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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.

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  • $\begingroup$ Both sets have the cardinality of the real line.. $\endgroup$ – Eduardo Longa Sep 11 '18 at 0:33
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You came up with the right answer, let me explain the last part (about the surjectivity).

We have a map $f:\mathbb{R}^n\rightarrow \mathbb{R}^N$ taking $(x_1,x_2,...,x_n)\mapsto g_{(x_1,x_2,...,x_n)}$ our goal is to show that this map is surjective by definition this means

For every $g\in \mathbb{R}^N$ there exists $x=(x_1,...,x_n)\in\mathbb{R}^n$ such that $f(x)=g$ (recall that $f(x)=g_x$).

Given $g:N\rightarrow\mathbb{R}$ we can take $x=(g(1),g(2),g(3),...,g(n))$ we claim that $g_x=g$. Indeed, for every $1\leq i\leq n$ we have that $g_x(i)= x_i=g(i)$ hence $g_x$ and $g$ gives the same value for every $i\in N$ and so they're equal.

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  • $\begingroup$ Alright that makes sense. Thanks. $\endgroup$ – AColoredReptile Sep 11 '18 at 0:52
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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 $f$.

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I would just appeal to cardinal arithmetic. We have $$\mathfrak c = |\Bbb R|=2^{\aleph_0}=2^{|\Bbb N|}\\ \mathfrak c = |\Bbb R|\le |\Bbb R^n|\le |\Bbb R^{|\Bbb N|}|=(2^{|\Bbb N|})^{|\Bbb N|}=2^{|\Bbb N|{|\Bbb N|}}=2^{|\Bbb N|}=\mathfrak c$$

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