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 1


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).

  • $\begingroup$ Can you link to a picture of the construction? $\endgroup$
    – Todd Trimble
    Jan 3, 2015 at 16:35
  • $\begingroup$ @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. $\endgroup$ Jan 3, 2015 at 16:48
  • 2
    $\begingroup$ I think your description is very clear and transparent. $\endgroup$ Jan 3, 2015 at 17:06
  • 1
    $\begingroup$ Having read it over more slowly, I now agree with Christian. It's a very nice construction! $\endgroup$
    – Todd Trimble
    Jan 3, 2015 at 17:17
  • 1
    $\begingroup$ @WłodzimierzHolsztyński thank you very much for your elegance answer:) $\endgroup$ Jan 6, 2015 at 20:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.