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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 from a cirlcle, and from 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).

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).

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 homeomorphich 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 from a cirlcle, and from 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).

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 from a cirlcle, and from 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).

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 homeomorphich 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.

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 from a cirlcle, and from 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).

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 homeomorphich 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 from a cirlcle, and from 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).