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PaperclipPerfector  6d ago (text post)   28539 thread views

Culture War Roundup for the week of February 17, 2025

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Grok 3 is a whelming outcome. The only thing notable about it is how consistent it is with most predictions, including scaling laws.

Unfortunately, its time as SOTA is going to be short, or nonexistent, because the full-fat o3 from OpenAI already ekes out a win in benchmarks. Of course, o3 is not technically available to the public, so among released models that are pay to play, Grok reigns.

I've only played around with it a little bit, through lmarena.ai because I'm not in dire need of any paid plan. It seemed fine? There's one particular question that I ask LLMs, courtesy of my maths PhD cousin: "Is the one-point compactification of a Hausdorff space itself Hausdorff?" The correct answer is yes, or so I'm told. Grok 2 fails, Grok 3 succeeds. But so do GPT-4o, Gemini 2.0 Pro and the like.*

(I've asked this so many times that my keyboard automatically suggests the entire question, talk about machine learning)

In short, Grok 3 is a mild bull signal for LLMs, and a slightly stronger one for xAI. It doesn't seem to be astonishingly good, or break ground other models haven't reached. It also hasn't been made retarded by too much Reinforcement Learning from Elon's Feedback. He shared an excerpt showing it lambasting an outlet called The Information, but actual users get a far more measured response. I cynically suspect that a few xAI engineers probably set the sycophancy setting to the maximum when he's using it.

*I'm probably remembering the original explanation I was given wrong, and that my cousin had said it was a no except with additional qualifiers. Mea culpa.

Edit: On a second try with Grok 3:

In all relevant cases, we can find disjoint open sets separating any two distinct points in (X^). Therefore, the one-point compactification (X^) of a Hausdorff space (X) (assuming (X) is locally compact) is itself Hausdorff.

Which I believe is the correct answer.

"Is the one-point compactification of a Hausdorff space itself Hausdorff?" The correct answer is yes, or so I'm told.

It’s not true. You need to also assume that the original space is locally compact. For example, if you one-point compactify the space of rational numbers (which is obviously Hausdorff), the resulting space is non-Hausdorff. That’s because the only compact subsets of rationals are discrete, and thus finite, so open subsets that contain the added point at infinity are exactly of the form Q plus the extra point minus some finite subset. This means that it intersects every other nonempty open subset (because all open subsets of that space are infinite). Thus, you cannot separate the point at infinity from any other point by two disjoint open sets, because there are no disjoint open sets from the ones that contain the point at infinity.

I'm going to pre-face this by saying I absolutely do not have the maths for this, or at least I've forgotten the explanation my cousin gave.

It's been quite a while since I first asked him, and I don't have him at hand. I don't recall if he pointed out if this holds true if and only if the initial Hausdorff space is also locally compact. I think he might have done so, but I can't be sure. In that case, the error would be mine, because I was only looking for a yes or no answer.

It's an interesting quandary I find myself in, given that I have authoritative sources on both sides as well as LLMs disagreeing with each other. I fed the initial question into the best LLMs I have access to, and if they said yes, then I submitted your rebuttal. They acknowledge that you are correct, though there is some quibbling in the reasoning traces I am not qualified to judge.

I have had mixed results asking reasoning models to answer. The ones I am most satisfied with is:

Final Answer: No. The one-point compactification of a Hausdorff space is Hausdorff if and only if the space is locally compact Hausdorff. Since there exist Hausdorff spaces that are not locally compact, the one-point compactification of such spaces are not Hausdorff. Therefore, the one-point compactification of a Hausdorff space is not necessarily Hausdorff.

Courtesy of Gemini 2.0 Flash Thinking.

ChatGPT with reasoning on:

Conclusion: The one-point compactification of a Hausdorff space is Hausdorff if and only if is locally compact. If is not locally compact, then there exists at least one point in that does not have a compact neighborhood, and consequently, the separation of that point from fails in .

Thus, the answer is: The one-point compactification of a Hausdorff space is Hausdorff if and only if the space is locally compact.

I can't really preserve the formatting with just markdown, so anything Latex would be missing. I'd be happy to share other responses in an offsite text repository, since I don't want to clutter up the thread.

My tentative conclusion is that both you and my cousin are correct, and even LLMs are (most of the time). My error was in expecting a yes or no answer due to my fallible memory.

There's one particular question that I ask LLMs, courtesy of my maths PhD cousin: "Is the one-point compactification of a Hausdorff space itself Hausdorff?" The correct answer is yes, or so I'm told.

Are you just asking it as a yes/no question? This is a standard question that a first-year undergrad could be asked to check that they understood the definitions, and it's unlikely that the answer wouldn't be in the training set somewhere. For example, I quickly fed it to a Q4_K_M quantised Qwen2.5-3B (that's on the level that you could run on a smartphone nowadays), and it completed

Q: Is the one-point compactification of a Hausdorff space itself Hausdorff?

A:

with

Yes, the one-point compactification of a Hausdorff space is itself Hausdorff.<|endoftext|>

edit: See @wlxd's discussion for why the correct answer is actually "No". In fact, Qwen2.5-3B is almost perfectly on the edge: the log-odds of the first token of an alternative answer that goes "No, the one-point compactification of a Hausdorff space is not necessarily Hausdorff.<|endoftext|>" is only about 0.21 lower, so the probability of it being chosen is about e^-0.21 or 0.81 times the probability that it would pick the "Yes...". (Think 45% No to 55% Yes.)

I've replied to him. I think there's been error on my part, my cousin's explanation probably included an elaboration on the additional qualifications of the answer.

Most of the reasoning models seem to be saying yes (I didn't keep track of precise numbers), and if they said no, then @wlxd 's response is one they acknowledge as correct.

The point is that this answer is just incorrect. There are non Hausdorff one point compactifications of Hausdorff spaces. You need additional assumption of local compactness for it to be true.

You are right, but I don't think that was "the point", given that @self_made_human apparently was led to believe that it is yes (and seemed to treat that answer being given as a success criterion).

(I was actually in the process of writing up another response as I had realised it is not true, after I fed the question to DeepSeek-R1's Qwen7B distill to reason through and found that it choked as it tried to conjure up compact neighbourhoods that I didn't see the grounds for existing, but I hadn't gotten to the point of having a good counterexample yet)

It's likely PEBKAK on my part. My cousin had explained the reasoning almost a year back (eons in LLM time) and I'd likely forgotten the finer qualifications and only remembered the yes or no bit.

I give a counter example in my other comment.

Ah, thanks, that works.

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