Decomposing the real line into everywhere isomorphic suborders
We show that if $\mathbb{R} = A \cup B$ is a partition of $\mathbb{R}$ into two suborders $A$ and $B$, then there is an open interval $I$ such that $A \cap I$ is not order-isomorphic to $B \cap I$. The proof depends on the completeness of $\mathbb{R}$, and we show in contrast that there is a partiti...
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| creator | Ervin, Garrett |
| description | We show that if $\mathbb{R} = A \cup B$ is a partition of $\mathbb{R}$ into
two suborders $A$ and $B$, then there is an open interval $I$ such that $A \cap
I$ is not order-isomorphic to $B \cap I$. The proof depends on the completeness
of $\mathbb{R}$, and we show in contrast that there is a partition of the
irrationals $\mathbb{R} \setminus \mathbb{Q} = A \cup B$ such that $A \cap I$
is isomorphic to $B \cap I$ for every open interval $I$. We do not know if
there is a partition of $\mathbb{R}$ into three suborders that are isomorphic
in every open interval. |
| doi_str_mv | 10.48550/arxiv.2303.11532 |
| format | Article |
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two suborders $A$ and $B$, then there is an open interval $I$ such that $A \cap
I$ is not order-isomorphic to $B \cap I$. The proof depends on the completeness
of $\mathbb{R}$, and we show in contrast that there is a partition of the
irrationals $\mathbb{R} \setminus \mathbb{Q} = A \cup B$ such that $A \cap I$
is isomorphic to $B \cap I$ for every open interval $I$. We do not know if
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two suborders $A$ and $B$, then there is an open interval $I$ such that $A \cap
I$ is not order-isomorphic to $B \cap I$. The proof depends on the completeness
of $\mathbb{R}$, and we show in contrast that there is a partition of the
irrationals $\mathbb{R} \setminus \mathbb{Q} = A \cup B$ such that $A \cap I$
is isomorphic to $B \cap I$ for every open interval $I$. We do not know if
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two suborders $A$ and $B$, then there is an open interval $I$ such that $A \cap
I$ is not order-isomorphic to $B \cap I$. The proof depends on the completeness
of $\mathbb{R}$, and we show in contrast that there is a partition of the
irrationals $\mathbb{R} \setminus \mathbb{Q} = A \cup B$ such that $A \cap I$
is isomorphic to $B \cap I$ for every open interval $I$. We do not know if
there is a partition of $\mathbb{R}$ into three suborders that are isomorphic
in every open interval.</abstract><doi>10.48550/arxiv.2303.11532</doi><oa>free_for_read</oa></addata></record> |
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| subjects | Mathematics - Logic |
| title | Decomposing the real line into everywhere isomorphic suborders |
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