288: For Injective Monotone Continuous Operation from Ordinal Numbers Collection into Ordinal Numbers Collection and Image of Subset of Domain, Union of Image Is in Range

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A description/proof of that for injective monotone continuous operation from ordinal numbers collection into ordinal numbers collection and image of subset of domain, union of image is in range

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Topics

About: set

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof
• 3: Note

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Starting Context

• The reader knows a definition of injection.
• The reader knows a definition of monotone operation.
• The reader knows a definition of continuous operation on the ordinal numbers collection.
• The reader knows a definition of set.
• The reader admits the proposition that any unbounded collection of ordinal numbers is not any set.
• The reader admits the proposition that any bounded collection of ordinal numbers has the supremum.
• The reader admits the proposition that for any 2 ordinal numbers, one is a proper subset of the other if and only if the former is a member of the latter.

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Target Context

• The reader will have a description and a proof of the proposition that for any injective monotone continuous operation from the ordinal numbers collection into the ordinal numbers collection and the image of any nonempty subset of the domain, the union of the image is in the range.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Description

For any injective monotone continuous operation, $f:O\to O$, where $O$ is the ordinal numbers collection, any nonempty subset of the ordinal numbers set, $I\subseteq O$, and $S=\{f(o)|o\in I\}\subseteq f(O)$, the union of $S$ is in the range, which is $\mathrm{\exists }{o}^{\prime }\in O,\cup S={\cup }_{o\in I}f(o)=f(o^{\prime })$.

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2: Proof

$I$ is bounded by the proposition that any unbounded collection of ordinal numbers is not any set. There are 2 possibilities: 1) $I$ has the maximum, $m$ 2) $I$ has the non-maximum supremum, $m$, by the proposition that any bounded collection of ordinal numbers has the supremum.

For 1), ${\cup }_{o\in I}f(o)=f(m)$, because $m\in I$ and for any $o\in I$, ($o\in m$ and $f(o)\in f(m)$) or ($o=m$ and $f(o)=f(m)$), which implies $f(o)\subseteq f(m)$, by the proposition that for any 2 ordinal numbers, one is a proper subset of the other if and only if the former is a member of the latter, while there is an $o\in I$ such that $f(o)=f(m)$.

For 2), $m$ is really a limit ordinal number, because otherwise, $m$ would be a successor ordinal number ($m$ cannot be $0$), but then, $m=m^{\prime +}$ for an $m^{\prime }$ and $m^{\prime }$ would have to be in $I$ because otherwise, $m^{\prime }$ would be the supremum instead of $m$, but $m^{\prime }$ could not be in $I$ because $m^{\prime }$ would be the maximum, a contradiction. So, $f(m)={\cup }_{o\in m}f(o)$ by the continuousness. In fact, ${\cup }_{o\in I}f(o)={\cup }_{o\in m}f(o)$, because for any $p\in {\cup }_{o\in I}f(o)$, there is an $o^{\prime }\in I$ such that $p\in f(o^{\prime })$, but $o^{\prime }\in m$ because $m$ is the supremum, so, $p\in {\cup }_{o\in m}f(o)$; for any $p\in {\cup }_{o\in m}f(o)$, there is an $o^{\prime }\in m$ such that $p\in f(o^{\prime })$, but as $m$ is the supremum, there is an $o^{″}\in I$ such that $o^{\prime }\in o^{″}\in m$ (because otherwise, o' would be the supremum), then, $f(o^{\prime })\subset f(o^{″})$, so, $p\in f(o^{\prime })\subset f(o^{″})$; so, $p\in {\cup }_{o\in I}f(o)$.

By the replacement axiom, the requirement that $I$ is a set is equivalent with the requirement that $S$ is a set: if $I$ is a set, $S$ will be a set; if $S$ is a set, as $f$ is injective, the inverse, $f^{-1}$, from $S$ will constitute the legitimate formula for the replacement axiom, so, $I$ will be a set.

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3: Note

The term, "operation", instead of 'map' or 'function' is used here because it is not from a set into a set.

While my definition of 'injection' is about a map, the term, "injective", is used here although $f$ is not really any map, as it will not cause any practical confusion,

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References

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