I've shown there exist continuous onto map from $\mathbb{R}$ to $\mathbb{R}^{n}$ for any finite $n$. Now my question can we find a continuous surjection from $\mathbb{R} \to \mathbb{R}^{\omega}$ ? Where $\mathbb{R}^{\omega}$ is the usual notation and it has product topology. (I can see its not possible with box topology because $\mathbb{R}^{\omega}$ with box topo is not connected)
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Given continuous surjections $f_n\colon[0,1]\to[0,1]^n$, you can probably stitch them togther to cover $\Bbb R^\omega$, which is the union of a sequence of subsets homeomorphic to $[0,1]^n$ (for example, the $n$th subset is $[-n,n]^n \times {0}^\omega$). – Greg Martin Jun 19 '16 at 06:37
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2@Greg Martin . That would fail to include any sequence $(x_n)_{n\in \omega}$ $\in R^{\omega}$ for which ${n: x_n\ne 0}$ is infinite, – DanielWainfleet Jun 19 '16 at 08:28
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Good point, darn. – Greg Martin Jun 19 '16 at 20:38
