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Can we say N is strictly a subset of P(N)?

  1. Oct 26, 2012 #1
    Essentially, do we know [itex]\mathbb{N}[/itex][itex]\subset[/itex][itex]{P}\mathbb{(N)}[/itex]?
     
  2. jcsd
  3. Oct 26, 2012 #2

    micromass

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    A lot depends on your definition of [itex]\mathbb{N}[/itex]. But if we take the usual set theoretic definition, then the inclusion is indeed true.

    What is the set theoretic definition of [itex]\mathbb{N}[/itex]? It goes as follows:

    • Define [itex]0=\emptyset[/itex]
    • If [itex]0,...,n[/itex] are defined, then define [itex]n+1=\{0,...,n\}[/itex]
    • Define [itex]\mathbb{N}=\{0,1,2,...\}[/itex]

    Now we can check that [itex]\mathbb{N}\subseteq \mathcal{P}(\mathbb{N})[/itex]. First of all, [itex]\emptyset[/itex] is a subset of [itex]\mathbb{N}[/itex] and thus [itex]\emptyset\in \mathcal{P}(\mathbb{N})[/itex]. Furthermore, [itex]\{0,...,n\}[/itex] is a subset of [itex]\mathbb{N}[/itex], and thus [itex]n+1=\{0,...,n\}\in \mathcal{P}(\mathbb{N})[/itex].
    This shows that [itex]\mathbb{N}\subseteq \mathcal{P}(\mathbb{N})[/itex]. The inclusion is strict since [itex]\{2\}[/itex] is a subset of [itex]\mathbb{N}[/itex], but not an element.
     
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