Let $N=\{1,2,3,…,n\}$, show there is a bijection between $R^n\ and\ R^N$
Define $f: R^n \rightarrow R^N$ by $f(x_1,..,x_n)=g_{(x_1,…x_n)}$ where the n-tuple is an element of $R^n$ and $g_{(x_1,…x_n)}:N \rightarrow \mathbb{R}$ is defined by:
$g_{(x_1,…x_n)}(i)=x_{i}$ for $i \in N$
What I want is g to map the elements in N to the corresponding real number in the n-tuple. But I'm not sure that I can define g in this way.
But if I can here is my proof this is injective.
Suppose $(x_1,…x_n) \neq (y_1,…y_n)$ then there is at least one $i \in N$ such that $x_i \neq y_i$.
Then the function $g_{(x_1,…x_n)}(i)=x_i \neq y_i=g_{(y_1,…y_n)}$ for some $i \in N$
Since $f(x_1,..,x_n)=g_{(x_1,…x_n)}$ then $f(x_1,…,x_n)\neq f(y_1,…,y_n)$
My issue and why I think this doesn't work is because when I want to prove this is surjective, I want to prove that for any g, there is an n-tuple s.t $g=g_{(x_1,…,x_n)}$ but the way I've defined g seems to make it depend upon this tuple already existing.
Best Answer
You don't have to worry too much about the tuple existing. You're looking at $\mathbb{R}^n$, which is the set of all $n$-tuples with real numbers in every entry. You are completely free to choose which real numbers to put in all $n$ entries.
If you look at the map $g \mapsto (g(1), g(2), \ldots, g(n))$, then this is a well-defined map from $\mathbb{R}^N$ to $\mathbb{R}^n$, and I think you'll find that it's the inverse of $f$.