Space of bounded functions and $\mathbb{R}^N$ Let $B(X)$ $=$ $\{$ $f:X\rightarrow \mathbb{R}$ $:$ $f$ is bounded $\}$.
Show that if $X=$ $\{$ $1,2...,n$ $\}$. Then $B(X)$ $=$ $\mathbb{R}^n$.
My attempt:
If $g$ is a bounded function from $X$ to $\mathbb{R}$ then observe that $g(1)=x_1$, $g(2)=x_2$...., $g(n)=x_n$. Hence $g$ coincides with the vector in $\mathbb{R}^n$  with components $x_1,x_2...,x_n$. On the other hand, if $x \in \mathbb{R}^n$ then $x=(x_1,x_2...,x_n)$. Clearly then $x=f$ where $x_i=f(i)$. Since the domain is finite, $x\in B(X)$.
Is this correct?
 A: The general idea of your answer is correct. But there is slightly confusing abuse of notation. Let $A(X)=\{f:X\to\mathbb{R}: f \text{ is bounded}\}$. Then $B(X)=\{Im(f): f\in A(X)\}$. Then we wish to show $B(X)=\mathbb{R}^n$.
Suppose there is a function $g$ such that $Im(g)\in B(X)$. Since $g\in A(X)$, we know $g$ is bounded so $Im(g)=(g(1),g(2),...,g(n))\in\mathbb{R}^n$. Since $g$ is arbitrary, we conclude $B(X)\subset \mathbb{R}^n$.
Conversely, suppose $y\in \mathbb{R}^n$. Define the function $g$ by $g(i)=y_i$ for all $i=1,...,n$. Then $g\in A(X)$ so $Im(g)\in B(X)$. Therefore, $\mathbb{R}^n\subset B(X)$.
A: I would like to elaborate on my solution.
Definition: Let $n\in\mathbb{N}$ be fixed. Given a set $X$. An $n-tuple$ of $X$ is defined by a function $\textbf{x}:$ $\{$ 1,2....,n $\}$ $\rightarrow X$.  The $n-tuple$ is denoted by the following:
$\textbf{x}=(x_1,x_2....,x_n)$ where $x_i$ $=$ $\textbf{x}(i)$
From the above definition ,it is immediate that:
If $X,Y$ are sets. Then  
$X\times Y$ $=$ $\{$ $(x,y)$ $:$ $x\in X$ and $Y\in Y$ $\}$ $=$ $\{$ $\textbf{x}$ $:$ $\{$ 1,2 $\}$ $\rightarrow $ $X\cup Y$ : $\textbf{x}(1)\in X$ and $\textbf{x}(2)\in Y$ $\}$
Returning to my problem, I claim that the following two sets are equal:
$\mathbb{R}^2$ $=$ $\{$ $(x,y)$ $x\in \mathbb{R}$ and $y\in \mathbb{R}$ $\}$
and $B( ${$ $1,2$ $})$ =$ $\{$ $f$ $:$ $\{$ $1,2$ $\}$ $\rightarrow$ $\mathbb{R}$ $:$ $f$ is bounded $\}$ are equal (note, I set n=2 for simplicity)
Indeed, if $(x,y)\in \mathbb{R}^2$ Then it is indeed an $2-tuple$ by the definition above. It is therefore bounded because the domain is bounded. If $f$ $:$ $\{$ $1,2$ $\}$ $\rightarrow$ $\mathbb{R} \cup \mathbb{R} $ is bounded, then notice that  $f(1),f(2)\in \mathbb{R}$ and so $f=(f(1),f(2))$ by the above definition, is a member of $\mathbb{R}^2$.
