I posted a related question earlier about the same problem in Chapter 5 of N. L. Carothers Real Analysis.
Show that $ \mathbb{N}$ is homeomorphic to the set $\{e^{(n)} : n\geq 1\}$ when considered as a subset of one of the spaces $c_0, \ell _1,\ell _2, \ell_3$. If we instead take the discrete metric on $\mathbb{N}$ show that the map $n\rightarrow e^{(n)}$ is an isometry into $c_0$.
I have tried to prove this. First I showed that $\{e^{(n)} : n\geq 1\}$ was a subset of $c_0, \ell _1,\ell _2, \ell_3$ then I claimed that a homeomorphism $f:\mathbb{N} \rightarrow \{e^{(n)} : n\geq 1\}$ existed since every subset of $\mathbb{N}$ is closed and $f(n) = e^{(n)}$ produces closed sets in $\mathbb{R}^n.$ But each of $c_0, \ell _1,\ell _2, \ell_3$ are subsets of $\mathbb{R}^n$ and so $f$ produces closed sets in each of these subspaces.
Is that true though? If a function produces closed sets in $\mathbb{R} ^{n}$ does it produces closed sets in subspaces of $\mathbb{R} ^{n}$?
Thanks! (ps previous post was deleted)