I had a similar construction in mind but tried to formulate it in terms of a dense and complete order that forks at every point when going upward, to get a tree like structure like yours, but it didn't work
I think that should work because every time we need to add a new segment, @AlessandroCodenotti, there's a nonzero distance between the point at which it'd be added and everything constructed thus far
But the idea is "We just won the lottery!" may mean you and I just won lots of money, or it may mean me and my buddy have won lots of money and you haven't.
Similarly, "Nice weather we're having" if the communication is remote (like this one is)
It's rather fascinating, actually, that the $n=2$ case is simple, but the $n=3$ case is not. There's something fundamental there I feel like. Idk what it is
I have no restriction on hypotheses, even a non $T_0$ space would be enough, but I'm also curious to see how well-behaved it can be
the next question will be about what spaces satisfying the $n=2$ case look like, I mean if they are in some sense similar to $\mathbb R$ or if they can get very messy
@AlessandroCodenotti, the $n=4$ case can be gotten from my example by having the new segments cut through the existing segment as opposed to having their endpoint on the existing segment
I'm not entirely convinced that your example works yet, but if that's the case I agree concerning $n=4$ (I mean, it feels like it should work, but I need to think about it!)
if the binary expansions were unique then $f\mapsto\sum\limits_{i=0}^\infty\frac{f(i)}{2^{i+1}}$ would be a bijection
that's not the case though so we introduce an equivalence relation $f\sim g\iff\sum\limits_{i=0}^\infty\frac{f(i)}{2^{i+1}}=\sum\limits_{i=0}^\infty\frac{g(i)}{2^{i+1}}$
And now if you can prove that $|2^{\mathbb N}|=|2^\mathbb{N}/\sim|$ you're done (Hint: how many functions can there be in the same equivalence class?)
okay guys I need your aesthetic expertise. How should I make the horizontal asymptotes look on this graph? I can change type of dots and dashes, alpha, thickness, color...
these occur at $y = \pm \dfrac \pi 2$
this is $y = \operatorname{Si}\left({t}\right)$ if you're curious
yes, it can be done in a short sentence and you'll probably see it if you think about what makes $(0,1)$ and $[0,1]$ different as far as order is concerned
@MikeMiller Yeah, I think you're right. Call each space in the construction $S_i$ so that $S_0=\Bbb R$ and $S_{i+1}=S_i\cup\rm extra~lines$. Every point in $S_n\subset S_{n+m}$ splits $S_{n+n}$ into three when removed, so that should do it
@Akiva No. Pick a neighborhood of any of the starting points. Then for any of the attached [0,infty) in the neighborhood in S_0, it can have whatever open neighborhood of 0 in the tendrils
Look at the bad point in Hawaiian earring. Any nbhd contains a copy of the Hawaiian earring - it's terribad. But nothing like that happens in the wedge.
@MikeMiller I see. But I think there's a modification of it that is first-countable and also solves the problem. It's harder to fix second-countability.
I wonder if it's impossible for second-countable spaces to be "3-splittable" at every point.
Upvoted. Very interesting - I suppose it's very possible that $X$ is a manifold? The current examples are not, but that's not really much of an issue anyway.
