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Let $\newcommand{\N}{\mathbb{N}}\newcommand{\B}{\mathbf{B}}\B(\N)$ be the collection of all bounded functions $f:\N\to\N$. (A function $f:\N\to\N$ is bounded if there is $M\in\N$ such that $f(k) < M$ for all $k\in\N$.) We define a metric $d$ on $\B(\N)$ by setting $$d(f,g) = \sum_{n\in\N}\frac{1}{2^n}|f(n)-g(n)|$$ for all $f,g\in \B(\N)$. This metric induces a topology on $\B(\N)$.

Is there an injective continuous map $\varphi:\B(\N)\to\mathbb{R}$ endowed with the Euclidean topology? Is there a surjective continous map $\psi$ from $\B(\N)$ onto $\mathbb{R}$?

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    $\begingroup$ Finding a surjective continuous map is a straightforward exercise. For instance $f\mapsto \sum 2^{-n}f(n)$. $\endgroup$
    – YCor
    Commented Nov 19 at 23:52
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    $\begingroup$ Finding an injective map is also a straightforward exercise. Map injectively $\mathbf{N}$ into the Cantor $K$. Then $\mathbf{N}^\mathbf{N}$ maps injectively continuously into $K^\mathbf{N}\simeq K$, which embeds into $\mathbf{R}$. $\endgroup$
    – YCor
    Commented Nov 19 at 23:53
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    $\begingroup$ @YCor Alternatively, $\mathbb{N}^\mathbb{N}$ is the Baire space, which is known to be homeomorphic to the space of irrational numbers. $\endgroup$
    – David Gao
    Commented Nov 19 at 23:57
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    $\begingroup$ Ah, this is not exactly the product topology but a finer one. So to ensure continuity (namely Lipschitz), take $f\mapsto\sum 3^{-n}f(n)$ instead. For the injection, just observe that the identity map to ($\mathbf{N}^\mathbf{N}$, product topology) is continuous and conclude as in my previous comment. $\endgroup$
    – YCor
    Commented Nov 20 at 0:21
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    $\begingroup$ @YCor What do you mean? $f \mapsto \sum 2^{-n} f(n)$ is continuous. $|\sum 2^{-n} f(n) - \sum 2^{-n} g(n)| \leq \sum 2^{-n} |f(n) - g(n)| = d(f, g)$. $\endgroup$
    – David Gao
    Commented Nov 20 at 0:30

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For every $K$ the subspace $X_K$ of all functions bounded by $K$ is homeomorphic to the Cantor set, as $X_K=\{1,2,\ldots,K\}^\mathbb{N}$ and the metric induces the product topology. It follows that $\mathbf{B}(\mathbb{N})$ is a $\sigma$-compact separable metric that is also zero-dimensional. Hence it is can be embedded into the Cantor set, and so also into $\mathbb{R}$.

For every $n\in\mathbb{N}$ let $A_n=\{f:f(1)=n$ and $f(k)\le2$ if $k>1\}$. The family $\{A_n:n\in\mathbb{N}\}$ is a discrete family of copies of the Cantor set, hence its union, $A$, is also closed. For each $n$ map $A_n$ continuously onto the interval $[-n,n]$. This yields a continuous surjection $s:A\to\mathbb{R}$; extend it to all of $\mathbf{B}(\mathbb{N})$ by the Tietze-Urysohn theorem.

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