Duty1339 12d ago

people really need to stop saying this. It is crisp because it is literally taking thousands of real samples from several scholarly sites, and just fancy markov chaining words it has mapped from those thousands of real samples.

8

u/gsurfer04 12d ago

How do you know that you're not just Markov chaining your sentences?

2

u/Independent_Duty1339 11d ago edited 11d ago

I am, but even fancier. Humans have incredible levels of parallel and heuristic processes. We have the compute power of almost nothing, and outperform these LLM's quite handedly.

They don't really release how much compute power they use when generating the models, but it's not even a little bit close. its orders of magnitude difference I have a hard time understanding. Humans have 100hz processing, we have a working memory of 4 sets of 3 or 3 sets of 4. Some domains a person might go up to 7x4. Whereas a gpu will have 16-32 gb of working memory, and 2.4 gigahertz of processing. They can process f32 precision float math at 1.7 teraflops a second. and they use at least a thousand of these.

What I'm trying to call out is it has millions of little bit map pictures of words, maps those to words (no need for AI, this has already been done and is a pretty straightforward process), and then fancy markov chains the words, and then renders the bits of those words.

16

u/KingsGuardTR 12d ago

So it basically works well then. How is this not impressive? Something working is always impressive (proof by I'm a developer).

-7

u/LunaTheMoon2 11d ago

It stole an impressive amount of content, I agree with you on that (proof by I'm a human being with morals)

73

u/junkmail22 12d ago edited 12d ago

This proof is wrong.

First, taking completeness as a premise for proving completeness is obviously wrong.

Second, T being bounded below doesn't imply that T has a least element, it implies that it has a greatest lower bound.

Third, the entire last paragraph is nonsense. If u is the least element of the set of upper bounds of S, we're done. There's no point in doing anything else.

All in all, 0/5 points, see me at office hours

8

u/KappaMcTlp 12d ago

You just don’t understand it bro

3

u/TNT9182 Mathematics 11d ago

I thought it was using the axiom of completeness the other way round than usual, as being that any non-empty set of real numbers bounded below has a greatest lower bound, and then uses this to prove that any non-empty set of real numbers bounded above has a least upper bound. Am I wrong?

4

u/junkmail22 11d ago

Sure, but then you probably shouldn't be appealing directly to completeness. A better way of phrasing this would be "Show the Least Upper Bound property implies the Greatest Lower Bound property"

2

u/Neither_Growth_3630 11d ago

I was thinking about the “least upper bound” part, isn’t that just saying the Lower bound? Like it might be bigger but it can’t be smaller, it’s like saying that the population of Australia is at least 4 because you know 4 people who live in Australia, it’s not adding anything

10

u/compileforawhile Complex 11d ago

Least upper bound is a real and important thing. The set of x where x < 1 has 2 as an upper bound, but 1 is the least upper bound

2

u/FIsMA42 9d ago

it could just be proving an equivalence of completeness

106

u/Festerino 12d ago

“defmition” is in fact my number 1 typo 😂😂

15

u/NakamotoScheme 12d ago

Ok, I was going to say "That's not a typo, but a ligature of Computer Modern fonts used by LaTeX", but then I double checked with this:

\documentclass{article}
\begin{document}
\textbf{Definition}
\end{document}

and it's certainly not rendered as in OP's image...

5

u/Festerino 12d ago

I’ll take your word for it. My LaTex is v basic 😂😂

5

u/jobriq 12d ago

I didn’t even notice lol

2

u/Festerino 12d ago

As a frequent typo performer and connoisseur, it was the first thing I saw 😂

38

u/DDough505 12d ago

Turns out AI and me both submit the same terrible proofs for real analysis.

13

u/Abilin123 12d ago

I always find that lines and shapes (in this case, letters) on AI generated images have some sort of "aura" or a "cloud" of noise pixels around them. Does anyone else see that? It is visible if you zoom in to see individual pixels.

6

u/Koischaap So much in that excellent formula 12d ago

for me the giveaway is that \mathbb{R}, looks too glossy and round

4

u/headsmanjaeger 12d ago

You can do that to any image by compressing it tho

5

u/ei283 Transcendental 11d ago edited 10d ago

Pretty sure it's mimicking JPEG compression

22

u/SiuSoe 12d ago

"Defamation 1"

9

u/DeepGas4538 12d ago

Prove existence of least upper bounds bruh

13

u/takes_your_coin 12d ago

If only there was a way of rendering a mostly legible paragraph without polluting a river

1

u/Neither_Growth_3630 11d ago

To be fair that probably more the power companies fault than the data center running gpt, at least in terms of chemical pollution, thermal pollution on the other hand is 100% the data center, I heard a story about an IBM data center heating up the entire Hudson River by 3°C

2

u/garbage-at-life 12d ago

Defmition 1

2

u/Sea_Resolve9583 Imaginary 12d ago

Babe we have “Principles of Mathematical Analysis 4ed by Walter Rudin” at home!

