Ordinal: Difference between revisions

43 bytes added ,  10 months ago
Provable over ZFC (and I'd expect some weaker theories)
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(Provable over ZFC (and I'd expect some weaker theories))
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* If \(\beta\) is a limit ordinal, \(\alpha^\beta = \bigcup\{\alpha^\gamma: \gamma < \beta\}\)
 
There are helpful visual representations for these, namely with sets of lines. For some basic intuition, [https://www.youtube.com/watch?v=SrU9YDoXE88 see this video]. For example, \(\alpha + \beta\) can be visualized as \(\alpha\), followed by a copy of \(\beta\). Note that our definition gives \(1 + \omega = \bigcup\{1+n: n < \omega\} = \omega\), and this makes sense, since a single line, followed by infinitely many lines, is no more than just infinitely many lines. Meanwhile, \(\omega + 1 = \omega \cup \{\omega\}\): you have infinitely many lines, followed by a single one after all of them. This intuition is formally provedformalized by the following statement, which is provable over ZFC: "if \(X\) and \(Y\) are well-ordered sets with order-types \(\alpha\), \(\beta\), respectively, then \(X\), concatenated with a copy of \(Y\), has order-type \(\alpha + \beta\)".
 
Also, \(\alpha \cdot \beta\) can be imagined as \(\beta\), with each individual line in \(\beta\) replaced with a copy of \(\alpha\). For example, \(\omega \cdot 2\), is two lines, with each individual line replaced with a copy of \(\omega\), i.e: 2 copies of \(\omega\), or \(\omega + \omega\).
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