There are some issues with your explanation of the long line. I know you said it's not a proof and more of an intuitive explanation, but some of the intuition you invoked isn't right. The first thing is when you said that the distance from 0 would have to be an increasing function on the long line. There's actually no reason it has to be strictly increasing. Look at the normal line R. d(x,y) = min(|y - x|, 1) is a valid metric, and its metric topology is the same as the standard topology on R. In general, if you have a metric space, capping the metric at 1 doesn't change the topology. The unintuitive idea here is there's no reason a metric has to be "as big" as the space it's representing. If it were possible to define distance on the long line using some extended number system that has numbers beyond the reals, you could just cap that distance at 1 to turn it into an ordinary real metric. The long line being longer than the reals isn't really what makes it not metrizable. You also say "pick a point on the long line that's farther away than the real numbers can reach". Such a point actually doesn't exist. Any closed interval between two points in the long line is homeomorphic to a closed interval in the reals, no matter how far apart the original points are. It is true that, for any particular increasing function from the reals to the long line, there have to be points of the long line the function doesn't reach. It's just that, given any particular point in the long line, there is a map from the reals that reaches it. The reals can reach as far along the long line as you want, but they'll never cover the whole thing. The topological proof that the long line is non-metrizable you've probably seen is that the long line is sequentially compact but not compact, and a sequentially compact metric space has to be compact, so the long line can't be a metric space. The proofs of those implications are a little long, but this is the intuitive sketch for the long line. - Every sequence of points in the long line is bounded. This is because a sequence can only go past a countable number of countable ordinals, which have to be bounded by a countable ordinal, which is in the long line. - Every sequence is bounded inside a closed interval in the long line, and any closed interval in the long line is homeomorphic to a closed interval of the reals. A closed interval is sequentially compact, i.e. every sequence in it has a convergent subsequence. So the long line itself must be sequentially compact. - If there were a metric on the long line, it would be possible to construct a sequence of points which, according to the metric, definitely don't bunch up anywhere. But because the long line is sequentially compact, they have to bunch up somewhere. That's the contradiction. I don't fault you for making these mistakes because the long line is deceptively really hard to actually explain. I think it's one of the coolest concepts in math though, and I'm happy you're one of the very few people who have talked about it online.
There are some issues with your explanation of the long line. I know you said it's not a proof and more of an intuitive explanation, but some of the intuition you invoked isn't right.
The first thing is when you said that the distance from 0 would have to be an increasing function on the long line. There's actually no reason it has to be strictly increasing. Look at the normal line R. d(x,y) = min(|y - x|, 1) is a valid metric, and its metric topology is the same as the standard topology on R. In general, if you have a metric space, capping the metric at 1 doesn't change the topology. The unintuitive idea here is there's no reason a metric has to be "as big" as the space it's representing. If it were possible to define distance on the long line using some extended number system that has numbers beyond the reals, you could just cap that distance at 1 to turn it into an ordinary real metric. The long line being longer than the reals isn't really what makes it not metrizable.
You also say "pick a point on the long line that's farther away than the real numbers can reach". Such a point actually doesn't exist. Any closed interval between two points in the long line is homeomorphic to a closed interval in the reals, no matter how far apart the original points are. It is true that, for any particular increasing function from the reals to the long line, there have to be points of the long line the function doesn't reach. It's just that, given any particular point in the long line, there is a map from the reals that reaches it. The reals can reach as far along the long line as you want, but they'll never cover the whole thing.
The topological proof that the long line is non-metrizable you've probably seen is that the long line is sequentially compact but not compact, and a sequentially compact metric space has to be compact, so the long line can't be a metric space. The proofs of those implications are a little long, but this is the intuitive sketch for the long line.
- Every sequence of points in the long line is bounded. This is because a sequence can only go past a countable number of countable ordinals, which have to be bounded by a countable ordinal, which is in the long line.
- Every sequence is bounded inside a closed interval in the long line, and any closed interval in the long line is homeomorphic to a closed interval of the reals. A closed interval is sequentially compact, i.e. every sequence in it has a convergent subsequence. So the long line itself must be sequentially compact.
- If there were a metric on the long line, it would be possible to construct a sequence of points which, according to the metric, definitely don't bunch up anywhere. But because the long line is sequentially compact, they have to bunch up somewhere. That's the contradiction.
I don't fault you for making these mistakes because the long line is deceptively really hard to actually explain. I think it's one of the coolest concepts in math though, and I'm happy you're one of the very few people who have talked about it online.
could you do a video on hyperreals some day? I find it truly awesome
That's actually a great idea for a video. I'll add it to my list 🙂
@@CHALKND tysm!
Well, let's see what this is...
I don't have the foundation to properly understand it yet.