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I know that the cardinality of $\mathbb{R}^n$ is equal to $\mathbb{R}$. I also know that the cardinality of $\mathbb{N}^n$ is equal to $\mathbb{N}$, but how do I prove that the cardinality of $\mathbb{R}^\mathbb{N}$ is smaller than the cardinality of $\mathbb{N}^\mathbb{R}$.

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    $\begingroup$ You can start first by showing $\mathbb{R}^{\mathbb N}$ is actually $\mathbb{R}$. $\endgroup$ – Quang Hoang Jan 28 at 17:10
  • $\begingroup$ Your choice of fonts is very inconsistent. $\endgroup$ – Asaf Karagila Jan 28 at 17:10
  • $\begingroup$ $|\mathbb{R}^{\mathbb{N}}| = |\mathbb{R}|^{|\mathbb{N}|} = (2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0\aleph_0} = 2^{\aleph_0}$. $|\mathbb{N}^{\mathbb{R}}| = |\mathbb{N}|^{|\mathbb{R}|} = (\aleph_0)^{2^{\aleph_0}} \geq 2^{2^{\aleph_0}}$. $\endgroup$ – Arturo Magidin Jan 28 at 18:17
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$$|\mathbb{N}^\mathbb{R}|\ge |2^\mathbb{R}|>|\mathbb{R}|=|\mathbb{R}^n|(=|\mathbb{R}^\mathbb{N}|).$$

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