Motivated by this [question](https://mathoverflow.net/questions/191120/the-letters-of-the-word-art) we ask:



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> 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)
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A question in general topology
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).



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> 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).
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