I am trying to solve an exercise from Munkres:
Is there a continuous surjective map $f : \mathbb{R} \rightarrow \mathbb{R} ^\infty$ in the product topology? What about uniform and box topologies?
The answers I have found so far all use the idea of "gluing" some functions together. Specifically, notice that $\mathbb{R}^\infty = \cup_{n = 1} ^ \infty [-n, n] ^ n$ and we may define a continuous surjective map $f_n$ from $[n, n + 1]$ to $[-n, n] ^ n$ for each $n \in \mathbb{Z}_+$. Using some tricks I may choose the functions $f_n$ so that they agree on end points. However, what I do not understand is why the function $f$ obtained by gluing all these functions together is continuous. The pasting lemma should only work when the topological space is a finite union of closed subspaces right? Thanks in advance!
Update: It turns out that I can use local finiteness to finish the proof for product topology. However, I am stuck on proving or disproving the claim for uniform and box topogies on $\mathbb{R} ^ \infty$.
Clarification $\mathbb{R} ^ \infty$ is used to denote the space of all real sequences that are eventually $0$.
Best Answer
The pasting lemma works in the particular case you mentioned. This is basically because if $\{A_i\}$ is a locally finite collection, then $\cup \overline{A_i}$ = $\overline {\cup {A_i}}$.
Therefore, for each closed $F \subseteq \mathbb{R}^\infty$, $f^{-1}(F) = \cup _n f_n^{-1}(F)$ is closed, since the collection $\{f_n ^{-1} (F) \}_n$ is locally finite.