# A question in general topology

Motivated by this question we ask:

Up to homeomorphism, are there only a finite number of connected locally compact hausdorff topological space $X$ such that $X$ has an open set $U$ homeomorphic to $\mathbb{R}$ such that $X-U$ is homeomorphic to $\mathbb{R}$, too? (Note that there is only one disconnected possibility, so we add connectedness assumption)

## 1 Answer

There are continuum pairwise non-homeomorphic closed $1$-dimensional subspaces $\ X\$ of $\ \mathbb R\times[0;\infty)\$ which contain $\ \mathbb R\times\{0\},\$ and such that $\,\ X\,\setminus\,(\mathbb R\times\{0\})\,\$ is homeomorphic to $\ \mathbb R$.

The construction: a line goes from up high gradually down to level $\ y=0,\$ between $\ x=-1\$ to $\ x=1\$ as follows: it goes one time from $\ x=0\$ to $\ x=-1,\$ and back to $\ x=0;\$ Then it goes a few times back and forth from $\ x=0\$ to $\ x=1;\$ then the same again and again except that the numbers of the repetitions on the right form an infinite sequence of non-negative integers $\ n_1\ n_2\ \ldots\$ Then such $\ (n_1\ n_2\ \ldots)\$ and $\ (m_1\ m_2\ \ldots)\$ curves are homeomorphic $\ \Leftrightarrow\ \exists_{\nu\ \mu}\forall_{k\ge 0}\ n_{\nu+k}=m_{\mu+k}.\$ Such two integer sequences are called (and really are :-)) equivalent. But there are continuum of inequivalent sequences of this kind.

Proof   One should consider the sequence, on one such curve, of points for which the x-coordinate is $\ \frac{-1}2,\$ and the sequence of points for which the x-coordinate is $\ \frac{1}2.\$ Consider a homeomorphism onto another such curve. Look at the image of the two mentioned sequences. Etc.

REMARK 0   To prove that there are a continuum of such non-homeomorphic curves it is enough to consider only 0-1 sequences or 1-2 sequences.

REMARK 1   A curve $\ (n_1\ n_2\ \ldots)\$ admits a self-homeomorphism which reverses the orientation of $\ \mathbb R\times\{0\}\ \Leftrightarrow\$ all integers $\ n_k\$ are equal 1 but for a finite number of exceptions, or when all $\ n_k\$ are equal $\ 0\$ but for a finite number.

REMARK 2   Instead of $\ \mathbb R\times\{0\},\$ one can take any closed (or open, or half-open -- either way) interval which contains $\ (-1;1)\times\{0\}\$ (but for the sake of REMARK 1, the symmetric nature of points $\ (-1\ 0)\$. and $\ (1\ 0)\$ should be preserved).

A similar or possibly virtually identical construction (but for trivial details) was invented by Zenon Waraszkiewicz (1909-1946); today his curves are called Waraszkiewicz spirals. Perhaps someone can find their description in literature. These spirals gave a continuum of non-homeomorphic compact curves in $\ \mathbb R^2$. EDIT If I remember well, Waraszkiewicz's spiral consists of a circle, and of a spiral which approaches the circle from (say) outside, and nearly perfectly cycles around the circle left (i.e. against the clock) a few times, then right around a few times, etc. You may even assume that there are only single cycles left, and multiple right (each time a positive number). The construction looks a little different but the idea of the proof is the same (at least in my rendition; however a proof is natural).

• Can you link to a picture of the construction? – Todd Trimble Jan 3 '15 at 16:35
• @ToddTrimble -- I wish. I don't have the ability to make computer graphics. I am sorry for my limitation. But the proof is really simple, and I hope that my outline is adequate. I am willing to go into all necessary $\epsilon$-$\delta$ (:-) if I am requested to do so. – Włodzimierz Holsztyński Jan 3 '15 at 16:48
• I think your description is very clear and transparent. – Christian Remling Jan 3 '15 at 17:06
• @ChristianRemling -- thank you. For the sake of MO it is sometimes preferable to post only an outline instead of a complete argument in the whole gory detail (I'd think). – Włodzimierz Holsztyński Jan 3 '15 at 17:15
• Having read it over more slowly, I now agree with Christian. It's a very nice construction! – Todd Trimble Jan 3 '15 at 17:17