Is there a continuous surjective map of $\mathbb{R}$ onto $\mathbb{R}^{\infty}$? a) If $\mathbb{R}^{w}$ is given the product topology, determine whether or not there is a continuous surjective map of $\mathbb{R}$ onto the subspace $\mathbb{R}^{\infty}$. 
b) What happens to a) if $\mathbb{R}^w$ is given the uniform topology or the box topology ?
Where $\mathbb{R}^{w}=\prod_{n=1}^{\infty} X_n $ with $X_n=\mathbb{R}$ and $R^{\infty}$ is a subset of $R^{w}$ consisting of all sequences that are "evantually zero", that is, all sequences $(x_1,x_2,...)$ such that $x_i \neq 0$ for only finitely many values of $i$. 
I don't have any idea for this problem. Thanks everybody. 
 A: For (a) the answer is positive: $\mathbb{R}^\infty$ is a continuous image of $\mathbb{R}$. First of all note that $\mathbb{R}^\infty$ can be written as
$$\mathbb{R}^\infty=\bigcup_{n=0}^{\infty} I_n$$
where each $I_n$ has some nice properties: it is compact, connected, locally connected and second countable (so that Hahn-Mazurkiewicz Theorem applies) and $0\in I_n$ for all $n$. For example $I_n=[-n,n]^n\times\{0\}^\infty$.
Now there's a space filling curve $f_n:[0,1]\to I_n$ for each $n$ by the Hahn–Mazurkiewicz Theorem. We can tweak each $f_n$ so that $f_n(0)=f_n(1)=0$ since each $I_n$ is path connected.
Now define
$$f:[0,\infty)\to\mathbb{R}^\infty$$
$$f(t)=f_n(t-n)\text{ if }t\in [n,n+1]$$
It's a simple glueing of all those curves. The problematic points of continuity are integers only. But these are guaranteed to be continuous by $f_n(0)=f_n(1)=0$ property. Obviously $f$ is surjective.
To finish the construction just compose $f$ with any continuous surjection $\mathbb{R}\to[0,\infty)$, e.g. $x\mapsto x^2$.

As for (b) I think that the answer stays the same because even though uniform, product and box topologies are different on $\mathbb{R}^\omega$ they coincide on $\mathbb{R}^\infty$. Right?