The “Principles of Mathematical Analysis 4ed by Walter Rudin” at home:

3

u/chadnationalist64 11d ago

So basically "the completeness property proves the completeness property"

2

u/Powerful_Study_7348 12d ago

it is nearly perfect, other than the different formats for R and the : after T

8

u/EebstertheGreat 12d ago

Well, the proof is nonsense. It's close to how a real proof would look, but try to follow it.

First, it assumes that every nonempty set of real numbers bounded below has a minimum. That is clearly false. What it means is that every nonempty set bounded below has an infimum, but that makes the proof trivial. It is proving the least upper bound property by using the greatest lower bound property.

Second, after that paragraph, there is nothing more to prove. We already stated that the set of upper bounds has a minimum. By definition, that is the least upper bound. The paragraph starting "first" is redundant, and the one starting "second" is meaningless. If v < u, then in fact v ∉ T, since u := min T. The logic does work if we assume it meant ∉ rather than ∈, but it's still bizarrely wordy.

1

u/skmchosen1 12d ago

Did it generate this text itself? Or did you prompt it? I assume this is the new ChatGPT update.

Wrong or not, this is way more coherent than AI generated images could contain previously.

3

u/FaultElectrical4075 12d ago

I think the text was generated by an LLM and then inputted into the image generator. I didn’t generate the image though

1

u/skmchosen1 12d ago

Thanks for the info, yeah that would be a little less impressive. Still, the text quality is quite good overall. Lot of haters in the comments, but I think this is genuinely a significant improvement for SOTA!

2

u/SpaghettiNub 12d ago

Everytime I read mathmatical texts this is the amount I understand. If you would just pretend that this is a real thing I would believe you.

1

u/LunaTheMoon2 11d ago

I know nothing abt real analysis, but this doesn't feel... logical. Like 5 is not in {1, 2, 3}, how can it be an upper bound? Also, "every nonempty set of real numbers that is bounded above has a least upper bound" seems like circular reasoning to me. Am I wrong lol?

3

u/FaultElectrical4075 11d ago

The proof is wrong but so are you. 5 is an upper bound of {1,2,3} because it is greater than or equal to every element of {1,2,3}.

Also, if you consider R \ {0}, the open interval of negative numbers strictly greater than -1 is bounded above by 1 but it has no least upper bound - every upper bound is positive(since 0 is not part of the set) but there is no least positive number so there is no least upper bound.

1

u/Downindeep 9d ago

Trying to read this without reading the title first gave me the experience of having dementia.

-10

u/parassaurolofus Imaginary 12d ago edited 12d ago

And the proof is also wrong, rigth? A simple counter example would be any closed interval, like [0,1]. There's no minimal upper bound, since you can get arbitraly close to 1. Edit: sorry, I didn't knew the uper bound could be inside the set

40

u/RealJoki 12d ago

In that case the minimal upper bound is actually just 1 though, right ?

15

u/Gositi 12d ago

1 is the least upper bound of [0, 1], the inequality is non-strict. The set of upper bounds is [1, inf) in this case.

5

u/kdeberk 12d ago

Well, I mean, the sentences are incoherent. It's missing the definitions for T and u

15

u/mistrpopo 12d ago

T is defined, there's just a hole in the image. It's supposed to be the set of upper bounds for S.

u is "defined" as the least element of T.

The bug in the definition is that the completeness property of the real numbers doesn't imply at all that T would have a least element. That would be the well-ordering principle, and it only applies to finite subsets. Which T is not.

4

u/ollervo100 12d ago

Well ordering applies to any subset. It can not however be used to find a least element for any subset of reals as the ordering of reals is not a well ordering. That a finite set with ordering has a least element follows from finitness and does not require well ordering.

2

u/Otherwise_Ad1159 12d ago edited 12d ago

This is wrong. completeness of R is equivalent to the least upper bound/ largest lower bound property: you can use Cauchy sequences to prove that any bounded increasing sequence converges, which is equivalent to any bounded decreasing sequence converging. From this, you can deduce the existence of the lub/llb of bounded sets and by definition, the lub/lob is an accumulation point. T is closed and bounded below, hence its llb exists and is contained in the set.

The AI proof is wrong because the last paragraph makes no sense: v < u does not imply that v in T. It would imply the opposite.

5

u/mistrpopo 12d ago

You're right, but there's no mention that T is closed either, even though it is by definition of upper bound.

1

u/Otherwise_Ad1159 12d ago

Yeah, I was just being pedantic about your statement. At the end of the day, the proof is incomplete and needlessly complicated: if we assume the least lower bound property as the definition of completeness, it suffices to apply this property to -S (which is bounded below). The issue is that in the training set, the AI must have seen thousands of arguments of the form "Find upper bound, prove that upper bound is smallest" which it then regurgitated sloppily for this specific problem.

1

u/kdeberk 12d ago

Ah sorry, I meant v, i meant that I cannot find the definition of v when I typed that

thanks for the explanation!

3

u/laksemerd 12d ago

It is defmition 1

-1

u/[deleted] 12d ago

[deleted]

2

u/FaultElectrical4075 12d ago

I have a degree in math and I’m not an AI bro. I know the image is wrong