# How to prove $\mathbb{R}^n < \mathbb{N}^\mathbb{R}$

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 $$\mathbb{N}^\mathbb{R}$$.

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

$$|\mathbb{N}^\mathbb{R}|\ge |2^\mathbb{R}|>|\mathbb{R}|=|\mathbb{R}^n|(=|\mathbb{R}^\mathbb{N}|).$$