Prove that $\bigcup\mathbb{N}=\mathbb{N}$.

Showing that $\mathbb{N}\subseteq \bigcup\mathbb{N}$ is simple.

However, I'm not seeing how to handle showing $\bigcup\mathbb{N}\subseteq\mathbb{N}$.

I know that $\mathbb{N}:=\bigcap\{z\in\mathcal{P}(x)|\text{ $x$ is inductive}\}$. Therefore,

$$\bigcup\mathbb{N}\Leftrightarrow \bigcup\big(\bigcap\{z\in\mathcal{P}(x)|\text{ $x$ is inductive}\}\big)$$

But I'm not sure what to make of this.


The definition of a natural number I'm working with is $\mathbb{N}=\{\mathbf{0},\mathbf{0^{+}},\mathbf{0^{++}},\mathbf{0^{+++}}\ldots\}$, where $\mathbf{0}=\emptyset$ and $\mathbf{0^+}=\mathbf{0} \cup {\mathbf{\{0\}}}$ and the set of natural numbers is as defined above

  • $\begingroup$ Do you mean $\bigcup_{n\in \mathbb{N}}\{n\}$? $\endgroup$ – QTHalfTau Jul 7 '16 at 2:58
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    $\begingroup$ No, I mean $\bigcup$ given by $\bigcup A := \{x: \exists X \in A: x \in X\}$ $\endgroup$ – user352541 Jul 7 '16 at 3:00
  • $\begingroup$ if you mean union with itself the answer is trivial $\endgroup$ – Zelos Malum Jul 7 '16 at 3:02
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    $\begingroup$ How do you define a number $n\in \mathbb{N}$? Do you define $n=\{0, 1, \cdots, n-1\}$ with $0=\emptyset$? It seems so... Otherwise I don't understand what you mean by $\bigcup \mathbb{N}$. $\endgroup$ – Hamed Jul 7 '16 at 3:10
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    $\begingroup$ $\mathbb{N}=\{\bf{0},\bf{0^+},\bf{0^{++}},\bf{0^{+++}}\ldots\}$, where $\bf{0}=\emptyset$ and $\bf{0^{+}}={\bf{0}\cup\{\bf{0}\}}$. $\endgroup$ – user352541 Jul 7 '16 at 3:21

If you define what you mean by $\mathbb{N}$ a little more precisely (e.g., $\mathbb{N}$ is the minimal set containing $0$ and closed under the map taking $x$ to $x \cup \{x\}$) then you should be able to show by induction on sets that every element of a natural number is a natural number; that is, if $n \in \mathbb{N}$ and $x \in n$, then $x \in \mathbb{N}$. Then $\bigcup \mathbb{N}$ includes only natural numbers, so is clearly a subset of $\mathbb{N}$.


The "..." in your def'n of $\mathbb N$ is not precise and conceals some assumptions. Without clarifying this def'n you may either get stuck or get into a circular argument.


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